Mensuration 453

CHAPTER 10

M ensur ation
It is one of the easiest chapters, which contributes almost 6-8% problems in Quantitative
Aptitude Section of CAT. Besides there are several other aptitude tests which include
plethora of questions from this topic itself.
Therefore it is advised that those students who are not so good in other sections such as
algebra or sort of logical questions they must emphasise on this chapter. Even the questions
asked from this chapter are not as much complex as they are in Geometry.

10.1 Mensuration
Definition : Mensuration is a science of measurement of the lengths of lines, areas of
surfaces and volumes of solids.
Planes : Planes are two dimensional i.e. these two dimensions are namely length and breadth.
These occupy surface.
Solids : Solids are three dimensional, namely length, breadth and height. These occupy
space.
Chapter Checklist
Mensuration
Conversion of Important Units 2-D Figures (Planes)
1 km = 10 hectometre =100 decametre Rectangles and Squares
=1000 metre =10,000 decimetre Triangles
=100
, ,000 centimetre =10,00,000 millimetre Parallelogram, Rhombus
1 hectare =10,000 square metre and Trapezium
1 are =100 square metre Circles
1 square hectometre =100 square decametre 3-D Figures (Solids)
1 square decametre =100 square metre Cuboid and Cube
1 square metre = 100 square decimetre Cylinder and Cone
1 square decimetre =100 square centimetre Sphere, Prism and Pyramid
1 square centimetre =100 square millimetre
CAT Test
2 = 1.414, 3 = 1.732, 5 = 2.236, 6 = 2.45
Weight = Volume × density
454 QUANTUM CAT
10.2 2-D Figures
Table 2-D Figures (Plane figures)
Name Figure Nomenclature Area Perimeter

Rectangle l → length l × b = lb 2l + 2b = 2(l + b)

b b → breadth
l

a
Square a → side (i) a × a = a2 a + a + a + a = 4a
d → diagonal d2
a d a d=a 2 (ii)
2

Triangle a, b and c are three sides (i)

1
×b×h a + b + c = 2s
(Scalene) a c of triangle and s the 2
h semiperimeter, where (ii) s (s − a)(s − b)(s − c)
 a + b + c (Hero’s formula)
s=  
b  2 
b is the base and h is the
altitude of triangle
Equilateral a → side (i)
1
×a×h 3a
triangle h → height or altitude 2
a a 3 3 2
h h= a (ii) a
2 4

Isosceles a → equal sides (i)

1
×b×h 2a + b
triangle b → base 2
a a h → height or altitude 1
h (ii) × b × 4a2 − b2
4a2 − b2 4
h=
2
b

Right angled b → base 1

×b×h b+ h+ d
triangle h
d h → altitude/height 2
d → diagonal
b d = b2 + h 2

Isosceles right a → equal sides 1 2

a 2a + d
angled a
d d → diagonal 2
triangle d=a 2
a

Quadrilateral C AC is the diagonal and 1

× AC × (h1 + h2 ) AB + BC + CD + AD
D
h1, h2 are the altitudes on 2
h1
AC, from the vertices
h2 D and B, respectively.
A B
Mensuration 455

Name Figure Nomenclature Area Perimeter

a
Parallelogram a and b are sides adjacent to a×h 2(a + b)
b b each other.
h
h → distance between the
a parallel sides
a
Rhombus D C a → each equal side of 1
× d1 × d2 4a
d1 rhombus 2
a a d1 and d2 are the diagonals
d2 d1 → BD
A B d2 → AC
a

Trapezium D b
C a and b are parallel sides to  a + b AB + BC + CD + AD
  ×h
each other and h is the  2 
h perpendicular distance
between parallel sides
A B
a

Regular hexagon a a → each of the equal side 3 3 2 6a

a
a a 2

a a

Regular octagon a a → each of equal side 2a2 (1 + 2) 8a

a a

a a

a a
a

Circle r → radius of the circle πr2 2πr (called as

22 circumference)
π= = 3.1416(approx)
r 7

Semicircle r r
r → radius of the circle 1 2 πr + 2r
πr
2

Quadrant r → radius 1 2
πr
1
πr + 2r
r 4 2

Ring or circular path R → outer radius π (R 2 − r2 ) (outer) → 2πR

(shaded region) r → inner radius (inner) → 2πr
R
r
456 QUANTUM CAT
Name Figure Nomenclature Area Perimeter

Sector of a circle O → centre of the circle  θ  l + 2r

(i) πr2  
r → radius  360°
O l → length of the arc 1
θ (ii) r × l
r θ → angle of the sector 2
 θ 
l = 2πr  
A B  360°
l

Segment of a circle θ → angle of the sector Area of segment ACB  πθ  θ 

2r  + sin   
r → radius (minor segment)  360 °  2 
O AB → chord  πθ sin θ 
= r2  − 
r θ ACB → arc of the circle  360° 2 

A B
C

Pathways running l l → length (l + b − w ) w 2(l + b) − 4w

across the middle of w b → breadth = 2[ l + b − 2w]
a rectangle w w w → width of the path (road)
w b

Outer pathways w l → length (l + b + 2w ) 2w (inner) → 2 (l + b)

l
b → breadth (outer)
b w
w → widthness of the path → 2 (l + b + 4w )

Inner path w l → length (l + b − 2w ) 2w (outer) → 2 (l + b)

b → breadth (inner)
w b
w → widthness of the path → 2 (l + b − 4w )

10.3 Rectangles and Squares Solution

(a) Perimeter = 2(15 + 8) = 46 m
1. Area of a rectangle = length × breadth = l × b
(b) Area of floor = 15 × 8 = 120 m2
1
2. Area of a square = (side) 2 = (diagonal) 2 (c) Length of diagonal = (15) 2 + ( 8) 2 = 17m
2
1 2 NOTE The maximum possible length of any rod that can be placed
= d =a 2
on a rectangular floor is equal to the diagonal of the floor of the room.
2
3. Diagonal of a rectangle d b
= (length) 2 + (breadth) 2 = l 2 + b 2
l
4. Diagonal of a square = side 2 + side 2 = side 2 = a 2 Here d > l
5. Perimeter of a rectangle = 2 (length + breadth) = 2( l + b)
Exp. 2) The side of a square shaped garden is
6. Perimeter of a square = 4 × side = 4a 20 m. Find the :
7. Area of four walls of a room = 2 ( l + b) × h (a) area of the garden
Exp. 1) The length and breadth of a rectangular room are (b) perimeter (or boundary) of the garden
15 m and 8 m respectively : (c) maximum possible distance between any two corners
(a) Find the perimeter of room. of the garden.
(b) Find the area of the floor of room. Solution (a) Area = (side) 2 = (20) 2 = 400 m 2 (square metre)
(c) Find the maximum possible length of the rod that can (b) Perimeter = 4 × side = 4 × 20 = 80 m
be put on the floor. (c) Diagonal = side 2 = 20 × 2 = 20 × 1.414 = 28.28 m
Mensuration 457

Exp. 3) One side of a rectangular lawn is 12 m and its Exp. 7) Find the cost of carpeting a room 17 m long and
diagonal is 13 m. Find the area of the field. 9 m wide with a carpet 60 cm broad at 40 paise per metre.
Solution d = l 2 + b2 ⇒ 13 = 122 + b 2 ⇒ b = 5 m Solution Area of carpet = Area of room
∴ Area = l × b = 12 × 5 = 60 m 2 l × 0.6 = 17 × 9 (60 cm = 0.6 m)
⇒ l = 255 m
Exp. 4) The length of a rectangle is 1 cm more than its
∴ Cost of carpeting the floor = rate × length of carpet
breadth. The diagonal is 29 cm. Find the area of the
rectangle. = 0.4 × 255 = ` 102.00
(a) 481 cm 2 (b) 841 cm 2 (c) 420 cm 2 (d) 870 m 2 Exp. 8) The dimensions of a lawn are in the ratio 4 : 1
Solution l = ( b + 1) and its area is 1/4 hectares. What is the length of the lawn?
and d = l 2 + b 2 = ( b + 1) 2 + b 2 = 29 Solution l × b = 4x × x =
1
× 10000
4
⇒ b 2 + b = 420 ⇒ b = 20
⇒ x = 25
∴ l = 21
⇒ length = 4x = 100 m
∴ Area = l × b = 420 cm2
Exp. 9) A room is 16 m long, 7 m broad and 8 m high.
Exp. 5) The length of a wall is 5/4 times of its height. If Find the cost of white washing the four walls of room at
the area of the wall be 180 m 2 . What is the sum of the ` 7.5 per m 2 , white washing is not to be done on the doors
length and height of the wall? and windows, which occupy 65 m 2 .
Solution Let the length be 5x and height be 4x. Solution Area of 4 walls of a room = 2(16 + 7) × 8 = 368 m2
Then, l × h = 180 = 5 x × 4x = 20x 2 ⇒ x = 3
Net area of 4 walls = 368 − 65 = 303 m2
∴ l + h = 15 + 12 = 27 m
∴ Cost of white washing = 303 × 7.5 = ` 2272.5
Alternatively 1.25 h2 = 180 ⇒ h = 12 m
∴ l = 15 m; l + h = 27 m Exp. 10) The ratio between the sides of a room is 5 : 3.
The cost of white washing the ceiling of the room at
Exp. 6) A rectangular grassy lawn is 18 m by 12 m. It has 50 Paise per square m is ` 270 and the cost of papering the
a gravel path 1.5 m wide all around it on the outside. What walls at 10 P per square metre is ` 48. The height of the
is the area of the path. room is :
Solution Area of path (outside the lawn) = ( l + b + 2w) 2w (a) 6 m (b) 8 m
= (18 + 12 + 3) 3 = 99 m2 (c) 5 m (d) 10 m
Total cost 270
Solution Area of ceiling = = = 540 sq m
12 Cost of 1 sq unit 0.5
1.5 m 18 Now since l : b = 5x : 3x
⇒ l × b = 15 x 2 = 540 m 2
Alternatively Area of lawn = 18 × 12 = 216 m 2 ⇒ l = 30 and b = 18 m
Total area (lawn + path) = (18 + 3) × (12 + 3) Now, Area of the 4 walls
= 21 × 15 = 315 m 2 Total cost 48
= = = 480 m 2
Cost of 1 sq unit 0.1
∴ Area of path (only) = 315 − 216 = 99 m2
480
∴ Height = =5 m
2( 30 + 18)
458 QUANTUM CAT
Introductory Exercise 10.1
1. A rectangular field has its length and breadth in the ratio 12. A square field of 2 sq km is to be divided into two equal
of 16 : 9. If its perimeter is 750 cm. What is its area? parts by a wall which coincides with a diagonal. Find
(a) 7500 cm2 (b) 32400 cm2 the length of the wall.
2
(c) 14400 cm (d) 14000 cm2 (a) 2 km (b) 1 km (c) 4.2 km (d) 2 km
2. A rectangular field costs ` 110 for levelling at 13. There are two square fields. Of the two square fields
50 paise per square metre. If the ratio of one contains 1 hectare area while the other is broader
length : breadth is 11 : 5. Find the length of the field. by 11 per cent. Find the difference in area expressed in
(a) 16 m (b) 21 m sq m.
(c) 22 m (d) none of these (a) 2321 sq m (b) 1210 sq m
3. A room is half as long again as it is wide. The cost of (c) 2121 sq m (d) 7700 sq m
carpeting it at 62 paise per square metre is 14. The expenses of carpeting a half of the floor were
` 2916.48. Find the cost of white washing the ceiling at
` 759, but if the length had been 6 m less than it was,
30 paise per metre.
the expenses would have been ` 561. What is the
(a) ` 2211.5 (b) ` 1114.2
length?
(c) ` 1411.2 (d) can’t be determined
(a) 21 m (b) 23 m (c) 45 m (d) 27 m
4. The length of a rectangular plot of ground is four times
15. If a roll of paper 1 km long has area 1/25 hectare, how
its breadth and its area is 4 hectares. How long will it
take to a dog to walk round it at the rate of 3 km/h? wide is the paper?
(a) 12 min (b) 20 min (c) 21 min (d) 18.5 min (a) 4 m (b) 40 cm (c) 40 dm (d) 25 cm
5. Find the length of the wire required to fence a square 16. The diagonal and one side of a rectangular plot are
field 6 times having its area 5 hectares and 76 ares. 289 m and 240 m respectively. Find the other side.
(a) 5760 m (b) 6760 m (c) 52500 m (d) 11760 m (a) 237 m (b) 181 m (c) 161 m (d) 159 m
6. A room is 19 m long and 3.50 m broad. What will be 17. How many tiles 20 cm by 40 cm will be required to
the cost of covering its floor with a carpet of 70 cm pave the floor of a prayer hall of a room 16 m long and
wide at 95 paise per metre? 9 m wide?
(a) ` 90.25 (b) ` 99.25 (a) 18000 (b) 2700 (c) 1800 (d) 14400
(c) ` 90.75 (d) none of these
18. If the area of a square be 22050 sq cm. Find the length
7. Find the cost of paving a courtyard 316.8 m × 65 m of diagonal.
with stones measuring1.3 m × 1.1 m at ` 0.5 per stone. (a) 201 cm (b) 220 cm
(a) ` 1440 (b) ` 7200 (c) 211 cm (d) 210 cm
(c) ` 72,000 (d) none of these
19. If requires 90 g paint for painting a door 12 cm × 9 cm,
8. A rectangular garden 63 m long and 54 m broad has a how much paint is required for painting a similar door
path 3 m wide inside it. Find the cost of paving the 4 cm × 3 cm?
path at ` 37/2 per square metre. (a) 30 g (b) 27 g (c) 10 g (d) 45 g
(a) ` 12321 (b) ` 11100
20. The area of a rectangular football field is 24200 sq m.
(c) ` 74000 (d) none of these
It is half as broad as it is long. What is the approx
9. If the length of a ractangular field is doubled and its minimum distance a man will cover if he wishes to go
breadth is halved (i.e. reduced by 50%). What is from one corner to the opposite one?
percentage change in its area? (a) 283 m (b) 246 m (c) 576 m (d) 289 m
(a) 0% (b) 10% (c) 25% (d) 33.33%
21. The area of the four walls of a room is 2640 sq m and
10. A path of uniform width runs all around the inside of the length is twice the breadth and the height is given
rectangular field 116 m by 68 m and occupies as 11 m. What is the area of the ceiling?
720 sq m. Find the width of the path. (a) 2800 m2 (b) 3200 m2
(a) 1 m (b) 1.5 m (c) 2 m (d) 4 m (c) 320 m 2
(d) none of these
11. A drawing room is 7.5 m long 6.5 m broad and 22. If the perimeter of a square and a rectangle are the
6 m high. Find the length of paper 2.5 dm wide to same, then the areas A and B enclosed by them would
cover its walls allowing 8 sq m for doors. satisfy the inequality :
(a) 368 m (b) 640 m (c) 625 m (d) 888 m (a) A > B (b) A ≥ B (c) A < B (d) A ≤ B
Mensuration 459

23. If the perimeter of a rectangle and a square each is 28. The area of a rectangular field is 52000 m2. This
equal to 80 cm and the difference of their areas is 100 rectangular area has been drawn on a map to the scale
sq cm, the sides of the rectangle are : 1 cm to 100 m. The length is shown as 3.25 cm on the
(a) 25 cm, 15 cm (b) 28 cm, 12 cm map. The breadth of the rectangular field is :
(c) 30 cm, 10 cm (d) 35 cm, 15 cm (a) 210 m (b) 150 m
24. The number of square shaped tin sheets of side 25 cm (c) 160 m (d) 123 m
that can be cut off from a square tin sheet of side 1 m, is : 29. If the length of diagonal BD of a square ABCD is
(a) 4 (b) 40 (c) 16 (d) 400 4.8 cm, the area of the square ABCD is :
25. The length and breadth of a rectangular field are (a) 9.6 cm2 (b) 11.52 cm2
2
120 m and 80 m respectively. Inside the field, a park (c) 12.52 cm (d) 5.76 cm2
of 12 m width is made around the field. The area of the 30. If the side of square is increased by 20%, then how
park is : much per cent does its area get increased?
(a) 2358 m2 (b) 7344 m2 (c) 4224 m2 (d) 3224 m2 (a) 40% (b) 20% (c) 44% (d) 24%
26. A 5m wide lawn is cultivated all along the outside of a 31. The ratio of the area of a square to that of the square
rectangular plot measuring 90m × 40 m. The total drawn on its diagonal is :
area of the lawn is : (a) 1 : 1 (b) 1 : 2 (c) 1 : 3 (d) 1 : 4
(a) 1441 m2 (b) 1400 m2 (c) 2600 m2 (d) 420 m2
32. The length and breadth of a square are increased by
27. The length of a rectangle is 2 cm more than its 60% and 40% respectively. The area of the resulting
breadth. The perimeter is 48 cm. The area of the rectangle exceeds the area of the square by :
rectangle (in cm2) is : (a) 224% (b) 24%
(a) 96 (b) 128 (c) 143 (d) 144 (c) 124% (d) 100%

10.4 Triangles
A
1 ratio of 2 : 1 from the vertex to the base.
1. Area of a triangle = × base × height (General formula)
2 AO BO CO 2 r2
∴ = = = = Q P
2. Area of a scalene triangle = s ( s − a ) ( s − b) ( s − c) OR OP OQ 1 r1
O
where s is the semiperimeter of the triangle and OP = OQ = OR , all are the inradii
B C
a, b and c are the three sides of the triangle, OA = OB = OC , R
 a + b + c all are the circumradii
s =  A
 2  NOTE Radii means more than one radius.
Scalene triangle : A triangle whose all sides are O is the centre of two circles. Circle PQR is Q r1 P
different in length called as incircle (touching the sides) and
O
1 circle ABC is called as circumcircle (touching r2
C
3. Area of a right angle triangle = × base × height the vertices)
B R
2
Right angle triangle : A triangle in which two sides are side side
∴ Inradius = and Circumradius =
perpendicular 2 3 3
Also, Hypotenuse = (base) 2 + (height) 2
Exp. 1) The base of a right angled triangle is 8 cm and
4. Area of an equilateral triangle hypotenuse is 17 cm. Find its area.
 3  Solution Hypotenuse = (base) 2 + (altitude) 2
1 3
= × side ×  side  = × (side) 2
2  2  4 (17) = 64 + ( x) 2
⇒ 289 = 64 + x 2 15 cm 17 cm
Equilateral triangle : A triangle in which all the three
⇒ x 2 = 225 ⇒ x = 15 cm
sides are equal also all the three internal angles are equal
1 90°
3 ∴ Area = × base × altitude
Height of an equilateral triangle ( h) = × side 2 8 cm
2 (height is called altitude also)
An important property : In an equilateral ∆ perpendiculars 1
= × 8 × 15 = 60 cm 2
drawn from all the three vertices intersect each other in the 2
460 QUANTUM CAT
Exp. 2) Find the area of an equilateral triangle whose Exp. 4) Find the area and perimeter of a triangle whose
side is 4 3 cm. sides are 17 cm, 8 cm and 15 cm long.
3 Solution Since, 82 + 15 2 = 17 2
Solution Area of an equilateral triangle = × (side)2
4 ⇒ 64 + 225 = 289
3 3 289 = 289 8 17
= × (4 3)2 = × 48 = 12 3 cm2
4 4 Hence, it is a right angled triangle.
1 90°
Exp. 3) The area of a right angled triangle is ∴ Area = × b × h
2 15
24 cm 2 and the length of its hypotenuse is 10 cm. The 1
= × 15 × 8 = 60 cm2
length of the shorter leg is : 2
(a) 5 cm (b) 4 cm and perimeter = 8 + 15 + 17 = 40 cm
(c) 6 cm (d) 3 cm
Solution H 2 = A 2 + B 2 = 100 cm2
Exp. 5) Find the area and perimeter of an isosceles
triangle whose equal sides are 5 cm each and base is 6 cm.
1
Area = × A × B Solution Area of an isosceles triangle
2
b
1 A H = 4a 2 − b 2
⇒ A × B = 24 cm2 4
2 a a
6
⇒ A × B = 48 cm2 = 4 × 25 − 36
4
∴ ( A + B) 2 = A 2 + B 2 + 2AB B
3
= × 64 = 12 cm 2 b
( A + B) 2 = 196 ⇒ A + B = 14 …(i) 2
Again ( A − B) 2 = A 2 + B 2 − 2AB Alternatively We know that the altitude CD bisects the
⇒ ( A − B) 2 = 4 ⇒ A − B = 2 …(ii) base AB in the isosceles triangle ABC.
∴ AD = BD = 3 cm C
Therefore by solving eqs. (i) and (ii) ,we get
Using Pythagoras theorem in ∆ ADC
A=8 and B=6
we have, 5 cm 5 cm
Therefore, the shorter leg is 6 cm.
CD 2 = AC 2 − AD 2
Alternatively Go through options.
∴ CD = 4 cm
1
Q × A × B = 24 ⇒ A × B = 48 1
∴Area of triangle = × base × height A D B
2 2 6 cm
48
Let us assume B = 6, then A = =8 1
= × 6 × 4 = 12 cm 2
6 2
Now, A 2 + B 2 = 100, ( 8) 2 + ( 6) 2 = 100 ⇒100 = 100
Also, the perimeter of triangle = 5 + 5 + 6 = 16 cm
Hence, choice (c) is correct.

Introductory Exercise 10.2

1. What is the area of the triangle whose sides are 3. The sides of a triangle are 25 m, 39 m and 56 m
84 m, 80 m and 52 m? respectively. Find the perpendicular distance from the
(a) 1620 sq. m (b) 2016 sq. m vertex opposite to the side 56 m.
(c) 1818 sq. m (d) none of these (a) 15 m (b) 16.5 m
2. Two poles 15 m and 30 m high stand upright in a play (c) 18.6 m (d) 21 m
ground. If their feet be 36 m apart, find the distance 4. ABC is a triangle and D , E , F are the mid-points of the
between their tops.
sides BC , CA, AB respectively. The ratio of the areas of
(a) 41 m
∆ ABC and ∆ DEF is :
(b) 36 m
(a) 4 : 1 (b) 5 : 1
(c) 39 m
(c) 3 : 1 (d) can’t be determined
(d) none of the above
Mensuration 461

5. The integral base of an isosceles triangle can be whose 8. If every side of a triangle is doubled, then increase in
area is 60 cm2 and the length of one of the equal sides area of the triangle is :
is 13 cm : (a) 200% (b) 300%
(a) 20 cm (b) 10 cm (c) 400% (d) none of these
(c) 16 cm (d) data insufficient
9. If the altitude of an equilateral triangle is 2 3, then its
6. A ladder is resting with one end in contact with the top
area is :
of a wall of height 60 m and the other end on the
(a) 4 3 cm2 (b) 12 3 cm2
ground is at a distance of 11 m from the wall. The
8
length of the ladder is : (c) cm2 (d) none of these
3
(a) 61 m (b) 71 m
(c) 87 m (d) none of these 10. If the perimeter of an equilateral triangle and a square
7. The base of a triangular field is three times its height. is same and the area of equilateral triangle is P and the
If the cost of cultivating the field is ` 36.72 per hectare is area of square is Q, then :
` 495.72, find the height and base of the triangular field. (a) P < Q (b) P ≤ Q
(a) 480 m, 1120 m (b) 400 m, 1200 m (c) P > Q (d) P ≥ Q
(c) 300 m, 900 m (d) 250 m, 650 m

10.5 Parallelogram, Rhombus and Trapezium

1. (a) Area of parallelogram = base ( b) × height ( h) ( AC = d1 and BD = d 2 )
(b) Area of parallelogram D C

= product of any two adjacent sides

θ2
× sine of the included angle θ1
2. Perimeter = 2 (sum of any two adjacent sides) A B
1 1 1
3. (a) Area of rhombus = × (product of diagonals) ∴ Area = d1 d 2 sin θ 1 = d1 d 2 sin θ 2
2 2 2
1
= × d1 d 2 1
(b) Area = × diagonal × sum of the perpendiculars
2 2
(b) Area of rhombus = product of adjacent sides drawn from the opposite vertices on it.
× sine of the angle included by them C
4. Perimeter of rhombus = 4 × side D
N
h1
1
5. Area of a trapezium = × sum of parallel sides × height h2
2 M
height → distance between the two parallel sides A B
6. Perimeter of trapezium = sum of all the four sides 1
1 =
d × ( h1 + h2 )
7. (a) Area of quadrilateral = × product of diagonal 2
2
where, MD = h1 and BN = h2 and AC = d
× sine of the angle between them
462 QUANTUM CAT
Introductory Exercise 10.3
1. The adjacent sides of a parallelogram are 6 cm and 8. Area of a rhombus is 256 cm2. One of the diagonal is
8 cm and the angle between them is 30°. What is the half of the other diagonal. The sum of the diagonals is :
area of the parallelogram? (a) 38 cm (b) 48 cm
(a) 24 cm2 (b) 12 cm2 (c) 28 cm (d) 56 cm
2
(c) 40 cm (d) 24 3 cm2 9. The lengths of two parallel sides of a trapezium are
2. A parallelogram has sides 30 cm and 20 cm and one of 30 cm and 50 cm and its height is 16 cm. Its area is :
its diagonal is 40 cm long. Then its area is : (a) 460 cm2 (b) 750 cm2
2
(a) 75 5 cm2 (b) 245 cm2 (c) 320 cm (d) 640 cm2
(c) 150 15 cm 2
(d) 300 cm2 10. ABCD is a trapezium in which AB || CD and AB = 2CD. If
3. The distance of a 24 cm long side of a parallelogram its diagonals intersect each other at O, then ratio of
from the opposite side is 22 cm. The area of the areas of triangles AOB and COD is :
parallelogram is : (a) 1 : 4 (b) 1 : 2
(a) 264 cm2 (b) 246 cm2 (c) 4 : 1 (d) 2 : 1
(c) 460 cm2 (d) 528 cm2 11. The area of a trapezium is 441 cm2 and the ratio of
4. The two adjacent sides of a parallelogram are 25 cm parallel sides is 5 : 9. Also the perpendicular distance
between them is 21 cm, the longer of parallel sides is :
and 40 cm respectively. The altitude drawn on the
(a) 36 cm (b) 27 cm
longer side is 18 cm, then the area of the
(c) 18 cm (d) 28 cm
parallelogram is :
(a) 450 cm2 (b) 720 cm2 12. The cross-section of a canal is in the shape of a
(c) 500 cm 2
(d) none of these trapezium and the area of cross-section is 360 m2.
If the canal is 12 m wide at the top and 8 m wide at the
5. In the adjoining figure, the ratio of the areas of the
bottom the depth of the canal is :
parallelogram ABCD and that of triangle ABN is :
(a) 36 m (b) 180 m
A D (c) 45 m (d) none of these
13. The area of a hexagon whose one side is 4 m, is :
30°

(a) 6 3 m2 (b) 24 3 m2
2
(c) 42 3 m (d) 24 m2
N
B x C 14. ABCD is a quadrilateral AC = 19 cm. The lengths of
3x
perpendiculars from B and D on AC are 5 cm and 7 cm
(a) 6 : 1 (b) 5 : 1
respectively. Then, the area of ABCD (in cm2) is :
(c) 4 : 1 (d) 8 : 1
(a) 162 (b) 144
6. If the perimeter of a rhombus is 4 p and lengths of its (c) 228 (d) 114
diagonals are a and b, then its area is : 15. ABCD is a square, AC = BD = 4 2 cm,
a ab
(a) (b) AE = DE = 2.5 cm. Find the area of
b 2 E
the adjoining figure ABCDE.
(c) ab/p (d) p (a2 + b2 )
(a) 19 cm2
7. The ratio of the lengths of the diagonal of a rhombus is (b) 22 cm2
2 : 5. Then, the ratio of the area of the rhombus to the (c) 17 cm2 A D
F
square of the shorter diagonal : (d) none of the above
(a) 5 : 4 (b) 5 : 2
(c) 2 : 5 (d) none of these B C
Mensuration 463

10.6 Circles
1. Area of a circle = πR 2 (R → radius of the circle) Exp. 3) A circular road runs round a circular garden. If
the difference between the circumference of the outer
2. Circumference of the circle = 2πR
circle and the inner circle is 44 m. Find the width of the
 θ 
3. Length of an arc = 2πR   road.
 360° 
Solution Let R and r be radii of outer circle and inner circle
 θ  1 respectively.
4. Area of a sector = πR 2   = (arc × R )
 360°  2 ∴ Width of the road = R − r
∴ 2πR − 2πr = 44m ⇒ 2π ( R − r) = 44 m
 θ  R
2
5. Area of segment = πR 2  − sin θ  22
 360°  ⇒ ( R − r) = 7 m Q π = 
2  7
Diameter = 2 × radius Exp. 4) The radius of a circle is 5 m. What is the radius
of another circle whose area is 25 times that of the first?
Solution Ratio of areas = (ratio of radii) 2
O
A B OB = OA → radius 25 5
= (ratio of radii) 2 ⇒ Ratio of radii =
AB → diameter 1 1
Therefore radius of another circle is 5 times.
Hence, the required radius = 25 m
Exp. 1) The radius of a circular wheel is 1 3 m. How
4 Exp. 5) What is the radius of a circle whose area is equal
many revolutions will it make in travelling to the sum of the areas of two circles whose radii are 20 cm
11 km? and 21 cm?
Solution Total distance (travelled) = 11 km = 11000 m Solution πR 2 = πr12 + πr22
Distance travelled in one revolution πR 2 = π (r12 + r22 )
= circumference of the wheel R 2 = ( 400 + 441)
22 7
= 2× π ×r = 2× × = 11 m R 2 = 841 ⇒ R = 29 cm
7 4
11000
∴ Number of revolutions in 11 km = Exp. 6) In a circle of radius 28 cm, an arc subtends an
11
angle of 108° at the centre.
= 1000 revolution
(a) Find the area of the sector.
Exp. 2) It takes 13.5 mL to paint the surface of the (b) Find the length of the arc.
 θ 
circular sheet of radius 17 cm. How much paint is Solution (a) Area of the sector = πr 2  
 360°
required to paint a similar circular sheet with double the
22 108°
radius? = × 28 × 28 ×
Solution Ratio of radii = 1 : 2 7 360°
3
∴ Ratio of areas = 1 : 4 = 22 × 4 × 28 × = 739.2 cm
10
 Area of C1 πr12 1 × 1 1 
 Since = = =   θ 
Area of C 2 πr22 2 × 2 4 (b) Length of the arc = 2πr  
  360°
∴ Quantity of required paint is 4 times. 22 108
=2× × 28 × = 52.8 cm
Thus we need 4 × 13.5 = 54 mL paint. 7 360°
464 QUANTUM CAT
Introductory Exercise 10.4
1. If the circumference of a circle is 704 cm, then its area 11. A figure consists of a square of side ‘a’ m with
is : semicircles drawn on the outside of the square.
(a) 49324 m2 (b) 39424 m2 The area (in m2) of the figure so formed will be :
(c) 3672 cm 2
(d) 39424 cm2  1
(a) a2 (π + 1) (b) a2  π + 
 4
2. If the circumference of a circle is 4.4 m, then the area
of the circle (in m2) is πa2
(c) a2 + (d) none of these
(a) 49/π (b) 49π 2
(c) 4.9π (d) none of these 12. The length of a rope by which a buffalo must be
3. A circular wire of radius 4.2 m is cut and bent in the tethered so that she may be able to graze a grassy
form of a rectangle whose longer side is 20% more area of 2464 sq m is :
than its shorter side. The longer side of the rectangle is (a) 35 m (b) 27 m
: (c) 24 m (d) 28 m
(a) 7.2 m (b) 72 cm 13. A circle of radius ‘a’ is divided into 6 equal sectors. An
(c) 8 m (d) none of these equilateral triangle is drawn on the chord of each
4. The inner circumference of a circular path around a sector to lie outside the circle. Area of the resulting
circular lawn is 440 m. What is the radius of the outer figure is :
circumference of the path, if the path is 14 m wide? (a) 3 a2 (π + 3 ) (b) 3 3 a2
(a) 96 m (b) 84 m 3 3 πa2
(c) 3 (a2 3 + π ) (d)
(c) 70 m (d) 88 m 2
5. The sum of the radius and the circumference of a 14. In the following figure, the area in (cm2 ) is :
circle is 51 cm. The area of the circle is :
(a) 151 cm (b) 152 cm
7cm
(c) 154 cm (d) data insufficient
6. The difference between the circumference and the
7cm 7cm
diameter of the circle is 15 m. What is the area of the
circle?
(a) 225 m2 (b) 165 m2 (a) 115.5 (b) 228.5
2
(c) 156 m (d) none of these (c) 154 (d) none of these
7. The radius of a circle is increased by 2 cm from 15. If a piece of wire 25 cm long is bent into an arc of a
5 cm to 7 cm. What is the percentage change in area of circle subtending an angle of 75° at the centre, then
the circle? the radius of the circle (in cm) is :
(a) 96% (b) 35% π 60
(a) (b)
(c) 70% (d) 74% 120 π
(c) 60π (d) none of these
8. If the circumference of a circle is increased by 20%,
then its area will be increased by : 16. Four horses are tethered at four corners of a square
(a) 44% (b) 32% plot of 42 m so that they just cannot reach one
(c) 40% (d) none of these another. The area left ungrazed is :
(a) 378 m2 (b) 438 m2
9. The area of a circular field is 124.74 hectares. The 2
(c) 786 m (d) none of these
cost of fencing it at the rate of 80 paise per metre is
(a) ` 3168 (b) ` 1584 17. The circumference of the following figure is :
(c) ` 1729 (d) none of these
10. Eldeco Housing Pvt. Ltd purchased a circular plot of
land for ` 158400 at the rate of 1400 per sq. metre.
The radius of the plot is : 20 cm
(a) 5 m (b) 6 m (a) (20 + 10 π ) (b) 20π
(c) 7 m (d) 14 m
(c) 10π (d) 30π
Mensuration 465

18. The area of a minor sector subtending the central 22. A rope by which a calf is tied is decreased from
angle at the centre 40° is 8.25 cm2. What is the area of 23 m to 12 m. What is the decrease in area to be
the remaining part (i.e. major sector) of the circle? grazed by it?
(a) 82.5 cm2 (b) 74.25 cm2 (a) 1110 m2 (b) 1210 m2
2
(c) 66 cm2 (d) none of these (c) 1120 m (d) 1221 m2

19. The area of a sector of a circle of radius 8 cm, formed 23. A wire is bent in the form of a square of side 66 m. It is
by an arc of length 5.6 cm, is : cut and again bent in the form of a circle. The diameter
of this circle is :
(a) 22.4 cm2 (b) 2.24 cm2 (a) 42 m (b) 84 m
(c) 56 cm2 (d) none of these (c) 21 m (d) none of these
20. How long will a man take to go, walking at 13.2 km/h, 24. A wire is in the form of a circle of radius 42 m is cut
round a circular garden of 700 m radius? and again bent in the form of a square. What is the
(a) 12 min (b) 30 min diagonal of the square?
(c) 20 min (d) none of these (a) 66 m (b) 66 3 m
21. What is the radius of circular field whose area is equal (c) 66 2 m (d) none of these
to the sum of the areas of three smaller circular fields 25. If the driving wheel of a bicycle makes 560 revolutions
of radii 8 m, 9 m and 12 m respectively? in travelling 1.1 km. Find the diameter of the wheel.
(a) 17 m (b) 20 m (a) 31.5 cm (b) 30.5 cm
(c) 21 m (d) 29 m (c) 62.5 cm (d) none of these

10.7 3-D Figures

3-D Figures are shown below
Name Figure Nomenclature Volume Curved/Lateral Total surface area
surface area

Cuboid l → length lbh 2(l + b) h 2 (lb + bh + hl)

h
b → breadth
h → height
b
l

Cube a → edge/side a3 4a2 6a2

a

a
a

Right circular r r → radius of base πr2h 2πrh 2πr (h + r)

cylinder h → height of the
h cylinder

Right circular r → radius 1 2

πr h πrl πr ( l + r )
cone h → height 3
l l → slant height
h l = r2 + h 2

r
466 QUANTUM CAT
Name Figure Nomenclature Volume Curved/Lateral Total surface
surface area area

Right triangular — area of base × height perimeter of base lateral surface

prism × height area + 2 (area of
base)
h

Base

Right pyramid — 1 1 lateral surface

× area of the base × perimeter of
3 2 area + area of
Slant × height the base × slant the base
height height

Sphere r → radius 4 3
πr — 4πr2
3
r

Hemisphere r r → radius 2 3
πr 2πr2 3πr2
3

Spherical shell r → inner radius 4

π [ R 3 − r3 ] — 4π [ R 2 + r2 ]
R → outer radius 3

R r

Frustum of a r
— π π (r + R ) l lateral surface
h (r2 + Rr + R 2 )
cone 3 area
h l
+ π [ R 2 + r2 ]

R
Mensuration 467

10.8 Cuboid and Cube (c) Diagonal = l 2 + b 2 + h2

G F = 162 + 182 + 242 = 1156 = 34 cm
Cuboid : A cuboid has 6 faces,
12 edges, 8 vertices and 4 diagonals. H E Exp. 2) Edge of a cube is 5 cm. Find :
Faces : h (a) volume (b) surface area (c) diagonal
C
ABCD, EFGH , ABEH , CFGD, D
b Solution Volume = a 3 = (5) 3 = 125 cm 3
BCFE, ADGH A l B Surface area = 6a 2 = 6 × (5) 2 = 150 cm 2
Edges : AB, BC, CD, AD, EF, FG, GH, EH, BE, CF, DG, AH Diagonal = a 3 = 5 3 = 8.660 = 8.66 cm
Vertices : A, B , C , D, E , F , G, H
Exp. 3) Three cubes of volumes, 1 cm 3 , 216 cm 3 and
Diagonals : AF, BG, CH, DE
512 cm 3 are melted to form a new cube. What is the
Formulae : Volume = l × b × h diagonal of the new cube?
(l → length, b → breadth, h → height) Solution Volume of new cube = 1 + 216 + 512 = 729 cm 3
Total surface area = 2 ( lb + bh + hl) ∴ Edge of new cube = 3 729 = 9 cm
Diagonal ( d ) = l 2 + b 2 + h 2 ∴ Surface area = 6a 2 = 6 × ( 9) 2 = 486 cm 2
G F ∴ Diagonal of the new cube = a 3 = 9 3
Cube : A cube has 6 equal faces, 12
=15.6 cm (approx.)
equal edges, 8 vertices and 4 equal H E
diagonals. a Exp. 4) The surface area of a cube is 864 cm 2 . Find its
Formulae : C volume.
D
Volume = ( a ) 3 a Solution 6a 2 = 864 ⇒ a 2 = 144 ⇒ a = 12 cm
A a B ∴ a 3 = (12) 3 = 1728 cm 3
Total surface area = 6( a ) 2
(a → edge of the cube) Exp. 5) Find the length of the longest pole that can be
Diagonal ( d ) = a 3 placed in a room 30 m long, 24 m broad and 18 m high.

Euler’s Theorem → (V + F ) = ( E + 2); where V → number Solution d = l 2 + b 2 + h2

of vertices, F → number of faces, E → number of edges d = 900 + 576 + 324 ⇒ d = 30 2 m

Exp. 1) The dimensions of a cuboid are 16 cm, 18 cm and Exp. 6) A brick measures 20 cm × 10 cm × 7.5 cm. How
24 cm. Find : many bricks will be required for a wall
(a) volume (b) surface area (c) diagonal 20 m × 2 m × 0.75 m?
Solution (a) Volume = l × b × h = 16 × 18 × 24= 6912 cm 3 total volume of a wall
(b) Solution Number of bricks =
volume of one brick
Surface area = 2 ( lb + bh + hl)
20 × 2 × 0.75 × 100 × 100 × 100
= 2 (16 × 18 + 18 × 24 + 24 × 16) = = 20000
20 × 10 × 7.5
= 2208 cm 2

Introductory Exercise 10.5

1. A cube of metal, each edge of which measures 4 cm, 3. The three co-terminus edges of a rectangular solid are
weighs 400 kg. What is the length of each edge of a 36 cm, 75 cm and 80 cm respectively. Find the edge of
cube of the same metal which weighs 3200 kg? a cube which will be of the same capacity.
(a) 64 cm (b) 8 cm (a) 60 cm (b) 52 cm
(c) 2 cm (d) none of these (c) 46 cm (d) none of these
2. The length of a tank is thrice that of breadth, which is 4. A tank 10 m long and 4 m wide is filled with water. How
256 cm deep and holds 3000 L water. What is the base many litres of water must be drawn off to make the
area of the tank? (1000 L = 1 cubic metre) surface sink by 1 m? (1000 L = 1 cubic metre)
(a) 111775 m2 (b) 1171.875 m2 (a) 20 kilolitre (b) 40 kilolitre
2
(c) 1.171875 m (d) none of these (c) 50 kilolitre (d) none of these
468 QUANTUM CAT
5. How many cubes each of surface area 24 sq. dm can 15. Three cubes each of edge 3 cm long are placed
be made out of a metre cube, without any wastage? together as shown in the adjoining figure. Find the
(a) 75 (b) 250 surface area of the cuboid so formed :
(c) 125 (d) 62
6. Three cubes of metal, whose edges are 3 cm, 4 cm and
5 cm respectively are melted to form a new cube. What
3cm

m
is the surface area of the new cube?

3c
(a) 216 cm2 (b) 56 cm2
3cm 3cm 3cm
(c) 36 cm2 (d) none of these
(a) 182 sq cm (b) 162 sq cm
7. A lid of rectangular box of sides 39.5 cm by 9.35 cm is
(c) 126 sq cm (d) none of these
sealed all around with tape such that there is an
16. A room is 36 m long, 12 m wide and 10 m high. It has
overlapping of 3.75 cm of the tape. What is the length
6 windows, each 3 m × 2.5 m; one door 9.5 m × 6 m
of the tape used?
and one fire chimney 4 m × 4.5 m. Find the
(a) 111.54 cm (b) 101.45 cm
expenditure of papering its walls at the rate of
(c) 110.45 cm (d) none of these
70 paise per metre, if the width of the paper is 1.2 m.
8. A cistern from inside is 12.5 m long, 8.5 m broad and (a) ` 490 (b) ` 690
4 m high and is open at top. Find the cost of cementing (c) ` 1000 (d) none of these
the inside of a cistern at ` 24 per sq m.
17. A school hall has the dimensions 30 m, 12 m by 6 m.
(a) ` 6582 (b) ` 8256
Find the number of children who can be
(c) ` 7752 (d) ` 8752
accommodated, if each child should get 8 m3 of
9. 250 men took a dip in a water tank at a time, which is space.
80 m × 50 m. What is the rise in the water level if the (a) 240 (b) 270
average displacement of 1 man is 4 m3 ? (c) 250 (d) 150
(a) 22 cm (b) 25 cm (c) 18 cm (d) 30 cm 18. When each side of a cube is increased by 2 cm, the
10. The edge of a cube is increased by 100%, the surface volume is increased by 1016 cm3 . Find the side of the
area of the cube is increased by : cube. If each side of it is decreased by 2 cm, by how
(a) 100% (b) 200% (c) 300% (d) 400% much will the volume decrease?
11. The edge of a cube is doubled. What will be the new (a) 12 cm, 729 cm3
volume? (b) 8 cm, 512 cm3
(a) 2 times (b) 3 times (c) 9 cm, 729 cm3
(c) 4 times (d) 8 times (d) 12 cm, 728 cm3
12. The external dimensions of a wooden box closed at 19. Three equal cubes are placed adjacently in a row. Find
both ends are 24 cm, 16 cm and 10 cm respectively the ratio of the total surface area of the resulting cuboid
and thickness of the wood is 5 mm. If the empty box to that of the sum of the total surface areas of the three
weighs 7.35 kg, find the weight of 1 cubic cm of wood. cubes.
(a) 10 g (b) 12.5 g (a) 5 : 7 (b) 7 : 9
(c) 27 g (d) 15 g (c) 9 : 7 (d) none of these
13. The internal dimensions of a tank are 12 dm, 8 dm and 20. A hollow square shaped tube open at both ends is
5 dm. How many cubes each of edge 7 cm can be made of iron. The internal square is of 5 cm side and
placed in the tank with faces parallel to the sides of the the length of the tube is 8 cm. There are 192 cm3 of
tank. Find also, how much space is left unoccupied. iron in the tube. Find its thickness.
(a) 35; 113 dm3 (b) 1313; 31.13 dm3 (a) 2 cm (b) 0.5 cm
3
(c) 1309; 31.013 dm (d) 1309; 13.31 dm3 (c) 1 cm (d) can’t be determined
14. The length, breadth and height of box are 2 m, 1.5 m 21. A cube of 11 cm edge is immersed completely in a
and 80 cm respectively. What would be the cost of rectangular vessel containing water. If the dimensions
canvas to cover it up fully, if one square metre of canvas of base are 15 cm and 12 cm. Find the rise in water
costs ` 25.00? level in the vessel.
(a) ` 260 (b) ` 290 (a) 6.85 cm (b) 7 cm
(c) ` 285 (d) none of these (c) 7.31 cm (d) 7.39 cm
Mensuration 469

22. A rectangular tank 25 cm long and 20 cm wide 27. Which of the following pairs is not correctly matched?
contains water to a depth of 5 cm. A metal cube of side Geometrical objects Number of vertices
10 cm is placed in the tank so that one face of the cube (A) Tetrahedron 4
rests on the bottom of the tank. Find how many litres (B) Pyramid with rectangular base 5
of water must be poured into the tank so as to just (C) Cube 6
cover the cube. (D) Triangle 3
(a) 1 L (b) 1.5 L (a) (A) (b) (B)
(c) 2 L (d) 2.5 L (c) (C) (d) (D)
23. A rectangular block has length 10 cm, breadth 8 cm 28. If the length of diagonal of a cube is 6 3 cm, then the
and height 2 cm. From this block, a cubical hole of length of its edge is :
side 2 cm is drilled out. Find the volume and the (a) 2 cm (b) 3 cm
surface area of the remaining solid. 36
(c) 6 cm (d) cm
(a) 152 cm3 , 512 cm2 (b) 125 cm3 , 215 cm2 3
3 2
(c) 152 cm , 240 cm (d) 125 cm3 , 512 cm2 29. The length of longest pole that can be placed on the
24. A rectangular tank of dimensions 24 m × 12 m × 8 m floor of a room is 12 m and the length of longest pole
is dug inside a rectangular field 600 m long and 200 m that can be placed in the room is 15 m. The height of
broad. The earth taken out is evenly spread over the the room is :
field. By how much will the level of the field rise? (a) 3 m (b) 6 m
(a) 1.925 cm (b) 0.02 m (c) 9 m (d) none of these
(c) 0.2 cm (d) none of these 30. The sum of length, breadth and depth of a cuboid is 12
25. How many bricks (number near to next hundred) will cm and its diagonal is 5 2 cm. Its surface area is
be required to build a wall 30 m long, 30 cm thick and (a) 152 cm2 (b) 94 cm2
5 m high with a provision of 2 doors, each 4 m × 2.5 m (c) 108 cm2 (d) 60 2 cm2
and each brick being 20 cm × 16 cm × 8 cm when 31. The volume of a wall, 3 times as high as it is broad and
one-ninth of the wall is filled with lime? 8 times as long as it is high, is 36.864 m3 . The height
(a) 13500 bricks (b) 13600 bricks of the wall is :
(c) 20050 bricks (d) 18500 bricks (a) 1.8 m (b) 2.4 m
26. A rectangular water reservoir is 15 m by 12 m at the (c) 4.2 m (d) none of these
base. Water flows into it through a pipe whose 32. If the areas of 3 adjacent sides of a cuboid are x, y, z
cross-section is 5 cm by 3 cm at the rate of
respectively, then the volume of the cuboid is :
16 m per second. Find the height to which the water
(a) xyz (b) xyz
will rise in the reservoir in 25 min.
(c) 3xyz (d) none of these
(a) 0.2 m (b) 2 cm
(c) 0.5 m (d) none of these

10.9 Cylinder and Cone

Cylinder : Total surface of the cylinder
Volume = base area × height = πr h
2
= curved surface area + 2 (base area)
Curved surface area = 2πrh + 2πr 2 = 2πr ( h + r )
= perimeter × height = 2πrh Volume of a hollow cylinder = πh ( R 2 − r 2 )
Radius
R r

h
Height

Base
470 QUANTUM CAT
Cone : 22
= πr 2 h = × 2 × 2 × 14
7
= 176 m 3
h (height) l (lateral/slant height)
Exp. 3) A hollow cylindrical tube open at both ends is
made of iron 2 cm thick. If the external diameter be 50 cm
and the length of the tube is 210 cm, find the number of
cubic cm of iron in it.
r (radius of the base) Solution External radius (R) = 25 cm
1 Internal radius (r) = ( 25 − 2) = 23 cm
Volume (V ) = × base area × height Volume of iron = πh ( R 2 − r 2 )
3
22
1 = × 210 × ( 25 2 − 23 2 )
= πr 2 h 7
3 = 63360 cm 3
Curved surface area = πrl
Total surface area = πrl + πr 2 = πr ( l + r ) Exp. 4) A well with 14 m inside diameter is dugout
15 m deep. The earth taken out of it has been evenly
Frustum of a cone : spread all around it to a width of 21 m to form an
Volume of the frustum of a cone embarkment. What is the height of the embarkment?
π Solution Area of embarkment × height of embarkment
= h ( r 2 + Rr + R 2 )
3 = volume of earth dugout
π ( R 2 − r 2 ) × h = π × 7 × 7 × 15
⇒ ( 282 − 7 2 ) h = 7 × 7 × 15
⇒ ( 35 × 21) × h = 7 × 7 × 15
r
⇒ h =1m
h l
Exp. 5) A cylindrical cistern whose diameter is
21 cm is partly filled with water. If a rectangular block of
R iron 14 cm in length, 10.5 cm in breadth and 11 cm in
Curved/Lateral surface area of the frustum of cone thickness is wholly immersed in water, by how many
centimetres will the water level rise?
= πl ( r + R )
Solution Volume of the block = 14 × 10.5 × 11 cm 3
Exp. 1) The base radius of a cylinder is 14 cm and its 21
Radius of the tank = = 10.5 cm
height is 30 cm. Find : 2
(a) volume (b) curved surface area 22 21 21
Volume of the cylinder = πr 2 h = × × ×h
(c) total surface area 7 2 2
Solution (a) Volume of cylinder 22 21 21
∴ × × × h = 14 × 10.5 × 11
22 7 2 2
= πr 2 × h = × 14 × 14 × 30
7 14 2
h= = 4 cm
= 18480 cm 3
3 3
22
(b) Curved surface area = 2πrh = 2 × × 14 × 30 Exp. 6) If the radius of cylinder is doubled, but height is
7
= 2640 cm 2 reduced by 50%. What is the percentage change in
(c) Total surface area = 2πr ( h + r) volume?
22 r1 r h1 h
=2× × 14 ( 30 + 14) Solution = and =
7 r2 2r h2 h/2
22 ∴ Actual volume = πr 2 h
=2× × 14 × 44 = 3872 cm2
7 h
New volume = π ( 2r) 2 × = 2πr 2 h
Exp. 2) How many cubic metres of earth must be dug to 2
make a well 14 m deep and 4 m in diameter? Therefore new volume is the twice of the original volume.
Solution Earth to be dugout from the well 2−1
Hence the change in volume = × 100 = 100%
= volume of the cylindrical well 1
Mensuration 471

1
Exp. 7) The radius of the base of a right cone is (d) Volume = πr 2 h
3
35 cm and its height is 84 cm. Find :
1 22
(a) slant height = × × 35 × 35 × 84 = 107800 cm 3
3 7
(b) curved surface area
(c) total surface area Exp. 8) Find the area of the iron sheet required to
(d) volume prepare a cone 20 cm high with base radius 21 cm.
Solution (a) Slant height( l) = r 2 + h2 Solution r = 21 cm, h = 20 cm
(r → radius of the circular base) ∴ l = r 2 + h2 = 29 cm
= 35 + 84
2 2
(h → height of the cone) ∴ Area of the sheet = total surface area of the cone
= 1225 + 7056 = πrl + πr 2 = πr ( l + r)
22
= 8281 = 91 cm = × 21 [29 + 21] = 3300 cm 2
7
22
(b) Curved surface area = πrl = × 35 × 91
7 Exp. 9) A solid metallic cylinder of base radius
= 10010 cm 2 3 cm and height 5 cm is melted to make n solid cones of
height 1 cm and base radius 1 mm. Find the value of n.
(c) Total surface area = lateral surface area + base area
volume of cylinder
= πrl + πr 2 = πr ( l + r) Solution n=
volume of one cone
22 π × 3 × 3 ×5
= × 35 ( 91 + 84) = = 13500
7 1 1 1
= 110 × 175 = 19250 cm 2 π× × ×1
3 10 10

Introductory Exercise 10.6

1. How many cubic metres of water fill a pipe which is 6. The amount of concrete required to build a cylindrical
3500 m long and 0.08 m in diameter? pillar whose base has a perimeter of 8.8 m and whose
(a) 17.5 m3 (b) 17.6 m3 (c) 21 m3 (d) 35 m3 curved surface area is 17.6 m2 :
(a) 12.32 m3 (b) 12.23 m3
2. A cube of metal, whose edge is 10 cm, is wholly 3
(c) 9.235 m (d) 8.88 m3
immersed in water contained in cylindrical tube whose
diameter is 20 cm. By how much will the water level 7. If the diameter of the base of a closed right circular
rise in the tube? cylinder be equal to its height ‘h’, then its whole
3 surface area is :
(a) 3.3 cm (b) 6 cm
11 2 3 3
(a) πh2 (b) πh3 (c) πh2 (d) πh3
2 3 2 2
(c) 3 cm (d) none of these
11 8. A right circular cylindrical tunnel of diameter 4 m and
3. Find the height of the cylinder whose volume is 511 m3 length 10 m is to be constructed from a sheet of iron.
and the area of the base is 36.5 m2 : The area of the iron sheet required :
280
(a) 7 m (b) 10.5 m (a) (b) 40 π
(c) 14 m (d) none of these π
(c) 80 π (d) none of these
4. The lateral surface area of a cylinder is 1056 cm2 and
its height is 16 cm. What is its volume? 9. The ratio between curved surface area and total
(a) 5566 cm3 (b) 4455 cm3 surface area is 2 : 3. If the total surface area be
(c) 5544 cm 3
(d) none of these 924 cm2, find the volume of the cylinder :
(a) 2156 cm3 (b) 1256 cm3
5. There is a cubical block of wood of side 2 cm. If the 3
(c) 1265 cm (d) none of these
cylinder of the largest possible volume is curved out
from it. Find the volume of the remaining wood : 10. If the volume and curved surface area of a cylinder are
7 12 269.5 cm3 and 154 cm2 respectively, what is the height
(a) cm2 (b) cm3 of the cylinder?
12 7
5 (a) 6 (b) 3.5
(c) 5 cm3 (d) none of these
7 (c) 7 (d) can’t be determined
472 QUANTUM CAT
11. If the curved surface area of a cylinder is 1320 cm2 and 21. A conical vessel has a capacity of 15 L of milk. Its
its base radius is 21 cm, then its total surface area is : height is 50 cm and base radius is 25 cm. How much
(a) 4092 cm2 (b) 2409 cm2 milk can be contained in a vessel in cylindrical form
(c) 4920 cm 2
(d) none of these having the same dimensions as that of the cone?
(a) 15 L (b) 30 L
12. The ratio between the radius of the base and the height
(c) 45 L (d) none of these
of a cylindrical pillar is 3 : 4. If its volume is 4851 m3 ,
the curved surface area of the pillar is : 22. The height of a cone is 30 cm. A small cone is cut off
(a) 924 m2 (b) 1617 m2 at the top by a plane parallel to the base. If its
2 1
(c) 425 m (d) none of these volume be of the volume of the given cone, at
27
13. If the ratio of total surface area to the curved surface what height above the base is the section made?
area of a cylinder be 4 : 1, what is the ratio of radius to (a) 20 cm (b) 18 cm (c) 27 cm (d) 15 cm
the height?
23. A tent is in the form of right circular cone 10.5 m high,
(a) 4 : 1 (b) 2 : 3
the diameter of the base being 13 m. If 8 men are in
(c) 3 : 2 (d) 3 : 1
the tent, find the average number of cubic metres of
14. The circumference of the base of a right cylinder is air space per man.
33 cm and height is 330 cm. What is the volume of this 3 9 3
cylinder? (a) 32 (b) 59.75 (c) 36 (d) 58
58 13 32
(a) 28586.25 cm3 (b) 3344 cm3
3 24. The radius and height of a right circular cone are in the
(c) 4433 cm (d) 3456 cm3 2
ratio of 5 : 12. If its volume is 314 m3 , its slant height
15. The radius of an iron rod decreased to one-fourth. If its 7
volume remains constant, the length will become : is :
(a) 2 times (b) 8 times (a) 26 m (b) 19.5 m
(c) 4 times (d) 16 times (c) 13 m (d) none of these
16. The total surface area of the cylinder is 2640 m2 25. The volume and height of a right circular cone are
and the sum of height and radius of base of cylinder is 1232 cm3 and 24 cm respectively, the area of its
30 m. What is the ratio of height and radius of the curved surface (in cm2) is :
cylinder? (a) 1100 (b) 225 (c) 616 (d) 550
(a) 7 : 9 (b) 9 : 7 26. The circumference of the base of a right circular cone
(c) 8 : 7 (d) 3 : 7 is 220 cm3 and its height 84 cm. The curved surface
area of the cone is :
17. The radii of two cylinders are in the ratio of 3 : 5 and
(a) 20020 cm2 (b) 2020 cm2
their heights are in the ratio 4 : 3. The ratio of their
(c) 2200 cm2 (d) 10010 cm2
volumes is :
(a) 12 : 25 (b) 13 : 25 27. How many metres of cloth 10 m wide will be required
to make a conical tent with base radius of 14 m and
(c) 4 : 5 (d) 5 : 4
height is 48 m?
18. The heights of two cylinders are in the ratio of 3 : 1. If (a) 110 m (b) 55 m (c) 77 m (d) 220 m
the volumes of two cylinders be same, the ratio of their
28. A cone of height 2.8 cm has a lateral surface area
respective radii are :
23.10 m2. The radius of the base is :
(a) 3 : 1 (b) 1 : 3
(a) 3.5 cm (b) 2 cm (c) 2.1 cm (d) 4 cm
(c) 1 : 9 (d) none of these
29. The radii of two cones are equal and their slant heights
19. The ratio of heights of two cylinders is 3 : 2 and the
are in the ratio 3 : 2. If the curved surface area of the
ratio of their radii is 6 : 7. What is the ratio of their smaller cone is 300 cm2, then the curved surface area
curved surface areas? of the bigger cone (in cm2) is :
(a) 9 : 7 (b) 1 : 1 (a) 250 (b) 450 (c) 150 (d) 200
(c) 7 : 9 (d) 7 : 4
30. The ratio of the volume of a right circular cylinder and
20. A hollow garden roller 42 cm wide with a girth of a right circular cone of the same base and height will
132 cm is made of iron 3 cm thick. The volume of the be :
iron of the roller is : (a) 2 : 3 (b) 1 : 3 (c) 3 : 1 (d) 9 : 1
(a) 15544 cm3
31. If the diameter of the base of right circular cone is
(b) 15444 cm3 equal to 8 cm and its slant height is 5 cm, then the
(c) 15545 cm3 area of its axial section is :
(d) none of the above (a) 9 cm2 (b) 12 cm2 (c) 20 cm2 (d) 40 cm2
Mensuration 473

32. If the base radius and the height of a right circular 37. A conical tent has 60° angle at the vertex. The ratio of
cone are increased by 40%, then the percentage its radius and slant height is :
increase in volume (approx) is : (a) 3 : 2 (b) 1 : 2
(a) 175% (b) 120% (c) 1 : 3 (d) can’t be determined
(c) 64% (d) 540%
38. Water flows at the rate of 5 m per min from a cylindrical
33. From a circular sheet of paper of radius 25 cm, a pipe 16 mm in diameter. How long will it take to fill up a
sector area 4% is removed. If the remaining part is conical vessel whose radius is 12 cm and depth is 35
used to make a conical surface, then the ratio of the cm?
radius and height of the cone is : (a) 315 s (b) 365 s
(a) 16 : 25 (b) 9 : 25
(c) 5 min (d) none of these
(c) 7 : 12 (d) 24 : 7
39. A reservoir is in the shape of a frustum of a right
34. If the radius of the base is doubled, keeping the height
circular cone. It is 8 m across at the top and 4 m
constant, what is the ratio of the volume of the larger
across at the bottom. It is 6 m deep its capacity is :
cone to the smaller cone?
(a) 224 m3 (b) 176 m3
(a) 2 : 1 (b) 3 : 1 (c) 4 : 1 (d) 4 : 3 3
(c) 225 m (d) none of these
35. A largest possible cone is cut out from a cube of
40. A conical vessel whose internal radius is 10 cm and
volume 1000 cm3 . The volume of the cone is :
height 72 cm is full of water. If this water is poured into
(a) 280 cm3 (b) 261.9 cm3
3 a cylindrical vessel with internal radius 30 cm, the
(c) 269.1 cm (d) 296.1 cm3 height of the water level rises in it is :
36. If the height and the radius of a cone are doubled, the 2 2
(a) 2 cm (b) 3 cm
volume of the cone becomes : 3 3
(a) 2 times (b) 8 times 2
(c) 5 cm (d) none of these
(c) 16 times (d) 4 times 3

10.10 Sphere, Prism and Pyramid

Sphere Circles in the Sphere
(i) Great Circle: If a plane Large Circle

r (radius)
cuts the sphere at the
centre, we get two
hemispheres of equal
4 3 volume and equal surface
Volume = πr , Surface area = 4πr 2
3 area. In this case, the circles that we get are called great
circles, since these are the circles with the greatest
Hemisphere : r (radius) possible chords. The longest chord (or diameter) of the
2
Volume = πr 3 great circle is same as the longest chord (or diameter) of
3 the sphere.
Curved surface area = 2πr 2 (ii) Small Circle: If a plane cuts the spheres in such a way
that the cut is not made through the centre of the sphere,
Total surface area = 3πr 2 = (2πr 2 + πr 2 )
then we get two frustums of unequal volume and
Spherical shell : unequal lateral surface area, but the area of the two
4 circles would be same. However, these circles would be
Volume = π (R 3 − r 3 ) called as small circles, since the diameter of these circles
3
is shorter than the diameter of the sphere.
Total surface area = 4π ( R 2 + r 2 ) Small Circle

R
r
474 QUANTUM CAT
Spherical cap of a Sphere Sector of a Sphere
If a plane cuts the sphere into two portions then each portion A sector of a sphere is the solid subtended at the centre of the
is known as a cap. sphere by a segment. Essentially, a sector of a sphere is the
The larger portion is known as major cap and the smaller combination of the spherical cap and the cone, where the
portion is known as minor cap. base of spherical cap and sphere are the same and the vertex
of the cone is the centre of the sphere. It is just like an
ice-cream cone.
R

R
Sphere Major Cap Minor Cap

Let us consider the radius of sphere as R and the

radius of the small circle as r and the height of the
cap as h.
πh 2 Let us consider the radius of the sphere as R and height of the
Volume of a Spherical cap = ( h + 3r 2 ) spherical cap of the sphere as h.
6
(If r is known) Volume of the Spherical Sector = Volume of the spherical
πh 2 2πR 2 h
Volume of a Spherical cap = (3R − h) cap + Volume of the cone =
3 3
(If R is known) Total Surface Area of the Spherical Sector = Lateral surface
Lateral Surface Area of a Spherical Cap = perimeter of area of the spherical cap + Lateral surface area of the cone
the sphere×height of the cap = 2πRh = πR (2h + r )

Total Surface Area of a Spherical Cap = 2πRh + πr 2 Prism

= π (2r 2 + h 2 ) Volume = Base area × height

Zone or Frustum of a Sphere

The portion of the surface of a sphere included between two
h (height)
parallel planes, which intersect the sphere, is called a zone.
The distance between the two planes is called height or Base
thickness of the zone.
Lateral surface area
r1 = perimeter of the base × height

r2

h (height)

Let us consider the radius of the sphere as R, height of the Base

zone as h and the radii of the two small circles as r1 and r2 .
πh 2 Pyramid
Volume of the zone = ( h + 3r12 + 3r22 ) 1
6 Volume = × base area × height
Lateral Surface Area of the zone 3
= perimeter of the sphere × height of the zone = 2 πR h 1
Lateral surface area = × Slant height (l)
Total Surface Area of the zone 2
= 2π (rh + r12 + r22 ) perimeter of the base
× slant height Base
Mensuration 475

Total surface area = lateral surface area + base area Curved surface area = 2πr 2
22 7 7
=2× × × = 77 cm 2
l (slant height) 7 2 2
l (slant height) Total surface area = 3 πr 2
22 7 7
=3× × ×
7 2 2
= 115.5 cm 2

Base
Base
Exp. 3) How many bullets can be made from a sphere of
Exp. 1) Find the volume and surface area of a sphere of 8 cm radius. The radius of each bullet must be 0.2 cm.
radius 3.5 cm. volume of sphere
4 3 Solution Number of bullets =
Solution Volume = πr volume of 1 bullet
3 4
4 22 π ×8×8×8
= × × 3.5 × 3.5 × 3.5 = 179.66 cm 3 = 3 = 64000
3 7 4
π × 0.2 × 0.2 × 0.2
Surface area = 4πr 2 3
22
=4× × 3.5 × 3.5 = 154 cm 2
7 Exp. 4) A sphere has the same curved surface as a cone
of height 12 cm and base radius 5 cm. Find the radius to the
Exp. 2) Find the volume, curved surface area and total
nearest cm.
surface area of a hemisphere of diameter 7 cm.
Solution 4πr 2 = π × 5 × 13
2 3
Solution Volume = πr 65
3 ⇒ r2 =
2 22 7 7 7 4
= × × × × = 89.833 cm 3 ⇒ r = 4 cm (approx.)
3 7 2 2 2

Introductory Exercise 10.7

1. A spherical ball of lead 6 cm in radius is melted and 5. The volume of a spherical shell whose external and
recast into three spherical balls. The radii of two of internal diameters are 14 cm and 10 cm respectively
these balls are 3 cm and 4 cm. What is the radius of 872
(a) 42 π cm3 (b) π cm3
the third sphere? 3
(a) 4.5 cm (b) 5 cm (c) 118 π cm3 (d) 86 π cm3
(c) 6 cm (d) 7 cm 6. A solid metal ball of diameter 16 cm is melted and
2. The radius of a copper sphere is 12 cm. The sphere is cast into smaller balls, each of radius 1 cm. The
melted and drawn into a long wire of uniform circular number of such balls is :
cross-section. If the length of the wire is (a) 256 (b) 2048 (c) 512 (d) 4096
144 cm, the radius of wire is : 7. If a hemispherical dome has an inner radius 21 cm
(a) 1 cm (b) 2 cm then its volume (in m3 ) is :
(c) 4 cm (d) none of these (a) 4910 m3
3. A hemispherical bowl of internal radius 6 cm contains (b) 18354 m3
alcohol. This alcohol is to be filled into cylindrical (c) 19404 m3
shaped small bottles of diameter 6 cm and height (d) none of the above
1 cm. How many bottles will be needed to empty the 8. A sphere of radius 9 cm is dropped into a cylindrical
bowl? vessel partly filled with water. The radius of the vessel
(a) 36 (b) 27 (c) 16 (d) 4 is 12 cm. If the sphere is submerged completely, then
4. A hemisphere of lead of diameter 14 cm is cast into a the surface of the water rises by :
right circular cone of height 14 cm. The radius of the (a) 27.5 cm
base of the cone is : (b) 27 cm
(a) 7 cm (b) 14 cm (c) 12 cm
(c) 21 cm (d) none of these (d) 6.75 cm
476 QUANTUM CAT
9. If the height of a cone is half the radius of a sphere Directions (for Q. Nos. 15 to 17) Answer these questions
then the radius of the base of the cone, which has the based on the following information.
same volume as a sphere of radius 7 cm is : A sphere of 17 cm radius is cut by two parallel planes
14
(a) 14 m (b) cm which are 7 cm apart but on the same side of the centre of
2
the sphere. The radius of one of the ends of the zone (or
(c) 14 2 cm (d) none of these
frustum) is 8 cm.
10. From a solid sphere of radius 15 cm, a right circular
15. What is the volume of the zone?
cylindrical hole of radius 9 cm whose axis passing 3206
through the centre is removed. The total surface area (a) 786π (b) π
3
of the remaining solid is : 4096
(c) 1786π (d) π
(a) 1188 π cm2 (b) 1080 π cm2 3
(c) 1152 π cm 2
(d) 1440 π cm2
16. What is the volume of the larger spherical cap of the
11. If a plane cuts the hemisphere parallel to the great two?
circle of the hemisphere, the portion of hemisphere 15650 14400
(a) π (b) π
that has the great circle is called as 7 3
(a) Cone (b) sector 16250
(c) π (d) none of these
(c) zone (d) spherical cap 3
Directions (for Q. Nos. 12 to 14) Answer these questions 17. What is the total surface area of the smaller spherical
based on the following information. cap of the two?
A sphere of 5 cm radius is cut by two parallel planes which (a) 72π (b) 81π
are 7 cm apart but on the opposite sides of the centre of the (c) 68π (d) none of these
sphere. The radius of one of the ends of the zone is 3 cm. 18. Find the volume of the sector of a sphere of radius 25 cm.
The radius of the base of the conical base is 7 cm.
12. What is the volume (in cu. cm.) of the zone?
1280
434 112 (a) 344 2π (b) π
(a) π (b) π 3
3 3
1250
541 (c) 120 3π (d) π
(c) π (d) 121π 3
3
13. What is the volume (in cu. cm.) of the smaller 19. The volume of a pyramid of base area 25 cm2 and
spherical cap of the two? height 12 cm is :
21 17 14 (a) 200 cm3 (b) 100 cm3
(a) π (b) π (c) π (d) 6π
5 4 3 (c) 400 cm3 (d) 800 cm3
14. What is the lateral surface area (in sq. cm.) of the 20. If the base of right rectangular prism remains constant
larger spherical cap of the two? and the measures of the lateral edges are halved, then
(a) 10π (b) 16π its volume will be reduced by :
(c) 25π (d) 20π (a) 50% (b) 33.33%
(c) 66.66% (d) none of these
Mensuration 477

CAT-Test
Questions Helping you bell the CAT

Multifaceted Exercise
1 If the side of an equilateral triangle is r, then the area of the 9 If the surface areas of two spheres are in the ratio
triangle varies directly as : 4 : 9, then the ratio of their volumes is :
(a) r (b) r (c) r2 (d) r3 (a) 8 : 25 (b) 8 : 26
2 The length of the minute hand of the clock is 6 cm. The area (c) 8 : 27 (d) 8 : 28
swept by the minute hand in 30 minutes is : 10 The side of a rhombus are 10 cm and one of its diagonal is
1 1 16 cm. The area of the rhombus is :
(a) cm2 (b) cm2
36π 18π (a) 96 cm2 (b) 95 cm2
2
(c) 18π cm2 (d) 36π cm2 (c) 94 cm (d) 93 cm2
11 In the adjoining figure PQRS is a
3 If the diagonals of a rhombus are 18 cm and 24 cm S
respectively, then its perimeter is : rectangle 8 cm × 6 cm, inscribed in the P
(a) 15 cm (b) 42 cm (c) 60 cm (d) 70 cm circle. The area of the shaded portion
will be : Q R
4 If the ratio of diagonals of two squares is 3 : 2 then the ratio
(a) 48 cm2
of the areas of two squares is :
(b) 42.50 cm2
(a) 4 : 5 (b) 6 : 5
(c) 32.50 cm2
(c) 9 : 4 (d) 3 : 2 (d) 30.5 cm2
5 In the given figure, ABCD is a 8m 12 In the adjoining figure AB = CD = 2BC = 2BP = 2CQ . In the
trapezium in which the parallel D C
sides AB, CD are both middle, a circle with radius 1 cm is drawn. In the rest figure
perpendicular to BC. Find the all are the semicircular arcs. What is the perimeter of the
m

area of the trapezium. whole figure?

17

(a) 140 m2
(b) 168 m2 E
B
2 A
(c) 180 m 16m
A P B O C Q D
(d) 156.4 m2
6 One cubic metre piece of copper is melted and recast into a (a) 4π (b) 8π
square cross-section bar, 36 m long. An exact cube is cut off
(c) 10π (d) none of these
from this bar. If cubic metre of copper cost ` 108, then the
cost of this cube is : 13 In a shower 10 cm of rain fall the volume of water that falls
(a) 50 paisa (b) 75 paisa on 1.5 hectares of ground is :
(c) 1 rupee (d) 1.50 rupee (a) 1500 m3 (b) 1400 m3
7 If ‘h’ be the height of a pyramid standing on a base which is (c) 1200 m3 (d) 1000 m3
an equilateral triangle of side ‘a’ units, then the slant height 14 The base of a prism is a right angle triangle and the two
is : sides containing the right angle are 8 cm and 15 cm. If its
(a) h 2 + a2/4 (b) h 2 + a2/8 height is 20 cm, then the volume of the prism is :
(a) 1600 cc (b) 300 cc
(c) h 2 + a2/3 (d) h 2 + a2
(c) 1200 cc (d) 600 cc
8 The area of the square base of a right pyramid is 64 cm2. If 15 A conical circus tent is to be made of canvas. The height of
the area of each triangle forming the slant surface is 22
the tent is 35 m and the radius of the base is 84 m. If π = ,
20 cm2, then the volume of the pyramid is : 7
128 then the canvas required is :
(a) 64 cm3 (b) cm3
3 (a) 24000 m2 (b) 24004 m2
64
(c) 3 cm3 (d) 64 2 cm3 (c) 24014 m2 (d) 24024 m2
3
478 QUANTUM CAT
16 The radius of base and the volume of a right circular cone 25 If BC passes through centre of the circle, then the area of
are doubled. The ratio of the length of the larger cone to the shaded region in the given figure is :
that of the smaller cone is : a2
(a) (3 − π ) B
(a) 1 : 4 (b) 1 : 2 2
(c) 2 : 1 (d) 4 : 1 π 
(b) a2  − 1 a
17 A cone and a hemisphere have equal base diameter and 2 
a
equal volumes. The ratio of their heights is : (c) 2a2 (π − 1) A C
(a) 3 : 1 (b) 2 : 1 a2  π 
(d)  − 1
(c) 1 : 2 (d) 1 : 3 2 2 
18 A hollow sphere of internal and external diameters 4 cm 26 A river 3 m deep and 60 m wide is flowing at the rate of 2.4
and 8 cm respectively is melted to form a solid cylinder of
km/h. The amount of water running into the sea per
base diameter 8 cm. The height of the cylinder is
minute is :
approximately :
(a) 6000 m3 (b) 6400 m3
(a) 4.5 cm (b) 4.57 cm
(c) 6800 m3 (d) 7200 m3
(c) 4.67 cm (d) 4.7 cm
19 The perimeter of the figure given below correct to one 27 A cone whose height is 15 cm and radius of base is 6 cm is
decimal place is : trimmed sufficiently to reduce it to a pyramid whose base is
an equilateral triangle. The volume of the portion removed
2m
is :
(a) 330 cm3 (b) 328 cm3
10 m 3
(c) 325 cm (d) 331 cm3
28 If a solid right circular cylinder is made of iron is heated to
2m
increase its radius and height by 1% each, then the volume
2m 2m of the solid is increased by :
20m (a) 1.01% (b) 3.03%
(a) 56 m (b) 56.6 m (c) 2.02% (d) 1.2%
(c) 57.2 m (d) 57.9 m 29 The base of a prism is a regular hexagon. If every edge of
20 The sum of the radii of the two circle is 140 cm and the the prism measures 1 m, then the volume of the prism is :
difference between their circumference is 88 cm. The 3 2 3 3 3 3
(a) m (b) m
radius of the larger circle is : 2 2
(a) 60 cm (b) 70 cm 6 2 3 5 3 3
(c) 63 cm (d) 77 cm (c) m (d) m
5 2
21 If the lateral surface of a right circular cone is 2 times its 30 If the side of a square is 24 cm, then the circumference of its
base, then the semi-vertical angle of the cone must be : circumscribed circle (in cm) is :
(a) 15° (b) 30° (c) 45° (d) 60° (a) 24 3π (b) 24 2π (c) 12 2π (d) 24π
22 There is a pyramid on a base which is a regular hexagon of 31 An isosceles right angled triangle has area 112.5 m2. The
(5a)
side 2a. If every slant edge of this pyramid is of length , length of its hypotenuse (in cm) is :
2
(a) 21.213 (b) 21.013
then the volume of this pyramid must be :
(c) 21.113 (d) 21.313
(a) 3a3 (b) 3a3 2
3 32 Two circles of unit radii, are so drawn that the centre of
(c) 3a 3 (d) 6a3
each lies on the circumference of the other. The area of the
14 region, common to both the circles, is :
23 The slant height of a conical tent made of canvas is m.
3 (4π − 3 3) (4π − 6 3)
The radius of tent is 2.5 m. The width of the canvas is 1.25 (a) (b)
12 12
m. If the rate of canvas per metre is ` 33, then the total cost
of the canvas required for the tent (in `) is : (4π − 3 3) (4π − 6 3)
(c) (d)
(a) 726 (b) 950 6 6
(c) 960 (d) 968 33 If the right circular cone is separated into three solids of
24 A hemispherical basin 150 cm in diameter holds water one volumes V1, V2 and V3 by two planes which are parallel to
hundred and twenty times as much a cylindrical tube. If the the base and trisects the altitude, then V1 : V2 : V3 is :
height of the tube is 15 cm, then the diameter of the tube (a) 1 : 2 : 3
(in cm) is : (b) 1 : 4 : 6
(a) 23 (b) 24 (c) 1 : 6 : 9
(c) 25 (d) 26 (d) 1 : 7 : 19
Mensuration 479

34 Water flows at the rate of 10 m per minute from a 45 Find the area of the shaded region in 6
D C
cylindrical pipe 5 mm in diameter. A conical vessel whose the given figure of square ABCD :
diameter is 40 cm and depth 24 cm is filled. The time taken (a) 128 cm2
to fill the conical vessel is : (b) 192 cm2 16
(a) 50 min (b) 50 min. 12 sec. 2
(c) 148 cm 8
(c) 51 min. 12 sec (d) 51 min. 15 sec
(d) 168 cm2 A B
35 The length of four sides and a 10
7 cm 46 In the following figure
diagonal of the given 5 cm
quadrilateral are indicated in the AB = BC and AC = 84 cm.
6 cm
diagram. If A denotes the area The radius of the inscribed
and l the length of the other circle is 14 cm. B is the centre
diagonal, then A and l are 7 cm 5 cm of the largest semi- circle.
A B C
respectively : What is the area of the
(a) 12 6, 4 6 (b) shaded region?
12 6, 5 6 (a) 335 cm2 (b) 770 cm2
2
(c) 6 6, 4 6 (d) 6 6, 5 6 (c) 840 cm (d) 650 cm2
36 If a regular square pyramid has a base of side 8 cm and 47 A tank 4 m long and 2.5 m wide and 6 m deep is dug in a
height of 30 cm, then its volume is : field 10 m long and 9 m wide. If the earth dugout is evenly
(a) 120 cc (b) 240 cc spread over the field, the rise in level of the field will be :
(a) 80 cm (b) 75 cm (c) 60 cm (d) 30 cm
(c) 640 cc (d) 900 cc
37 A cylinder circumscribes a sphere. The ratio of their
48 An open box is made of wood 2 cm thick. Its internal length
is 86 cm, breadth 46 cm and height is 38 cm. The cost of
volumes is
painting the outer surface of the box at
(a) 1 : 2 (b) 3 : 2 (c) 4 : 3 (d) 5 : 6
` 10 per m2 is :
38 In triangle ABC, BC = 8 cm, AC = 15 cm and AB = 17 cm. (a) ` 18.5 (b) ` 8.65
The length of the altitude drawn from B on AC is : (c) ` 11.65 (d) ` 17.50
(a) 6 cm (b) 7 cm 49 A rectangular tin sheet is 22 m long and 8 m broad. It is
(c) 8 cm (d) 10 cm rolled along its length to form a cylinder by making the
39 The area of the largest possible square inscribed in a circle opposite edges just to touch each other. The volume of the
of unit radius (in square unit) is : cylinder (in m3) is :
(a) 3 (b) 4 (c) 2 3π (d) 2 (a) 385 (b) 204
40 The area of the largest triangle that can be inscribed in a (c) 280π (d) 308
semicircle of radius r is : 50 The lateral surface of a cylinder is developed into a square
2
 r whose diagonal is 2 2 cm. The area of the base of the
(a) r2 cm2 (b)   cm2
 3 cylinder (in cm2) is :
(c) r 2 cm2 (d) 3 3r cm2 (a) 3π (b) 1/ π
(c) π (d) 6π
41 If a regular hexagon is inscribed in a circle of radius r, then
its perimeter is 51 If from a circular sheet of paper of radius 15 cm, a sector of
(a) 6 3r (b) 6r 144° is removed and the remaining is used to make a
(c) 3r (d) 12r conical surface, then the angle at the vertex will be :
 3  6
42 If a regular hexagon circumscribes a circle of radius r, then (a) sin −1   (b) sin −1  
 10  5
its perimeter is :
 3  4
(a) 4 3r (b) 6 3r (c) 2 sin −1   (d) 2 sin −1  
(c) 6r (d) 12 3r  5  5
43 In the adjoining figure there D 52 A right circular cone of radius 4 cm and slant height 5 cm is
are three semicircles in which carved out from a cylindrical piece of wood of same radius
BC = 6 cm and BD = 6 3 cm. and height 5 cm. The surface area of the remaining wood
What is the area of the shaded is :
region 90° (a) 84π (b) 70π
(in cm) : A B C (c) 76π (d) 50π
(a) 12π (b) 9π 53 If h, s, V be the height, curved surface area and volume of a
(c) 27 π (d) 28π cone respectively, then (3πVh 3 + 9V 2 − s 2h 2 ) is equal to
44 Area of a rhombus is 144 cm2 and the ratio of length of two (a) 0 (b) π
diagonals is 1 : 2. The sum of lengths of its diagonals are : V 36
(a) 72 cm (b) 40 cm (c) (d)
sh V
(c) 36 cm (d) 18 2 cm
480 QUANTUM CAT
54 If a cone is cut into two parts by a horizontal plane passing 57 A cylinder is circumscribed about a hemisphere and a cone
through the mid point of its axis, the ratio of the volumes of is inscribed in the cylinder so as to have its vertex at the
the upper part and the frustum is : centre of one end and the other end as its base. The
(a) 1 : 1 (b) 1 : 2 volumes of the cylinder, hemisphere and the cone are
(c) 1 : 3 (d) 1 : 7 respectively in the ratio of :
55 A cone, a hemisphere and a cylinder stand on equal bases (a) 3 : 3 : 2 (b) 3 : 2 : 1 (c) 1 : 2 : 3 (d) 2 : 3 : 1
of radius R and have equal heights H. Their whole surfaces 58 The base of a pyramid is a rectangle 40 m long and 20 m
are in the ratio : wide. The slant height of the pyramid from the mid-point of
(a) ( 3 + 1): 3 : 4 (b) ( 2 + 1): 7 : 8 the shorter side of the base to the apex is 29 m. What is the
(c) ( 2 + 1): 3 : 4 (d) none of these volume of pyramid?
56 If a sphere is placed inside a right circular cylinder so as to (a) 5600 m3 (b) 400 m3
3
touch the top, base and the lateral surface of the cylinder. If (c) 6500 m (d) 1753 110 m3
the radius of the sphere is R, the volume of the cylinder is : 59 A copper wire when bent in the form of a square, encloses
(a) 2πR 3 (b) 8πR 3 an area of 121 m2. If the same wire is bent to form a circle,
4
(c) πR 3 (d) none of these the area enclosed by it would be :
3 (a) 122 m2 (b) 112 m2 (c) 154 m2 (d) 308 m2

LEVEL 01 > BASIC LEVEL EXERCISE

1 The perimeter of a parallelogram with one internal angle 7 A square ABCD has an equilateral triangle drawn on the
150° is 64 cm. Find the length of its sides when its area is side AB (interior of the square). The triangle has vertex at
maximum. G. What is the measure of the angle CGB ?
(a) 16, 16 (b) 15, 17 (a) 60° (b) 80°
(c) 14, 18 (d) can’t be determined (c) 75° (d) 90°
2 A sphere of 30 cm radius is dropped into a cylindrical vessel 8 There are two concentric circles whose areas are in the
of 80 cm diameter, which is partly filled with water, then its ratio of 9 : 16 and the difference between their diameters is
level rises by x cm. Find x. 4 cm. What is the area of the outer circle?
(a) 27.5 cm (b) 22.5 cm (a) 32 cm2 (b) 64π cm2
(c) 18.5 cm (d) none of these (c) 36 cm2 (d) 48 cm2
3 Amit walked 12 m toward east, then he turned to his right
9 A square and rhombus have the same base. If the rhombus
and walked 18 m. He then turned to his right and walked
is inclined at 60°, find the ratio of area of square to the area
12 m. He again turned to his right and walked 28 m then he
of the rhombus.
again turned to his right and walked 24 m. At what
distance is he from the starting point and in which (a) 2 3 : 3 (b) 1 : 3
direction? (c) 3 : 2 (d) none of these
(a) 23 m north-east (b) 26 m north-east 10 Four isosceles triangles are cut off from the corners of a
(c) 26 m west (d) 34 m north-east square of area 400 m2. Find the area of new smaller square
4 Find the inradius of triangle if its area is 30 cm2 and (in m2).
200
hypotenuse is 13 cm. (a) 200 2 (b)
(a) 1 cm (b) 2 cm 2
(c) 2.5 cm (d) 2 2 cm (c) 200 (d) 100 2
5 Which of the following figure will have maximum area if 11 Altitude and base of a right angle triangle are ( x + 2) and
the perimeter of all figures is same? (2x + 3) (in cm). If the area of the triangle be 60 cm2, the
(a) Square (b) Octagon length of the hypotenuse is :
(a) 21 cm (b) 13 cm
(c) Circle (d) Hexagon
(c) 17 cm (d) 15 cm
6 ABCD is a trapezium with ∠ A = 90° and AB parallel to CD.
Then ∠ B is :
12 Find the area of a regular octagon with each side ‘a’ cm.
(a) 2a2 (1 + 2) (b) 2a (1 + π )
(a) 90° (b) 90° − ∠ C
(c) 360° − ∠ C (d) 180° − ∠ C (c) a2 ( 2 + 2) (d) none of these
Mensuration 481

13 ABCD is a square, 4 equal circles are just 20 A solid sphere is melted and recast into a right circular cone
touching each other whose centres are D C with a base radius equal to the radius of the sphere. What is
the vertices A, B, C , D of the square. the ratio of the height and radius of the cone so formed?
What is the ratio of the shaded to the (a) 4 : 3 (b) 2 : 3
unshaded area within square? A B (c) 3 : 4 (d) none of these
8 3 21 In the given figure there are 3 semicircles, the radii of each
(a) (b)
11 11 smaller circle is equal. If the radius of the larger circle be
5 6 22 cm, then the area of the shaded region is :
(c) (d)
11 11
14 ABCD is a trapezium, in which AD || BC , E and F are the
mid-points of AB and CD respectively, then EF is :
( AD + BC ) C
(a)
2 F π π
(a) 363 (b) 363 (c) 236.5 π (d) 363π
( AB + CD ) D 4 3
(b)
2 22 A rectangular lawn 60 m × 40 m has two roads each 5 m
DF × CF
(c) wide running in the middle of it, one parallel to length and
AE × BE the other parallel to breadth. The cost of gravelling the
A E B
AD + EF + BC roads at 80 paise per sq. m is :
(d)
2 (a) ` 380 (b) ` 385
4
15 A right circular cone resting on its base is cut at th its (c) ` 400 (d) none of these
5
height along a plane parallel to the circular base. The 23 There are two rectangular fields of same area. The length
height of original cone is 75 cm and base diameter is of first rectangular field is x% less than the length of the
42 cm. What is the base radius of cut out (top portion) cone? second field and breadth of the first field is (5x )% greater
(a) 4.2 cm (b) 2.8 cm (c) 3.5 cm (d) 8.4 cm than the breadth of the second field. What is the value of x?
16 l, b are the length and breadth of a rectangle respectively. If (a) 15 (b) 25
the perimeter of this rectangle is numerically equal to the (c) 50 (d) 80
area of the rectangle. What is the value of l − b (where l > b)? 24 In the adjoining figure ACB is a quadrant with radius ‘a’. A
(a) 1 (b) 2 semicircle is drawn outside the quadrant taking AB as a
(c) 3 (d) can’t be determined diameter. Find the area of shaded region.
17 In the adjoining figure ABC is an equilateral triangle and C 1
(a) (π − 2a2 )
is the centre of the circle, A and B lie on the circle. What is 4
the area of the shaded region, if the diameter of the circle is  1 A B
(b)   (πa2 − a2 )
28 cm?  4
 2  a2
(a) 102 − 49 3 cm2 A B
 3  (c) a 90° a
2
 2 
(b) 103 − 98 3 cm2 C (d) can’t be determined C
 3 
25 Ravi made an error of 5% in excess while measuring the
(c) (109 − 38 3) cm length of rectangle and an error of 8% deficit was made
(d) none of the above while measuring the breadth. What is the percentage error
18 l1, b1 and l2, b2 are the lengths and breadths of the two in the area?
rectangles respectively, but the areas of the rectangles are (a) − 3% (b) − 40%
same. l1 is increased by 25% and b1 is decreased by 25%. (c) − 3.4% (d) can’t be determined
Similarly l2 is decreased by 25% and b2 is increased by 25%.
If A1 and A2 is the new areas of the two rectangles 26 In the adjoining figure the
10 m
respectively, then : cross-section of a swimming pool is
(a) A1 > A2 (b) A1 < A2 shown. If the length of the
(c) A1 = A2 (d) can’t be determined swimming pool is 120 m, then the 6m
amount of water it can hold is :
19 An acute angle made by a side of parallelogram with other
(a) 5760 m3
pair of parallel sides is 60°. If the distance between these 6m
parallel sides is 6 3, the other side is : (b) 9600 m3
(a) 12 cm (b) 12 3 cm (c) 7200 m3
(c) 15 3 cm (d) none of these (d) none of the above
482 QUANTUM CAT
27 Around a circular garden a circular road is to be repair 36 A spherical steel ball was silver polished then it was cut into
which costs ` 22176 at the rate of ` 1 per sq m. If the inner 4 similar pieces. What is ratio of the polished area to the
radius is 112 m, find the width of the circular road. non polished area?
(a) 18 m (b) 28 m (a) 1 : 1 (b) 1 : 2
(c) 14 m (d) none of these (c) 2 : 1 (d) can’t be determined
28 An equilateral triangle is cut from its three vertices to form 37 What is the total surface area of the identical cubes of
a regular hexagon. What is the percentage of area wasted? largest possible size that are cut from a cuboid of size
(a) 20% (b) 50% 85 cm × 17 cm × 5.1 cm?
(c) 33.33% (d) 66.66% (a) 26010 cm2 (b) 21600 cm2
29 ABC is an equilateral triangle and PQRS C 2
(c) 26100 cm (d) none of these
is a square inscribed in the triangle in 38 125 identical cubes are cut from a big cube and all the
such a way that P and Q lie on smaller cubes are arranged in a row to form a long cuboid.
S R
AB and R , S lie on BC and AC What is the percentage increase in the total surface area of
respectively. What is the value of
A P the cuboid over the total surface area of the cube?
RC : RB ? Q B
2 1
(a) 1 : 2 (b) 1 : 3 (a) 234 % (b) 235 %
3 3
(c) 3 : 2 (d) 1 : 2 2
30 The area of a square and circle is same and the perimeter of (c) 134 % (d) none of these
3
square and equilateral triangle is same, then the ratio
39 In the adjoining figure a parallelogram ABCD is shown.
between the area of circle and the area of equilateral
AB = 24 cm and AO = BO = 13 cm. Find BC.
triangle is :
D C
(a) π : 3 (b) 9 : 4 3
(c) 4 : 9 3 (d) none of these O
31 Adjoining figure shows a square D C
A B
ABCD in which O is the point of
intersection of diagonals O 40 cm
(a) 8 cm (b) 10 cm
AC and BD. Four squares of (c) 11 cm (d) none of these
maximum possible area are formed A B 40 There are two circles
inside each four triangles AOB, 40 cm
intersecting each other.
BOC, COD and AOD. What is the total area of these Another smaller circle with
4 squares? centre O, is lying between the A O B
(a) 400 cm2 (b) 100 cm2 common region of two larger
2 circles. Centres of the circle
(c) 80 cm (d) none of these
32 What is the ratio of the area of circumcircle of equilateral (i.e., A, O and B) are lying on a
triangle to the area of square with the same side length as straight line. AB = 16 cm and the radii of the larger circles
the equilateral triangle? are 10 cm each. What is the area of the smaller circle?
(a) π : 3 (b) π : 3 (a) 4π cm2 (b) 2π cm2
(c) 3 : 2 (d) none of these 4 π
(c) cm2 (d) cm2
33 It is required to construct a big rectangular hall that can π 4
accommodate 400 people with 25 m3 space for each 41 ABCD is a square, inside which 4 circles D C
person. The height of the wall has been fixed at 10 m and with radius 1 cm, each are touching each
the total inner surface area of the walls must be 1300 m2. other. What is the area of the shaded
What is the length and breadth of the hall (in metres)? region?
(a) 30, 20 (b) 45, 20 (a) (2π − 3) cm2 A B
(c) 40, 25 (d) 35, 30 (b) (4 − π ) cm2
34 The perimeter of a rectangle and an equilateral triangle are (c) (16 − 4π ) cm2
same. Also, one of the sides of the rectangle is equal to the
(d) none of the above
side of the triangle. The ratio of the areas of the rectangle
and the triangle is : 42 ABCD is a square, E is a point on AB such that BE = 17 cm.
(a) 3 : 1 (b) 1 : 3 The area of triangle ADE is 84 cm2. What is the area of
(c) 2 : 3 (d) 4 : 3 square?
D C
(a) 400 cm2
35 If l, b, p be the length, breadth and perimeter of a rectangle 2
(b) 625 cm
and b, l, p are in GP (in order), then l/b is :
(c) 729 cm2
(a) 2 : 1 (b) ( 3 − 1): 1
(d) 576 cm2 A B
(c) ( 3 + 1): 1 (d) 2 : 3 E
Mensuration 483

43 If the volume of a sphere, a cube, a tetrahedron and a 50 Charles has a right circular cylinder which he inserted
octahedron be same then which of the following has completely into a right circular cone of height 30 cm. The
maximum surface area? vertical angle of the cone is 60° and the diameter of the
(a) Sphere (b) Cube cylinder is 8 3 cm. What is the volume of the cone?
(c) Octahedron (d) Tetrahedron 3000
(a) π cm3 (b) 3000π cm3
44 In a rectangle the ratio of the length is to breadth is same as 7
that of the sum of the length and breadth to the length. If (c) 4860π cm3 (d) can’t be determined
l and b be the length and breadth of the rectangle then
51 There are six faces in a cube. Rajeev fix one cube on each of
which of the following is true?
the faces. The dimensions of all the cubes are same. What is
l l2 b l+ b
(i) = 2 + 1 (ii) = the ratio of total surface area of the newly formed solid to
b b l−b l the area of a single cube?
(iii) lb = (l + b)(l − b) (a) 7 : 1 (b) 6 : 1 (c) 5 : 1 (d) 41 : 9
(a) only (i) is true 52 If the ratio of diagonals of two cubes is 3 : 2, then the ratio
(b) only (ii) is true of the surface areas of the two cubes respectively is :
(c) only (ii) and (iii) are true (a) 5 : 4 (b) 9 : 5
(d) only (i) and (ii) are true (c) 9 : 4 (d) can’t be determined
45 Three circles of equal radii touch each other as shown in 53 ABCDEF is a regular hexagon of side 6 E D
figure. The radius of each circle is 1 cm. What is the area of cm. What is the area of triangle BDF?
shaded region? (a) 32 3 cm2
 2 3 − π F C
(a)   cm2 (b) 27 3 cm2
 2 
(c) 24 cm2
 3 2 − π A B
(b)   cm2 (d) none of the above
 3 
Directions (for Q. Nos. 54 and 55) King Dashratha of Ayodhya
2 3
(c) cm2 had a rectangular plot of area 9792 m2 . He divided it into 4 square
π
shaped plots by fencing parallel fences to the breadth of the
(d) none of the above
rectangular plot. All the four sons got each square shaped plot.
46 How many spheres of radius 1.5 cm can be cut out of a However, some area of plot was still left which could not be
wooden cube of edge 9 cm? formed as a square shaped. So, four more square shaped plots
(a) 216 (b) 81 were formed by fencing parallel to the longer side of the original
(c) 27 (d) can’t be determined plot. The king gave one smaller square shaped plot to each of his
47 Kaurav and Pandav have a rectangular field of area 20000 sq. wives and one of the smaller square shaped plot retained with
m. They decided to divide it into two equal parts by dividing it himself and then nothing left to divide.
with a single straight line. Kaurav wanted to fence their land
54 What is the ratio of the area of larger square shaped plot to
immediately, so they incurred total expenses for the fencing all
the area of the smaller square shaped plot?
the four sides alone at ` 2 per metre. What is the minimum cost
(a) 17 : 1 (b) 25 : 9
that Kaurav had to incur?
(c) 16 : 1 (d) can’t be determined
(a) ` 800 (b) ` 1600 (c) ` 1200 (d) ` 600
55 What are dimensions of the original plot?
48 There is a cone of height 12 cm, out of which a smaller cone (a) 288 m, 34 m (b) 102 m, 96 m
(which is the top portion of the original cone) with the same (c) 306 m, 32 m (d) 204 m, 48 m
vertex and vertical axis is cut out. 56 A scalene triangle PQR, such that PQ = 25, QR = 51 and PR
What is the ratio of the volume of the larger (actual) cone to = 74, is rotated completely about PQ, what is the volume of
the remaining part (frustum) of the cone, if the height of the the solid formed in this way?
smaller cone is 9 cm? (a) 3300π cm3 (b) 4800π cm3
(a) 3 : 1 (b) 9 : 1 (c) 3600π cm3 (d) 4900π cm3
(c) 64 : 37 (d) 16 : 7
57 There are two squares ABCD and PQRS with the same area.
49 Radhey can walk along the boundary of a rectangular field
Square ABCD has the largest possible circle inscribed in it,
and also along the diagonals of the field. His speed is
while square PQRS has four circles, tangent to each other,
53 km/h. The length of the field is 45 km. Radhey started inscribed in it such that the maximum area is occupied by
from one corner and reached to the diagonally opposite them. What is the ratio of the area occupied by the circle in
corner in 1 hour. What is the area of the field? ABCD to the total area occupied by the circles in PQRS?
(a) 860 km2 (b) 1260 km2 (a) 1 : 2 (b) 1 : 2
2
(c) 1060 km (d) can’t be determined (c) 2 : 3 (d) 1 : 1
484 QUANTUM CAT
58 A square paper is cut into smaller pieces such that each 65 In the adjoining figure, 7 congruent circles
piece is exactly a square, not necessarily of the equal size. are placed in such a way that one circle is
From each such square a circle of the largest possible area at the centre and other 6 circles are
is cut out. Find the ratio of the area of all the possible circles tangent to it. A regular hexagon is drawn
to the area of the original square. such that its vertices are the centres of the
(a) 1 : 1 6 circles, as shown here. If the area of
(b) π : 2 circle is 1 sq. cm, what is the area of the hexagon?
(c) π : 4 7 3 2π 3
(a) sq cm (b) sq cm
(d) data insufficient π 7
59 The elder son of a carpenter picked up a 6 3
(c) sq cm (d) none of these
solid wooden cube from his garage and then π
he removed a few cubical blocks from it so 66 A right angle triangle whose
that he can sit on it with his friends hypotenuse is c and its perpendicular
comfortably and watch the football match in sides are a and b. Three semicircles a
his house. If the volume of the original cube are drawn along the three sides of b
is 216 cubic ft. and the volume of each this triangle in such a way that the
c
smaller cubic solid that has been removed is 8 cubic ft., what diameter of each semicircle is equal
is the surface area (in cu. ft.) of the remaining solid? to the length of the respective side of the triangle.
(a) 216 (b) 192 The semicircle drawn with the help of hypotenuse
(c) 180 (d) none of these intersects the other two semicircles as shown in the
60 Find the area of the rectangle ABCD, if P is any point on AB concerned diagram. Find the area of the shaded region.
such that DP = 18 mm, CP = 21 mm and ∠DPC = 90°. ab 2abπ
(a) (b)
(a) 378 mm2 ( b) 189 mm2 2 3
(c) 126 mm 2
(d) 136 mm2 abπ
(c) (d) data insufficient
61 An ant has to go from one corner to the farthest corner of a c
canister of size 6×6×12. What’s the minimum distance it 67 In the following diagram, the height
has to cover, (All the lengths are in cm)? and base of an isosceles triangle are 16 16
(a) 6( 2 + 2) (b) 6 3 16 cm each. Find the area of the
(c) 12 2 (d) 12( 2 + 1) circumscribing circle.
62 The ratio of semiperimeter of a parallelogram to its longer (a) 256 sq cm (b) 324 sq cm
side is same as the ratio of its longer side to its shorter side. (c) 314 sq cm (d) none of these
If the shorter side measures 2 cm, find the maximum 68 A large city is developed on a square plot
possible area of the parallelogram. that has two uniform roads connecting the
(a) 2( 5 + 2) (b) 2(1 + 5) opposite corners of the city, as shown in
(c) 4(1 + 5) (d) data insufficient the diagram. If the total area of the roads
63 A semi-circle and a circle with same radius inscribe a is half the area of the square plot, what is
square of greatest possible area. What is the ratio of area of the ratio of the length of the plot to the
square inscribed by semicircle to that by circle? width of the each road?
(a) 2:5 (b) 1:2 (a) 5/2 2 (b) (4 + 2)/ 3
(c) 1:4 (d) 4:5 (c) 2 2 + 2 (d) none of these
64 In the following figure the radii (OA 69 The diameter of a semicircle and the C
and OD) are 2 cm and ∆ABO and ∆DCO base of an equilateral ∆ABC coincide in
are right angle triangles. Also, OB = 2 such a way that the other two sides of D E
cm and OC = 1 cm. Find the area of the B √2 O 1 C the triangle intersect the circumference
shaded region in the circle. of the semicircle at the points D and E,
5π 2 2
as shown in the diagram. If the A
(a) 1 + sq cm B
6 A D perimeter of the triangle ABC is 6 cm,
5π 3 find the area of the shaded region.
(b) + sq cm
6 2  2π 3 π 3
(a)  −  cm2 (b)  −  cm2
5π 3  3 2 3 2
(c) 2 + − sq cm
6 2  2 − 3
5π 3 (c) π   (d) none of these
(d) 1 + + sq cm  3 
6 2
Mensuration 485

LEVEL 02 > HIGHER LEVEL EXERCISE

Directions (for Q. Nos. 1 to 3) Each edge of an equilateral
triangle is ‘a’ cm. A cone is formed by joining any two sides of the
triangle. 8 A hollow sphere with outer diameter 24 cm is cut into two
1 What is the radius and slant height of the cone? equal hemispheres. The total surface area of one of the
2
(a) a,
a
(b)
a a
, hemispheres is 1436 cm2. Each one of the hemispheres is
2π π 2 7
a a filled with water. What is the volume of water that can be
(c) ,a (d) 2a, filled in each of the hemispheres?
2π π
2 2
2 What is the volume of the cone? (a) 3358 cm3 (b) 3528 cm3
3 3
a2 a3
(a) 4 − π2 (b) 4π 2 − 1 2
(c) 2359 cm3
2
(d) 9335 cm3
24π 3 24π 2 3 3
3
a a 2  2
(c) 1 − 4π 2 (d) π 1 −  9 A big cube of side 8 cm is formed by rearranging together 64
8π 2 3  π small but identical cubes each of side 2 cm. Further, if the
3 If the cone is cut along its axis from the middle, the new corner cubes in the topmost layer of the big cube are removed,
shape we obtain after opening the paper is : what is the change in total surface area of the big cube?
(a) isosceles triangle (b) equilateral triangle (a) 16 cm2, decreases
(c) right angle triangle (d) none of these (b) 48 cm2, decreases
4 If the sum of the radius and the height of a closed (c) 32 cm2, decreases
cylinder is 35 cm and the total surface area of the cylinder (d) remains the same as previously
is 1540 cm2, then the circumference of the base of the 10 A large solid sphere of diameter 15 m is melted and recast
cylinder is : into several small spheres of diameter 3 m. What is the
(a) 66 cm (b) 44 cm percentage increase in the total surface area of all the
(c) 56 cm (d) can’t be determined smaller spheres over that of the large sphere?
5 An iron pillar has some part in the form of a right circular (a) 200% (b) 400%
cylinder and remaining in the form of a right circular cone. (c) 500% (d) can’t be determined
The radius of base of cone as well as cylinder is 21 cm. The 11 A cone is made of a sector with a radius of 14 cm and an
cylindrical part is 80 cm high and conical part is 16 cm angle of 60°. What is total surface area of the cone?
high. Find the weight of the pillar, if 1 cm3 of iron weighs (a) 119.78 cm2 (b) 191.87 cm2
2
8.45 g. (c) 196.5 cm (d) none of these
(a) 999.39 kg (b) 111 kg 12 Kishan Chand is a very labourious farmer. He erected a
(c) 1001 kg (d) 989 kg fence around his paddy field in a square shape. He used 26
6 ABCD is a square of side a cm. poles in each side, each at a distance of 4 m. What is the
AB, BC , CD and AD all are the chords D C area of his field?
of circles with equal radii each. If the (a) 1.6 hectare (b) 2.6 hectare
chords subtend an angle of 120° at (c) 5.76 hectare (d) 1 hectare
their respective centres, find the total
area of the given figure, where arcs 13 A rectangular lawn is surrounded by a path of width 2 m on
A B all sides. Now, if the length of the lawn is reduced by 2 m
are part of the circles :
 the lawn becomes a square lawn and the area of path
 πa2 a2  
(a) a2 + 4  −  becomes 13/11 times, what is the length of the original
  9 3 2 
lawn?
  πa2 a2   (a) 8 m (b) 9 m (c) 10 m (d) 12 m
(b) a2 + 4  − 
  9 4 3  Directions (for Q. Nos. 14 and 15) A cylinder with height and
(c) [ 9a2 − 4π + 3 3a2] radius 2 : 1 is filled with soft drink and then it is tilted so as to
(d) none of the above allow some soft drink to flow off to an extent where the level of soft
drink just touches the lowest point of the upper mouth.
7 A rectangle has a perimeter of 26. How many combinations
of integral valued length are possible? 14 If the 2.1 L soft drink is retained in the cylinder, what is the
(a) 4 (b) 8 capacity of the cylinder?
(c) 6 (d) 12 (a) 3.6 L (b) 4 L (c) 1.2 L (d) 4.2 L
486 QUANTUM CAT
D P C
15 If the quantity of soft drink is poured into a conical flask 21 ABCD and EFGA are the squares
whose height and base radius are same as that of the of side 4 cm, each. In square M Q
cylinder so as to fill the conical flask completely, the ABCD, DMB and PMQ are the
E B
quantity of soft drink left in the cylinder as a fraction of its arcs of circles with centres at A
total capacity is : A and C , respectively. In square
1 1 1 1 AEFG, the shaded region is
(a) (b) (c) (d) F G
3 6 9 10 enclosed by two arcs of circles
16 An elephant of length 4 m is standing at one corner of a with centres at A and F , respectively. What is the ratio of
rectangular cage 16 m × 30 m and facing towards the the shaded regions of the squares ABCD and AEFG,
diagonally opposite corner. If the elephant starts moving respectively?
towards the diagonally opposite corner it takes 15 s to 2 + π ( 2 − 2) (π − 2)
(a) (b)
reach the opposite corner. Find the speed of the elephant. (π − 2) 2( 2 + 1 − π)
(a) 1 m/s (b) 2 m/s 4
(c) (d) none of these
(c) 1.87 m/s (d) can’t be determined 3
17 What is the height of the cone which is formed by joining AB AD
22 In the adjoining figure = , where EBCF is a square.
the two ends of a sector of a circle with radius r and central BC DF
angle 60°? AE
Find the ratio of .
35 25 EF
(a) r (b) r B
6 6 A E
2
r
(c) (d) none of these
3
18 If a cube of maximum possible volume is cut off from a solid
sphere of diameter d, the volume of the remaining (waste) D C F
material of the sphere would be equal to :
(1 ± 7 ) (1 − 7 )
d3  d d3  π 1 (a) (b)
(a) π −  (b)  −  3 2
3  2 3 2 3
(1 + 5) (1 ± 5)
d2 (c) (d)
(c) ( 2 − π) (d) none of these 2 2
4
Directions (for Q. Nos. 23 to 25) In the following figure ABCD is
19 In the adjoining figure PQRS is a square and MS = RN and
a square. A circle ABCD is passing through all the four vertices of
the points A, P , Q and B lie on the same line. Find the ratio the square. There are two more circles on the sides AD and BC
of the total area of two circles to the area of the square. touching each other inside the square. AD and BC are the
Given that AP = MS. respective diameters of the two smaller circles. Area of the square
S R is 16 cm2 .
D C
M N
3
1 2 2 1
A P Q B
3
π 2π 3π 6 A B
(a) (b) (c) (d)
3 3 2 π 23 What is the area of region 1?
20 ABCD is a rectangle and there are four  π
R (a) 2.4 cm2 (b)  2 −  cm2
equilateral triangles. Area of ∆ ASD is  4
equal to area of ∆ BQC and area of D C
(c) 8 cm2 (d) (4π − 2) cm2
∆ DRC is equal to area of ∆ APB. The S Q
24 What is the area of region 2?
perimeter of the rectangle is 12 cm. A B
(a) 3 (π − 2) cm2 (b) (π − 3) cm2
Also the sum of the areas of the four
triangles is 10 3 cm2, the total area of P (c) (2π − 3) cm2 (d) 4 (π − 2) cm2

the figure thus formed is : 25 What is the area of region 3?

(a) 2(4 + 5 3) cm2 (a) (4 − 4π ) cm2

(b) 5 (4 + 2 3) cm2 (b) 4 (4 − π ) cm2

(c) 42 3 cm2 (c) (4π − 2) cm2

(d) none of the above (d) (3π + 2) cm2

Mensuration 487

26 In the adjoining figure ABCD is a square. D C 30 What is the perimeter of all the five squares?
Four equal semicircles are drawn in such (4 2 + 1) a (4 2 − 1) a
a way that they meet each other at ‘O’. (a) (b)
( 2 + 1) ( 2 + 1)
Sides AB, BC , CD and AD are the
5
respective diameters of the four O (c) a (d) (7 + 3 2) a
A B 6
semicircles. Each of the sides of the
square is 8 cm. Find the area of the shaded region. 31 What is the total area of all the five squares?
(a) 32(π − 2) cm2 (b) 16 (π − 2) cm2 (4 2 − 1) a2 (4 2 − 1) a2
(a) (b)
3  (4 2 − 1) 4( 2 − 1)
(c) (2π − 8) cm2 (d)  π − 4 cm2
4  31 2
(c) a (d) none of these
27 ABCD is a square. Another square EFGH G 16
with the same area is placed on the D C
Directions (for Q. Nos. 32 to 35) Each edge of a cube is equally
square ABCD such that the point of
H F divided into n parts, thus there are total n 3 smaller cubes. Let,
intersection of diagonals of square ABCD N 0 → Number of smaller cubes with no exposed surfaces
and square EFGH coincide and the sides A B N 1 → Number of smaller cubes with one exposed surfaces
of square EFGH are parallel to the E N 2 → Number of smaller cubes with two exposed surfaces
diagonals of square ABCD. Thus a new N 3 → Number of smaller cubes with three exposed surfaces
figure is formed as shown in the figure. What is the area
enclosed by the given figure if each side of the square is 32 What is the number of unexposed smaller cubes (N 0)?
4 cm? (a) (n − 2)3 (b) n3 (c) n ! (d) 8
3+ 2  33 What is the number of smaller cubes with one exposed
(a) 32 (2 − 2) (b) 16  
 2+ 2  surface (N1 )?
2+ 2 (a) 4 (n − 3)3 (b) 6 (n − 2)2
(c) 32   (d) none of these
3+ 2 (c) (n − 3)2 (d) (n + 1)2

28 A piece of paper is in the form of a right angle triangle in 34 What is the value of (N 2 )?
which the ratio of base and perpendicular is 3 : 4 and (a) 8 (n − 2)2 (b) 6 (n − 2)
hypotenuse is 20 cm. What is the volume of the biggest (c) 12 (n − 2) (d) 3 (n − 3)2
cone that can be formed by taking right angle vertex of the
35 What is the value of N 3?
paper as the vertex of the cone?
(a) 45.8 cm3 (b) 56.1 cm3 (a) (n − 1)! (b) (n − 2)2
(c) 61.5 cm 3
(d) 48 cm3 n (n + 1)
(c) (d) 8
29 In a particular country the value of diamond is directly 2
proportional to the surface area (exposed) of the diamond. 36 In a bullet the gun powder is to be filled up inside the
Four thieves steel a cubical diamond piece and then divide metallic enclosure. The metallic enclosure is made up of a
equally in four parts. What is the maximum percentage cylindrical base and conical top with the base of radius
increase in the value of diamond after cutting it? 5 cm. The ratio of height of cylinder and cone is 3 : 2.
(a) 50% (b) 66.66% A cylindrical hole is drilled through the metal solid with its
(c) 100% (d) none of these height two-third of the height of metal solid. What should
be the radius of the hole, so that the volume of the hole
Directions (for Q. Nos. 30 and 31) In the figure shown square II (in which gun powder is to be filled up) is one-third of the
is formed by joining the mid-points of square I, square III is volume of metal solid after drilling?
formed by joining the mid-points of square II and so on. In this way 88 55 55
(a) cm (b) cm (c) cm (d) 33π cm
total five squares are drawn. The side of the square I is ‘a’ cm. 5 8 8
37 A sector of the circle measures 19°
II I (see the figure). Usingonly a scale, a
IV III compass and a pencil, is it possible to split
the circle into 360 equal sectors of 1° 19°
V
central angle?
(a) Yes
(b) No
(c) Yes, only if radius is known
(d) can’t be determined
488 QUANTUM CAT
38 A circular paper is folded along its O B 44 What is the ratio of the area of inner circle to that of the
diameter, then again it is folded to outer circle?
form a quadrant. Then it is cut as (a) 3 : 4 (b) 9 : 16
shown in the figure. After it the paper (c) 3 : 8 (d) none of these
was reopened in the original circular 45 If there are some more circles and hexagons inscribed in
shape. Find the ratio of the original A the similar way as given above, the ratio of each side of
paper to that of the remaining paper? outermost hexagon (largest one) to that of the fourth
(The shaded portion is cut off from the quadrant. The (smaller one) hexagon is (fourth hexagon means the
radius of quadrant OAB is 5 cm and radius of each hexagon which is inside the third hexagon from the
semicircle is 1 cm). outside) :
(a) 25 : 16 (b) 25 : 9 (a) 9 : 3 2 (b) 16 : 9
(c) 20 : 9 (d) none of these (c) 8 : 3 3 (d) none of these
39 A cubical cake is cut into several smaller cubes by dividing
Directions (for Q. Nos. 46 and 47) Five spheres
each edge in 7 equal parts. The cake is cut from the top
are kept in a cone in such a way that each sphere
along the two diagonals forming four prisms. Some of them
touches each other and also touches the lateral
get cut and rest remained in the cubical shape. A complete surface of the cone. It is due to increasing radius of
cubical (smaller) cake was given to adults and the cut off the spheres starting from the vertex of the cone. The
part of a smaller cake is given to a child (which is not an radius of the smallest sphere is 16 cm.
adult). If all the cakes were given equally each piece to a
person, total how many people could get the cake? 46 If the radius of the fifth (i.e., the largest)
(a) 343 (b) 448 sphere be 81 cm, find the radius of the third (i.e., the
(c) 367 (d) 456 middlemost) sphere.
(a) 25 cm (b) 25 3 cm
Directions (for Q. Nos. 40 to 42) A square is inscribed in a circle
(c) 36 cm (d) data insufficient
then another circle is inscribed in the square
then. Another square is inscribed in the circle. 47 What is the least distance between the smallest sphere and
Finally a circle is inscribed in the innermost the vertex of the cone?
square. Thus there are 3 circles and 2 squares (a) 64 cm (b) 80 cm
as shown in the figure. The radius of the (c) 28 cm (d) none of these
outer-most circle is R. 48 Saumya has a pencil box of volume 60 cm3. What can be
40 What is the radius of the inner-most circle? the maximum length of a pencil that can be accommodated
R R in the box. Given that all the sides are integral (in cm) and
(a) (b)
2 2 different from each other?
(c) 2R (d) none of these (a) 7 2 cm (b) 905 cm
41 What is the sum of areas of all the squares shown in the (c) 170 cm (d) 3602 cm
figure? 49 There are two concentric hexagons. Each of
(a) 3R 2 (b) 3 2R 2 the side of both the hexagons are parallel.
3 2 Each side of an internal regular hexagon is
(c) R (d) none of these
2 8 cm. What is the area of the shaded region,
if the distance between corresponding
42 What is the ratio of sum of circumferences of all the circles
parallel sides is 2 3 cm?
to the sum of perimeters of all the squares?
(a) 120 3 cm2 (b) 148 3 cm2
(a) (2 + 3) πR (b) (3 + 2) πR
(c) 3 3 πR (d) none of these (c) 126 cm2 (d) none of these

Directions (for Q. Nos. 43 to 45) A regular 50 ABCD is a square. A circle is inscribed in D C

the square. Also taking A, B, C , D (the
hexagon is inscribed in a circle of radius R.
Another circle is inscribed in the hexagon. Now vertices of square) as the centres of four
another hexagon is inscribed in the second quadrants, drawn inside the circle,
(smaller) circle. which are touching each other on the
mid-points of the sides of square. Area A B
2
43 What is the sum of perimeters of both the hexagons? of square is 4 cm . What is the area of the shaded region?
(a) (2 + 3) R  3π 
3 (2 + 3) R (a)  4 −  cm
2
(b) (2π − 4) cm2
(b)  2
(c) 3 (3 + 2) R
(c) (4 − 2π ) cm2 (d) none of these
(d) none of the above
Mensuration 489

51 In a factory there are two identical solid blocks of iron.

When the first block is melted and recast into spheres of 56 Initially the diameter of a balloon is 28 cm. It can explode
equal radii ‘ r’, the 14 cc of iron was left, but when the when the diameter becomes 5/2 times of the initial
second block was melted and recast into sphere each of diameter. Air is blown at 156 cc/s. It is known that the
equal radii ‘ 2r’, the 36 cc of iron was left. The volumes of shape of balloon always remains spherical. In how many
the solid blocks and all the spheres are in integers. What is seconds the balloon will explode?
the volume (in cm3) of each of the larger spheres of radius (a) 1078 s (b) 1368 s
‘ 2r’? (c) 1087 s (d) none of these
(a) 176 57 The radius of a cone is 2 times the height of the cone.
(b) 12π A cube of maximum possible volume is cut from the same
(c) 192 cone. What is the ratio of the volume of the cone to the
(d) data insufficient volume of the cube?
52 There is a vast grassy farm in which there is a rectangular (a) 3.18 π
building of the farm-house whose length and breadth is 50 (b) 2.25 π
m and 40 m respectively. A horse is tethered at a corner of (c) 2.35
the house with a tether of 80 m long. What is the maximum (d) can’t be determined
area that the horse can graze? 58 Raju has 64 small cubes of 1 cm3. He wants to arrange all of
(a) 5425 π (b) 5245 π them in a cuboidal shape, such that the surface area will be
(c) 254 π (d) none of these minimum. What is the diagonal of this larger cuboid?
53 A cube of side 6 cm is painted on all its 6 faces with (a) 8 2 cm (b) 273
red colour. It is then broken up into 216 smaller identical (c) 4 3 cm (d) 129 cm
cubes. What is the ratio of N 0 : N1 : N 2. 59 The volume of a cylinder is 48.125 cm3, which is formed by
Where, N 0 → number of smaller cubes with no coloured rolling a rectangular paper sheet along the length of the
surface. paper. If a cuboidal box (without any lid i.e., open at the
N1 → number of smaller cubes with 1 red face. top) is made from the same sheet of paper by cutting out
N 2 → number of smaller cubes with 2 red faces : the square of side 0.5 cm from each of the four corners of
(a) 3 : 4 : 6 (b) 3 : 4 : 5 the paper sheet, then what is the volume of this box?
(c) 4 : 6 : 3 (d) can’t be determined (a) 20 cm3
54 Assume that a mango and its seed both are spherical. Now, (b) 38 cm3
if the radius of seed is 2/5 of the thickness of the pulp. The (c) 19 cm3
seed lies exactly at the centre of the fruit. What per cent of (d) none of the above
the total volume of the mango is its pulp?
3 Directions (for Q. Nos. 60 to 62) Consider a cylinder B
(a) 63 % (b) 97.67%
5 4
of height h cm and radius r = cm as shown in the
(c) 68 %
2
(d) none of these
π
3 figure. A string of certain length when wound on its
55 In the adjoining diagram ABCD is a square with side ‘ a’ cm. cylindrical surface, starting at point A, gives a
In the diagram the area of the larger circle with centre ‘ O ’ is maximum of n turns. A
equal to the sum of the areas of the remaining four circles
60 What is the vertical spacing (in cm) between two
with equal radii, whose centres are P , Q , R and S. What is
consecutive turns?
the ratio between the side of square and radius of a smaller
h h
circle? (a) (b)
n n
D C
h2
(c) (d) can’t be determined
S R n

O 61 If there is no spacing between any two courecutive turns

and the width of string be x cm, then the required length of
P Q the string is :
A B
8x 8h
(a) cm (b) cm
(a) (2 2 + 3) (b) (2 + 3 2) h x
(c) (4 + 3 2) (d) can’t be determined h
(c) 8hx cm (d) 2 2 cm
x
490 QUANTUM CAT
62 If the string is wound on the exterior four walls of a cube of 68 There are two cylindrical containers of equal capacity and
side a cm starting at point C and ending at point D exactly equal dimensions. If the radius of one of the containers is
above C, making equally spaced 4 turns. The side of the increased by 12 ft and the height of another container is
cube is : increased by 12 ft, the capacity of both the container is
2n (n)2 equally increased by K cubic ft. If the actual heights of each
(a) a = (b) a =
255 16 of the containers be 4 ft, find the increased volume of each
8n of the container.
(c) a = (d) a = 2 15n
257 (a) 1680 π cu ft (b) 2304 π cu ft
(c) 1480 π cu ft (d) can’t be determined
63 A blacksmith has a rectangular iron sheet 10 ft long. He has
to cut out 7 circular discs from this sheet. What is the 69 The four vertices of a unit square act as the
minimum possible width of the iron sheet if the radius of centers of four quarter-circles. The arcs of
each disc is 1 ft? these circles intersect each other within the
(a) 2 3 ft (b) (2 + 3) ft square as depicted below. If the area of the
(c) (3 + 2) ft (d) (2 + 2 3) ft region at the center of the square, which
itself looks like a bloated square, is cordoned off by all the
64 The perimeter of a square, a rhombus and a hexagon are π
same. The area of square, rhombus and hexagon be s, r, h, four arcs of circles is − 3 + 1, find the area of the shaded
3
respectively, which of the following is correct?
region.
(a) r > s > h (b) s > h > r 2π
(c) h > s > r (d) data insufficient (a) 4 − 3 −
3
65 In the following figure, ABC is an equilateral triangle 2π
(b) 4 − 2 3 −
inscribing a square of maximum possible area. Again in this 3
square there is an equilateral triangle whose side is same as π
(c) 4 − 3 −
that of the square. Further the smaller equilateral triangle 3
inscribes a square of maximum possible area. What is the (d) none of the above
area of the innermost square if the each side of the 70 Four small squares of equal size are cut off from each of the
outermost triangle be 0.01 m? four corners of a square sheet of area 576 sq cm. The
C
remaining sheet is then folded in such a way that it
becomes an open cuboidal box having maximum holding
capacity. Find the side of the smaller square that has been
cut off.
A B (a) 6 (b) 3 (c) 2.4 (d) 4
71 The following diagram depicts the
(a) (873 − 504 3) cm 2
(b) (738 − 504 3) cm2
Earth. For the sake of convenience
(c) (873 − 405 2) cm2 (d) none of these and simplicity let us consider that the
66 A blacksmith has a rectangular sheet of iron. He has to earth is completely spherical. The
make a cylindrical vessel of which both the circular ends topmost point (N), where all the
are closed. When he minimises the wastage of the sheet of great circles meet, is called the North
iron, then what is the ratio of the wastage to the utilised Pole. South Pole is denoted by the
area of sheet? point (S) is exactly opposite the
1 2 North Pole. All the vertical circles are passing through each
(a) (b)
11 17 pole are called longitudes and all the horizontal circles that
3 are parallel to each other are called as latitudes.
(c) (d) none of these
22 Let the centre of the earth be denoted by C and the radius
67 Barun needs an open box of capacity 864 m3. Actually, of the earth is 6370 km. A man has to reach from the North
where he lives the rates of paints are soaring high so he Pole to a point D on the latitude with radius3185 3 km in
wants to minimize the surface area of the box keeping the the Southern hemisphere of the earth. Find the minimum
capacity of the box same as required. What is the base area distance along the surface of the earth that he will have to
and height of such a box? traverse.
(a) 36 m2, 24 m (b) 216 m2, 4 m (a) 40040 km (b) 32026.67 km
2 (c) 13346.67 km (d) 14367.33 km
(c) 144 m , 6 m (d) none of these
Mensuration 491

72 A rectangle of dimensions 32×49 has two circles inscribed 77 There are four quadrants of 10 cm radius constructed
in it. What’s the maximum possible total area of the two within a square using the four vertices of that square. If the
circles? area of the square is 100 sq. cm, find the common area (in
(a) 337 π (b) 306π sq cm) among all the four quadrants, shown by the shaded
(c) 518π (d) 245π region.
73 A sheep and a goat are tethered at the diagonally opposite
ends of a square field. The length of each rope they are
attached by the post is 6m and the area of the field is 54 m2.
Find the approx. area that cannot be grazed by either of
them.
8
(a) 3m2 (b) 4m2 (c) 7 m2 (d) m2  2π  π 
π (a) 100  + 2 − 3 (b) 100  + 1 − 3
 3  3 
74 What is the area of the largest semicircle that can be π 
inscribed in the square whose diagonal is a 2 units? (c) 100  + 4 − 2 3 (d) none of these
3 
(a) 0.172πa2 (b) 0.125πa2
(c) 0. 225πa2
(d) 1. 25πa2 78 A thin opaque square glass sheet (30 cm × 30 cm) is cut
75 A semicircle of radius r is inscribed in a square such that the into four pieces of equal area, as shown in figure (i). When
diameter of the semicircle makes 60° angle with one of the these pieces are rearranged to form another square, as
sides of the square. What is the least possible area of the shown in figure (ii), it results in a
square that can inscribe such a semicircle? square hole (16 cm × 16 cm) at the
 7 + 4 3 2  11 + 6 2 2 centre of the new square. Find the
(a)   r (b)   r length of each side of the new
 4   4  square, which is larger than the Fig. (i) Fig. (ii)
 4 + 2 3 2 original one.
(c)   r (d) (5 + 7 )r2
 9  (a) 34 cm (b) 32 sq cm
(c) 42 sq cm (d) none of these
76 There are four quadrants of 10 cm radius
constructed within a square using the four 79 A rectangular paper is folded along its diagonal to form a
vertices of the square as their centers. These polygon as shown in the following figure. If the ratio of the
four quadrants share a common region, area of this polygon to that of the rectangle is 16/5, the
which is shown by the shaded region in the ratio of the length to the breadth of the original rectangle is
B′
adjoining figure. If the area of the given square is 100 sq
cm, find the area of the largest possible square that can be
D E C
inscribed in the common region occupied by four
quadrants.
(a) 100(4 − 3) sq cm A B

(b) 100(2 − 3) sq cm (a) 5/ 11 cm (b) 4/3 cm

(c) 100(4 − 2 3) sq cm (c) 21/11 cm (d) data insufficient
(d) data insufficient

Test of Your Learning

1 Three equal circles each of radius 1 cm 2 Three circular rings of equal radii of 1 cm
are circumscribed by a larger circle. Find each are touching each other. A string runs
the perimeter of the circumscribing circle. all around the set of rings very tightly.
3 What is the minimum length of string
(a) (2 − 3) π cm
2 required to bind all the three rings in the
 2 + 3 given manner?
(b)   cm 6
 3  (a) cm (b) 2 (3 + π ) cm
π
2 π
(c) (2 + 3) π cm (c) cm (d) can’t be determined
3 6
(d) none of the above
492 QUANTUM CAT
A
3 An equilateral triangle circumscribes 9 In a right angle triangle ABC,
all the three circles each of radius 1 what is the maximum possible
cm. What is the perimeter of the area of a square that can be b c
equilateral triangle? inscribed when one of its D
(a) 6 ( 3 + 1) cm vertices coincide with the
90°
(b) 3 (8 + 2) cm vertex of right angle of the C B
(c) 15 ( 3 − 1) cm triangle? a
(d) none of the above a ab
(a) (b)
4 Six circles each of unit radius are being b a+ b
2
circumscribed by another larger circle. a+ b  ab 
All the smaller circles touch each other. (c) (d)  
ab  a + b
What is the circumference of the larger
circle? 10 In the adjoining figure a square of B
 3 + 4 maximum possible area is
(a)   π cm circumscribed by the right angle
 2  c
triangle ABC in such a way that a
(b) 4 3π cm one of its side just lies on the
 4 + 3 hypotenuse of the triangle. What C 90°
(c) 2   π cm A
 3  is the area of the square? b

(d) can’t be determined 2

 abc  a2 + b2 + c2
5 There are six circular rings of iron, kept (a)  2  (b)
 a + b + ab
2
abc
close to each other. A string binds them as
tightly as possible. If the radius of each abc
(c) (d) none of these
circular iron ring is 1 cm. What is the a + b2 + c2
2

minimum possible length of string 11 In the adjoining figure PQRS is a square of maximum
required to bind them?
possible area which is circumscribed by the semicircle.
(a) 2 (6 + 3 3 + π ) cm (b) 6 (2 + 3) π cm
Points R and S lie on the diameter AB. What is the area of the
(c) 2 (6 + π ) cm (d) none of these square if the radius of the circle is ‘ r’?
6 An equilateral triangle C P Q
circumscribes all the six circles,
each with radius 1 cm. What is the
perimeter of the equilateral
A S C R B
triangle?
(a) 6 (2 + 3) cm 3 2 4 2 3 2 5 2
(a) r (b) r (c) r (d) r
(b) 3 ( 3 + 2) cm A B 4 5 5 4
(c) 12 ( 3 + 4) cm 12 In the adjoining figure a quadrant (of A
(d) none of the above circle) inscribes a square of maximum P R
7 A cube of maximum possible volume is cut from the sphere possible area. If the radius of the circle
of diameter 3 3 cm. What is the ratio of volume of the be ‘ r’ then what is the area of the square?
sphere to that of cube? B
r2 3r2 C Q
4 3 3 (a) (b)
(a) (b) π 2 5
π 2
r2
4 (c) (d) 2 6r
(c) π (d) none of these 3
3
8 A cube of maximum possible volume is cut from the solid 13 In the adjoining figure, AB is the
diameter of a semicircle which C
right circular cylinder. What is the ratio of volume of cube
inscribes a circle of maximum
to that of cylinder if the edge of a cube is equal to the height
possible area. If the radius of the A B
of the cylinder?
larger circle (i . e. , semicircle) is r, the area of the inscribed
11 π
(a) (b) 2 circle is :
7 7
7 5r2 2π 2 πr2
(c) (d) none of these (a) (b) r (c) (d) none of these
11 π 3 4
Mensuration 493

14 In a quadrant (of a circle) a circle of 16 A 12 cm long wire is bent to form a triangle with one of its
maximum possible area is given. If the radius angle as 60°. Find the sides of the triangle (in cm) when its
of the circumscribing quadrant be r, then area is largest?
what is the area of the inscribed circle? (a) 3, 4, 5
π r2 (b) 2, 4, 6
(a) (2 + 3 2) r2 (b)
(3 + 2 2) (c) 4, 4, 4
 3 + 2 2 (d) 2.66, 3.66, 4.66
(c)   π (d) none of these
17 Let S1, S2, … , Sn be the squares such that for each n > 1, the
 r2 
length of a side of Sn equals the length of the diagonal of
15 A cylindrical chocobar has its radius r unit and height
S( n + 1). If the length of a circle of S1 is 10 cm, then for which of
‘ h’ unit. If we wish to increase the volume by same unit
the following values of n the area of Sn is less than 1 square
either by increasing its radius alone or its height alone,
cm?
then how many unit we have to increase the radius or
(a) 4 (b) 8 (c) 6 (d) 7
height?
r2 + 2r r2 − 2rh
18 Area of a regular hexagon and a regular octagon is same.
(a) (b) Which one of the two has larger perimeter?
h h
(a) Hexagon (b) Octagon
2r2 − rh πr2
(c) (d) (c) can’t be determined (d) none of these
h2 2h

Answers
Introductory Exercise 10.1
1 (b) 2 (c) 3 (c) 4 (b) 5 (a) 6 (a) 7 (b) 8 (a) 9 (a) 10 (c)
11 (b) 12 (d) 13 (a) 14 (b) 15 (b) 16 (c) 17 (c) 18 (d) 19 (c) 20 (b)
21 (b) 22 (a) 23 (c) 24 (c) 25 (c) 26 (b) 27 (c) 28 (c) 29 (b) 30 (c)
31 (b) 32 (c)

Introductory Exercise 10.2

1 (b) 2 (c) 3 (a) 4 (a) 5 (d) 6 (a) 7 (c) 8 (b) 9 (a) 10 (a)

Introductory Exercise 10.3

1 (a) 2 (c) 3 (d) 4 (b) 5 (d) 6 (b) 7 (a) 8 (b) 9 (d) 10 (c)
11 (b) 12 (a) 13 (b) 14 (d) 15 (a)

Introductory Exercise 10.4

1 (d) 2 (d) 3 (a) 4 (b) 5 (c) 6 (d) 7 (a) 8 (a) 9 (a) 10 (b)
11 (c) 12 (d) 13 (b) 14 (a) 15 (b) 16 (a) 17 (a) 18 (c) 19 (a) 20 (c)
21 (a) 22 (b) 23 (b) 24 (c) 25 (c)

Introductory Exercise 10.5

1 (b) 2 (c) 3 (a) 4 (b) 5 (c) 6 (a) 7 (b) 8 (a) 9 (b) 10 (c)
11 (d) 12 (a) 13 (c) 14 (b) 15 (c) 16 (a) 17 (b) 18 (d) 19 (b) 20 (c)
21 (d) 22 (b) 23 (c) 24 (a) 25 (b) 26 (a) 27 (c) 28 (c) 29 (c) 30 (b)
31 (b) 32 (b)
494 QUANTUM CAT
Introductory Exercise 10.6
1 (b) 2 (c) 3 (c) 4 (c) 5 (b) 6 (a) 7 (c) 8 (b) 9 (a) 10 (c)
11 (a) 12 (a) 13 (d) 14 (a) 15 (d) 16 (c) 17 (a) 18 (b) 19 (a) 20 (b)
21 (c) 22 (a) 23 (d) 24 (c) 25 (d) 26 (d) 27 (d) 28 (c) 29 (b) 30 (c)
31 (b) 32 (a) 33 (d) 34 (c) 35 (b) 36 (b) 37 (b) 38 (a) 39 (b) 40 (a)

Introductory Exercise 10.7

1 (b) 2 (c) 3 (c) 4 (a) 5 (b) 6 (c) 7 (c) 8 (d) 9 (c) 10 (c)
11 (d) 12 (a) 13 (c) 14 (d) 15 (b) 16 (c) 17 (a) 18 (d) 19 (b) 20 (a)

Multifaceted Exercise
1 (c) 2 (c) 3 (c) 4 (c) 5 (c) 6 (a) 7 (c) 8 (a) 9 (c) 10 (a)
11 (d) 12 (c) 13 (a) 14 (c) 15 (d) 16 (b) 17 (b) 18 (c) 19 (b) 20 (d)
21 (b) 22 (c) 23 (d) 24 (c) 25 (d) 26 (d) 27 (d) 28 (b) 29 (b) 30 (b)
31 (a) 32 (c) 33 (d) 34 (c) 35 (a) 36 (c) 37 (b) 38 (c) 39 (d) 40 (a)
41 (b) 42 (a) 43 (c) 44 (c) 45 (a) 46 (b) 47 (b) 48 (c) 49 (d) 50 (b)
51 (c) 52 (c) 53 (a) 54 (d) 55 (c) 56 (a) 57 (b) 58 (a) 59 (c)

Level 01 Basic Level Exercise

1 (a) 2 (b) 3 (b) 4 (b) 5 (c) 6 (d) 7 (c) 8 (b) 9 (a) 10 (c)
11 (c) 12 (a) 13 (b) 14 (a) 15 (a) 16 (c) 17 (a) 18 (c) 19 (a) 20 (d)
21 (b) 22 (a) 23 (d) 24 (c) 25 (c) 26 (a) 27 (b) 28 (c) 29 (c) 30 (b)
31 (d) 32 (a) 33 (c) 34 (c) 35 (c) 36 (a) 37 (a) 38 (a) 39 (b) 40 (a)
41 (b) 42 (d) 43 (d) 44 (c) 45 (a) 46 (c) 47 (a) 48 (c) 49 (b) 50 (b)
51 (c) 52 (c) 53 (b) 54 (c) 55 (d) 56 (b) 57 (d) 58 (c) 59 (a) 60 (a)
61 (c) 62 (b) 63 (a) 64 (d) 65 (c) 66 (a) 67 (c) 68 (c) 69 (b)

Level 02 Higher Level Exercise

1 (c) 2 (b) 3 (c) 4 (b) 5 (a) 6 (b) 7 (c) 8 (a) 9 (d) 10 (b)
11 (a) 12 (d) 13 (c) 14 (d) 15 (b) 16 (b) 17 (a) 18 (b) 19 (b) 20 (a)
21 (a) 22 (c) 23 (c) 24 (d) 25 (b) 26 (a) 27 (a) 28 (b) 29 (c) 30 (d)
31 (c) 32 (a) 33 (b) 34 (c) 35 (d) 36 (b) 37 (a) 38 (a) 39 (b) 40 (a)
41 (a) 42 (d) 43 (b) 44 (a) 45 (c) 46 (c) 47 (a) 48 (b) 49 (a) 50 (b)
51 (a) 52 (a) 53 (c) 54 (b) 55 (b) 56 (a) 57 (b) 58 (c) 59 (a) 60 (a)
61 (b) 62 (c) 63 (b) 64 (c) 65 (a) 66 (a) 67 (c) 68 (b) 69 (a) 70 (d)
71 (c) 72 (a) 73 (b) 74 (a) 75 (a) 76 (b) 77 (b) 78 (a) 79 (a)

Test for Your Learning

1 (c) 2 (b) 3 (a) 4 (c) 5 (c) 6 (a) 7 (b) 8 (c) 9 (d) 10 (a)
11 (b) 12 (a) 13 (c) 14 (b) 15 (c) 16 (c) 17 (b) 18 (a)
Mensuration 495

Hints & Solutions

Introductory Exercise 10.1
1 2 (16 x + 9 x ) = 750; l = 16 x and b = 9 x  d2 
12 d 2 = 2 × area Q Area = 
110  2
2 Area = = 220 sq m
0.5 d 2 = 2 × 2 ⇒ d = 2 km
and 11 x × 5x = 220 13 100 × 100 = 10000 m2 = 1 hectare
3 Length : breadth = 3 : 2 111 × 111 = 12321 m2
Also, area of floor = area of roof
∴ Difference in area = 2321 m2
and 30 is almost 50% of 62. So you need not to solve it. Just
759 − 561
look out for the appropriate option. Thus 1411.2 is almost 14 Rate = = ` 33 per metre
50% of 2916.48 and rest of the options are not satisfactory. 6
1
4 1 hectare = 10000 m2, find length and breadth, then 15 l×b= × 10000 = 1000 × b ⇒ b = 0.4 m
25
perimeter
Perimeter = 1000 m = 1 km 16 Side = (289)2 − (240)2
distance 1 1600 × 900
∴ Time = = hrs = 20 min 17 = 1800
speed 3 40 × 20
5 5 hectares and 76 ares = 57600 m2, find perimeter then d2
18 Area =
multiply it by 6. 2
6 l × 0.7 = 19 × 3.5 ⇒ l = 95 m (70 cm = 0.7 m) 19
1 1 1
× =
3 3 9
Also (95)2 = 9025
1
So 95 × 0.95 = 90.25 ∴ 90 × = 10 g (It depends upon area)
9
Area of courtyard
7 Number of stones = = 14400 20 l : b = 2 : 1
Area of one stone
2x 2 = 24200, find diagonal.
Cost = Rate × Number of stones
= (0.5 × 14400) (110)2 + (220)2 = 110 12 + 22
8 Area = (63 + 54 − 6) × 6 [Q Area = (l + b − 2w ) 2w] = 110 × 5
Area = (111 × 6) = 110 × 2.236 = 246 m
37 37 × 3 111 21 2 (2x + x ) × 11 = 2640, find 2x × x = ?
and Rate = = =
2 2× 3 6
111 22 Consider some appropriate values.
∴ Cost = (111 × 6) ×
6 4 6
= 111 × 111 = 12321 e.g. 4 4 2 2
9 Original area = l × b 4 6
b
New area = 2l × = l × b 4a = 16 cm 2 (l + b) = 16 cm
2
but a = 16 cm
2 2
and l × b = 12 cm2
Hence, no change.
⇒ a >l×b
2
10 (116 + 68 − 2w ) 2w = 720
Remember, when a + b is constant, then the maximum
Solve through quadratic equation.
value of a × b is at when a = b.
Alternatively Go through options and put w = 2 from So, area of square (i . e. , l × b = a × a) is always greater than
option (c) you will find that both sides are equal. So the area of rectangle with same perimeter.
presumed choice is correct. Alternatively Let each side of a square be a and length
11 Net area = Total area of 4 walls − 8m2 and breadth of the rectangle be l and b, then
and Area of 4 walls = 2 (l + b) × h  l + b
4a = 2 (l + b) ⇒ a =  
 2 
496 QUANTUM CAT
∴ Area of the rectangle = l × b ⇒ l − b = 20
1
and Area of the square = a2 = (l + b)2 (Solving l + b = 40 and l − b = 20)
4 ∴ l = 30 and b = 10
But since we know that for any (different) certain values 100 × 100
24 = 16 (1 m = 100 cm)
AM > GM 25 × 25
l+ b
∴ > l×b 25 (120 + 80 − 24) × 24 = 176 × 24 = 4224
2
 l + b
2 26 (90 + 40 + 10) × 10 = 1400
⇒   >l×b
 2  27 2[( x + 2) + x] = 48 ⇒ x = 11 cm
⇒ Area of square > Area of rectangle ∴ ( x + 2) = 13 cm
23 Best way is to go through options and verify the result. 52000
28 = 160
Alternatively 2 (l + b) = 4a = 80 325
3.25 cm means 325 m, as per scale.
⇒ l + b = 40 and a = 20 ⇒ a2 = 400
d2
Also a2 − lb = 100 29 Area = ; d → diagonal
2
⇒ 400 − lb = 100
30 1.2 × 1.2 = 1.44 ⇒ 44% increase
⇒ lb = 300
Now (l − b)2 = (l + b)2 − 4lb 31 a2 : (a 2)2 = 1 : 2
(l − b)2 = 1600 − 1200 32 1.6 × 1.4 = 2.24 ⇒ increase = 1.24 ⇒ 124%

Introductory Exercise 10.2

1 Use Hero’s formula: (Since in ∆ ABC there are 4 similar triangles having same
Area of scalene triangle = s (s − a)(s − b)(s − c) area as ∆ DEF.)
A
2 D

F E
M
A 30 m
36 m
15 m B D C
b
5 60 = 4a2 − b2
B 36 m C 4
AM = BC = 36 m b
⇒ 60 = 676 − b2
4
MD = CD − MC = 30 − 15 (Q MC = AB = 15 m)
⇒ 57600 = b2 (676 − b2 )
∴ MD = 15 m
Now using Pythagoras theorem
AD 2 = AM 2 + MD 2, find AD By solving the above equation we can get the value of b.
Alternatively You can go through option.
3 Find the area, using Hero’s A
formula, then 6 Use pythagorus theorem
1 x 2 = (60)2 + (11)2
Area = ×b×h 25 m 39 m
2 Pythagorus theorem :  x 60
1  2 2 2
= × 56 × AD (Hypotenuse) =(base) + (height) 
2 D
B C 495.72
56 m 7 Area of field = 11
4 All the triangles are similar to 36.72
each other. = 13.5 hectare = 135000 m2
Area of ∆ ABC 4 Now, if h = x, b = 3x
∴ =
Area of ∆ DEF 1 1
∴ × b × h = 135000
2
Mensuration 497

1 10 Let each side of a square be 3x and each side of an

8 Area of triangle = ×b×h
2 equilateral triangle be 4 x
Let initially area of triangle = 1 × 1 = 1 unit then perimeter of square = 4 × 3x = 12x
Now, the area of triangle = 2 × 2 = 4 unit
and perimeter of equilateral triangle = 3 × 4 x = 12x
4 −1
Increase in area = × 100 = 300% Now area of square = (3x )2 = 9 x 2
1
(For your convenience assume any value of b and h.) 3
3 Area of equilateral triangle = × (4 x )2 = 4 3x 2
9 Height of an equilateral triangle = × side 4
2 Since 9 > 4 3 ⇒ 9 x 2 > 4 3x
3
∴ 2 3= × side ⇒ Side = 4 cm Hence area of square is greater than area of equilateral
2
3 triangle, when the perimeter of both is same.
∴ Area of an equilateral triangle = × (side)2
4 Alternatively Consider any suitable value and verify.
3
= × 4 × 4 = 4 3 cm2
4

Introductory Exercise 10.3

1 Area = 6 × 8 sin 30° 1
8 × x × 2x = 256 ⇒ x 2 = 256 ⇒ x = 16
1  1 2
= 6 × 8 × = 24 cm2  sin 30° = 
2  2 ∴ 2x = 32
2 To find the area of parallelogram, first find the area of Hence, x + 2x = 16 + 32 = 48
∆ ABC by Hero’s formula then double it. 1
9 Area = × (30 + 50) × 16 = 640 cm2
D C 2
1
40 × AB × OQ
20 Area of ∆ AOB 2
10 =
Area of ∆ COD 1 × CD × PO
A 30 B
2
Since, area of ABCD = 2 × Area of ∆ ABC D P C

= 2 × Area of ∆ ACD
O
3 Area = 22 × 24 = 528 cm2

A B
22 Q
AB × OQ 2CD × 2OP 4 Q AB = 2 CD
= = =  
24 CD × PO CD × OP 1  and OQ = 2 PO 
4 Area = 40 × 18 = 720 cm2 This is due to the similarity of triangles AOB and COD.
11 Area of trapezium = 441
25 18
1
⇒ (5x + 9 x ) × 21 = 441 ⇒ 14 x = 42 ⇒ x = 3
2
40 ∴ 9 x = 27 cm
Area of parallelogram ABCD BC × AN 2 × 4x 8 1
5 = = = 12 × (12 + 8) × h = 360 ⇒ h = 36 m
Area of triangle ABN 1 x 1 2
× BN × AN
2 3 3 3 3 3
1 13 6 × × (Side)2 = (Side)2 = × 4 × 4 = 24 3 m2
6 Area of rhombus = × product of diagonals 4 2 2
2
1
1
= ×a×b=
ab 14 Area of quadrilateral = × 19 × (5 + 7 ) = 114 cm2
2 2 2

10 x 2  Diagonal 
7 Area of rhombus =
1
× 2x × 5x = = 5x 2
15 AB = BC = CD = AD = 4 cm Q Side = 
 2 
2 2
and square of the shorter diagonal = (2x )2 = 4 x 2 and EF = 1.5 cm (By Pythagorus theorem)
5x 2
5 ∴ Area of ABCDE = Area of ABCD + Area of AED
∴ =
4x2 4 1
= (4)2 + × 4 × 1.5 = 19 cm2
2
498 QUANTUM CAT
Introductory Exercise 10.4
1 2πr = 704 ⇒ r = 112 11
22 a
∴ πr2 = × 112 × 112 = 39424 cm2 a a
7
a
7
2 2πr = 4.4 ⇒ r = m
10
∴ πr2 = 0.49π m2 Area of square = a2
2
22 1  a πa2
3 2πr = 2 × × 4.2 = 26.4 metre Area of circular parts = 4 × π  =
7 2  2 2
26.4 = 2 (6 x + 5x ) ⇒ 6 x = 7.2 m πa2  π
∴ Total area = a2 + = a2 1 + 
4 2πr = 440 ⇒ r = 70 m 2  2
12 πr2 = 2464 ⇒ r = 28 m
13 Basically there are 12 equilateral triangles each of side ‘a’.
70
14 m

84 m

∴ R = 70 + 14 = 84 m
 44 Fig. (i) Fig. (ii)
5 r + 2πr = 51 ⇒ r 1 +  = 51 ⇒ r = 7,
 7
3
Find area. ∴ 12 × × (a)2 = 3 3a2
4
 22  1 7 
2
6 (2πr − 2r) = 15 ⇒ 2r  − 1 = 15 1
7  14 × π × (7 )2 + 2  × π ×   
2  2  2 

7
⇒ r= Area of larger semicircle + 2 (area of smaller semicircle)
2
75 60
∴
22 7 7
πr2 =
× × = 38.5 m2 15 2π × R × = 25 ⇒ R =
7 2 2 360 π
New area π × 7 × 7 49 16 Ungrazed area = Area of square − 4 (area of quadrants)
7 = =
Original area π × 5 × 5 25 R
21 21
24 D C
Change in area = × 100 = 96%
25
New circumference 6 New radius
8 = = S Q
Original circumference 5 Original radius
2
New area  6 36
∴ =  =
Original area  5 25
A P B
11 42
∴ Change in area = × 100 = 44%
25 1
Alternatively = (42)2 − 4 × × π (21)2 = (21)2 [ 4 − π ] = 378
Change = (1.2 × 1.2) − (1 × 1) = 0.44 4
∴ Change = 44% 1
17 20 + × [ 2π × 10] = 20 + 10π
9 πr = 124.74 hectare
2 2
40 1 (360 − 40) 8
πr2 = 1247400 m2 ⇒ r = 630 m 18 = ∴ =
360 9 360 9
∴ 2πr = 3960
∴ Area of major sector = 8 × 8.25 = 66 cm2
∴ Cost = 3960 × 0.8 = ` 3168
1 1
158400 19 Area of a sector = × arc × radius = × 8 × 5.6 = 22.4 cm2
10 πr2 = ⇒ r2 = 36 ⇒ r = 6 m 2 2
1400
Mensuration 499

22 23 4a = 4 × 66 = 264 = 2πr ⇒ r = 42
20 2πr = 2 × × 700 = 4400 m = 4.4 km
7 ∴ d = 2r = 84 m
Distance 4.4 1
Time = = = h = 20 minute 24 r = 42 ∴ 2πr = 264 = 4a ⇒ a = 66
Speed 13.2 3
∴ d = a 2 = 66 2
21 πR 2 = π [ r12 + r22 + r32], find R. 1100 110
22 25 Circumference = = = 2πr
22 π [ 232 − 122] = × [ 529 − 144] = 1210 560 56
7

Introductory Exercise 10.5

1 Volume of original cube = (4)3 = 64 cm3 10 Let each edge 7of smaller cube = 1 m
and its weight = 400 kg ∴ Each edge of larger cube = 2 m
Since weight of the larger cube is 8 times the weight of and Surface area of smaller cube = 6 × (1)2 = 6 m2
smaller cube. Hence, the volume of new cube will be
∴ Surface area of larger cube = 6 × (2)2 = 24 m2
8 times the volume of smaller cube.
Hence volume of required cube = 8 × 64 = (8)3 24 − 6
∴ % increase in surface area = × 100 = 300%
6
∴ Edge of this cube = 8 cm NOTE It can be determined by using variables e.g., x
2 Volume of the tank = 3 m3 (edge of cube) instead of solving by assuming some numerals.
∴ Base area × height = 3 m3 S2  e2 
2
S2 4
3 Alternatively =  ⇒ =
⇒ Base area = = 1.171875 m2 S1  e1  S1 1
2.56
4 −1
[Q Volume of cuboid = (l × b) × h = (base area) × height] ∴ Percentage increase in surface area = × 100 = 300%
1
3 Volume of cube = Volume of cuboid where, S = surface area, e = edge of cube.
a3 = lbh 3
V2  e2 
11 =  {Q V = (e)3}
⇒ a3 = 36 × 75 × 80 = 216000 V1  e1 
{e = edge of cube}
⇒ a = 60 cm 3
V2  2 8
4 Base area × height = Volume 10 × 4 × 1 = 40 m 3 ∴ =  =
V1  1 1
But 1 m3 = 1000 litre = 1 kilolitre
∴ V2 = 8V1
∴ 40 m3 = 40, 000 litre = 40 kilolitre
12 External volume of the box = 24 × 16 × 10 = 3840 cm3
5 Volume of smaller (required) cube = 8 (dm)3 = 0.008 m3 Thickness of the wood = 5 mm = 0.5 cm
∴ Number of required cubes ∴ Internal length of box = 24 − 2 × 0.5 = 23 cm
Volume of larger cube 1 Internal breadth of box = 16 − 2 × 0.5 = 15 cm
= = = 125
Volume of each smaller cube 0.008 Internal height of box = 10 − 2 × 0.5 = 9 cm
6 Volume of new cube = (3)3 + (4)3 + (5)3 = 216 cm3 ∴ Internal volume of the box = 23 × 15 × 9 = 3105 cm3
∴ Each edge of the new cube = 6 cm ∴ Volume of the wood = 3840 − 3105 = 735 cm3
and hence, surface area = 6 × (a)2 = 216 cm2 Now, total weight of wood
7 Total length of tape = 2 (l + b) + 3.75 = Volume × weight of 1 cm 3 wood
= 2 (39.5 + 9.35) + 3.75 7350 = 735 × weight of 1 cm3 wood
= 101.45 cm ∴ Weight of 1 cm 3 wood = 10 gm
8 Area of surface to be cemented = 2 × (l + b) × h + (l × b) 13 Length of tank = 120 cm
i.e., area of four walls + area of floor 120 1
But since, = 17 , hence 17 cubes can be placed along
= 2 × (21) × 4 + (106.25) = 274.25 m2 7 7
∴ Cost of cementing = 24 × 274.25 = ` 6582 length and breadth of tank = 80 cm
80 3
9 Total volume of water displaced by 250 men But since, = 11 , hence 11 cubes can be placed along
7 7
= 250 × 4 = 1000 m3
breadth and height of tank = 50 cm.
Volume 1000
∴ Rise in water level (h) = = = 25 cm But since,
50 1
= 7 , hence only 7 cubes can be placed along
Base area 80 × 50 7 7
height of the tank.
500 QUANTUM CAT
∴ Total volume occupied by these cubes 23 Net volume = (10 × 8 × 2) − (2 × 2 × 2) = 152 cm3
= (7 ) × 17 × 11 × 7 = 448987 cm
3 3
Net surface area = 2 (10 × 8 + 8 × 2 + 2 × 10)
Total volume of the tank = 120 × 80 × 50 = 480000 cm 3 + 4 (2 × 2) − 2 (2 × 2)
∴ Area of unoccupied space = 480000 − 448987 = 240 cm2
= 31013 cm3 = 31.013 dm3 24 Available area for spreading the earth
14 Surface area of the cuboid = 2 (lb + bh + hl) = 11.6 m3 = (600 × 200) − (24 × 12) = 119712 m2
∴ Cost of canvas = 11.6 × 25 = ` 290 Volume of the earth = 119712 × rise in level
15 Required surface area = 2 (9 × 3 + 3 × 3 + 3 × 9) = 126 cm2 24 × 12 × 8 = 119712 × h
16 Area of 4 walls = 2 (36 + 12) × 10 = 960 m2 2304
⇒ h= = 0.01924 m = 1.92 cm
Total area of (windows + door + chimney) = 120 m2 119712
∴ Net area for papering = 960 − 120 = 840 m2 25 Net volume of the wall
840 = Total volume − Volume taken away due to doors
∴ Length of required paper = = 700 m
1.2 = (30 × 0.3 × 5) − 2 (4 × 2.5 × 0.3) = 39 m3
Hence, cost of papering = 700 × 0.7 = ` 490 8 1
Volume of the bricks = 39 × (Since part is lime in the wall)
17 Number of children × Space required for one child 9 9
= Volume of room 39 × 8
∴ Number of bricks =
30 × 12 × 6 9 × 0.2 × 0.16 × 0.08
∴ Number of children = = 270
8 = 13541.66 ≈ 13600
18 ( x + 2)3 − x 3 = 1016 26 Volume of water which flow in 25 minutes
⇒ x = 12 cm
= 25 × 60 × 0.05 × 0.03 × 16 = 36 m3
and x 3 − ( x − 2)3 = (12)2 − (10)2 = 728
36 1
19 You can see the figure shown in the question number 15. ∴ Rise in water level = = m = 0.2 m
15 × 12 5
Now let us consider that surface area of each face of the
cube 1 cm3. 27 A cube has 6 equal faces, 12 equal edges, 8 vertices and 4
equal diagonals.
∴ Total surface area of the cuboid = 14 cm2
(See the figure shown in orticle 10.8)
and Total surface area of the 3 cubes = 18 cm2
28 a 3 = 6 3 ⇒ a = 6 cm
Hence, required ratio = 14 : 18 = 7 : 9
NOTE In the arrangement of cuboid there are only 14 faces 29 l2 + b2 = 12 ⇒ l2 + b2 = 144
visible to us.
and l2 + b2 + h 2 = 15
20 Iron used in the tube
⇒ l2 + b2 + h 2 = 225
= Difference in external and internal volumes of the tube
⇒ h = 81 ⇒ h = 9 m
2

30 l + b + h = 12 cm, l2 + b2 + h 2 = 5 2

5 cm ⇒ l2 + b2 + h 2 = 50
m
8c Now, (l + b + h)2 = l2 + b2 + h 2 + 2 (lb + bh + hl)
7 cm
⇒ 144 = 50 + 2 (lb + bh + hl)
∴ 192 = 8 x 2 − 8 (5)2 ⇒ x = 7 cm
⇒ 2 (lb + bh + hl) = 94 cm2
7−5
Hence, the thickness of the tube = = 1 cm 31 h : b = 3 : 1 and l : h = 8 : 1
2
21 Base area of vessel × rise in water level = Volume of cube ⇒ l : h : b = 24 : 3 : 1
∴ 24 x × 3x × x = 36.864
15 × 12 × h = 11 × 11 × 11 ⇒ h = 7.39 cm
⇒ x 3 = 0.512 ⇒ x = 0.8
22 (Initial volume of water + required volume of water
∴ h = 3x = 2.4 m
+ volume of cube)
= Base area of vessel × 10 32 lb = x, bh = y, hl = z
∴ 25 × 20 × 5 + required volume of water+ 1000 ∴ lb × bh × hl = xyz
= 25 × 20 × 10 ⇒ (lbh)2 = xyz
⇒ Required volume of water = 1500 cm = 1.5 litre
3
⇒ lbh = xyz
Mensuration 501

Introductory Exercise 10.6

22 r 3x
1 Required volume of water = × (0.04)2 × 3500 12 = , πr2h = 4851 ⇒ x = 3.5
7 h 4x
= 17.6 m3 ∴ r = 10.5 m and h = 14 m
10 × 10 × 10 22
2 h= =3
2
cm ∴ 2πrh = 2 × × 10.5 × 14 = 924 m2
π × 10 × 10 11 7
2πr (h + r) 4
511 13 =
3 h= = 14 m 2πrh 1
36.5 h+ r 4 r 3
⇒ = ⇒ =
4 2π rh = 1056 cm2 h 1 h 1
33 Alternatively Go through options.
⇒ r= cm
π 21
2 14 2πr = 33 ⇒ r =
 33 4
∴ πr h = π ×   × 16
2
π 22  21
2
∴ Volume = πr2h = ×   × 330
⇒ Volume = 5544 cm3 7  4
= 28586.25 cm3
5 Diameter of the cylinder = edge of the cube = 2 cm
r1 4 x
Height of the cylinder = edge of the cube 15 = , but V1 = V2
r2 x
∴ Required volume = (2)3 − π × (1)2 × 2
∴ π (4 x )2 × h1 = π ( x )2 h2
 44 12
=8−  = cm3
7 7 ⇒ h2 = 16h1

2πrh 17.6 16 2πr (h + r) = 2640

6 h= = = 2m
2πr 8.8 ⇒ 2πr (30) = 2640 ⇒ r = 14 m
and 2πr = 8.8 ⇒ h = 16 m (Q r + h = 30 m)
⇒ r = 1.4 m ∴ h: r = 8:7
22 V1 π (r1 )2 h1 3 × 3 × 4 12
∴ πr h =
2
× (1.4)2 × 2 = 12.32 m3 17 = = =
7 V2 π (r2 )2 h2 5 × 5 × 3 25
h V1 1
7 2r = h ⇒ r = 18 =
2 V2 1
h h 3
∴ 2πr (h + r) = 2π ×  h +  = πh 2 2
 r1  h 1 r 
2
3 1
2 2 2   × 1 = ⇒  1 × =
 r2  h2 1  r2  1 1
8 2πrh = 2 × π × 2 × 10 = 40π m2
2
2πrh  r1  1 r1 1
9 =
2
⇒
h
=
2
⇒
h 2
= ⇒   = ⇒ =
2πr (h + r) 3 h+ r 3 r 1  r2  3 r2 3

∴ 2πr (h + r) = 924 2πr1h1 3 6 9

19 = × =
2 2πr2h2 2 7 7
∴ 2πrh = 924 × = 616 cm2
3 20 h = 42 cm, 2πR = 132 cm
⇒ 2 × π × x × 2x = 616 ⇒ R = 21 cm
⇒ x = 7 ∴ r = 7 cm and h = 14 cm ∴ r = 21 − 3 = 18 cm
22
∴ πr2h = × (7 )2 × 14 = 2156 cm3 ∴ Required volume = π (R 2 − r2 ) h
7
22
π × r × r × h 269.5 = [(21)2 − (18)2] × 42 = 15444 cm3
10 = ⇒ r = 3.5 cm 7
2× π × r × h 154
21 Since radius and height of the cylinder are same as that of
Now, 2π × 3.5 × h = 154 ⇒ h = 7 cm
cone. Therefore cylinder can contain 15 × 3 = 45 litre of
11 2πrh = 1320 ⇒ h = 10 cm milk.
22 1
∴ 2πr (h + r) = 2 × × 21 × 31 = 4092 cm2 Hint Volume of cone = πr 2 h and volume of cylinder = πr 2 h.
7 3
502 QUANTUM CAT
22 Best way is to go through option. Volume of cylinder πr 2 h 1 3
30 = = =
V1 1 (r )2 h 1 Volume of cone   2
1   1
1
Alternatively = ⇒ 1 2 1 =   πr h  
V2 27 (r2 ) h2 27  3  3
A
x13 (1)3 x 1 31 By Pythagorus theorem
⇒ = ⇒ 1 = 10 cm
3
x 2 (3)3
x2 3 AD = 3 cm
h1 1 ∴ Area of axial section 5 5
⇒ = 20 cm 1 3
h2 3 = × 8 × 3 = 12 cm2
2
h2 30 B C
∴ h1 = = = 10 cm 4 D 4
3 3 V (1.4)3 2.744
32 2 = =
Hence, cone is cut off at (30 − 10 = ) 20 cm above the base. V1 (1)3 1
 2.744 − 1 
Hint For more clarification of the concept seek help from geometry. ∴ % increase in volume =   × 100 = 174.4%
(Similarity of triangles)  1
1 22 13 13 33 Area of circular sheet = 625π
23 Volume of conical tent = × × × × 10.5
3 7 2 2 Since length of arc and area of sector are directly
Volume of tent 3 proportional to the central angle.
∴ Average space per man = = 58 Therefore, length of remaining arc
8 32
96
1 2 2200 = × 2 × π × 25 = 48π
24 × π × (5x )2 × (12x ) = 314 = 100
3 7 7
But the remaining arc is equal to the circumference of the
⇒ x = 1 ∴ r = 5 and h = 12
base of circular cone.
∴ l = 13 m
∴ 2πR = 48π ⇒ R = 24 cm
1 22
25 × × r2 × 24 = 1232 ⇒ r = 7 cm Now, since the slant height of cone is equal to the radius of
3 7
the original circular sheet.
22
∴ πrl = × 7 × 25 = 550 cm2 (Q l = r2 + h 2 ) Hence, l = 25 cm
7
h = 7 cm (Q l = r2 + h 2 )
26 2πr = 220 ⇒ r = 35 cm
Radius 24
∴ =
∴ l = r2 + h 2 = 91 cm Height 7
22 V2 (r2 )2 h2 (2r)2
∴ πrl =
× 35 × 91 = 10010 cm2 34 = = {Q h1 = h2}
7 V1 (r1 )2 h1 (r)2
22
27 Area of cloth = πrl = × 14 × 50 = 2200 m2 V2 4
7 ⇒ =
V1 1
2200
∴ Length of cloth = = 220 m
10 35 Each edge of cube= 10 cm
28 πrl = 23.10 ⇒ rl = 2.1 × 3. 5 Radius of base of cone = 5 cm
Height of cone = 10 cm
Now, we know that h = 2.8. So we can assume the value of
1 22
r from the given option. ∴ Volume of cone = × × 25 × 10 = 261.9 cm2
3 7
at r = 2.1, l = 3.5 (Q l = r2 + h 2 )
V2 r12 × h1 2 × 2 × 2 8
Alternatively r (r + h ) = (2.1 × 3.5)
2 2 2 2 36 = = =
V1 r22 × h2 1 × 1 × 1 1
⇒ r4 + (2.8)2 r2 = (7.35)2 ⇒ r4 + 7.84r2 = 54.0225 BD A
37 = cos 60°
⇒ k + 7.84k − 54.0225, k = r
2 2
AB
30°

BD 1
=
°
30

Now, solve the above quadratic equation, if you can else

AB 2
substitute the value of r from the given choices.
BD = CD, are the radii of the base
πr1l1 3
29 = (Q r1 = r2 ) and AB = AC are the slant heights 60° 60°
πr2l2 2 B
D
C
of the cone. A is the vertex and BC
A1 3 A1 3 is the base.
∴ = ⇒ =
A2 2 300 2
∴ A1 = 450 cm2 1
38 Volume of cone = π × 144 × 35
3
Mensuration 503

Volume of water flowing per second 39 Use the given formula :

500 π
= π × (0.8)2 × Volume = h (r2 + Rr + R 2 )
60 3
 π where π =
22
, h = 6, R = 4 and r = 2
  × 144 × 35
 3 7
∴ Required time = = 315 s
 500 1 8 2
π × 0.64 ×   40 π × (10)2 × 72 = π × (30)2 × h ⇒ h = = 2 cm
 60  3 3 3

Introductory Exercise 10.7

4 4
1 π (r13 + r23 + r33 ) = π (6)3 M
3 3 A B
E
⇒ 27 + 64 + r33 = 216
⇒ r33 = 125
G
⇒ r3 = 5 cm
4
2 πr2 × 144 = π × (12)3 (Q Volume remains constant) F
3
D C
⇒ r = 4 cm P

3 Number of bottles × Volume of each bottle In the diagram, AG = MG = PG = radius of sphere

= Volume of hemisphere Now, using Pythagoras Theorem in right angle ∆ AEG, we

2
n × π × (3) × 1 = π × (6)3 ⇒ n = 16
2 can find EG = AG 2 − AE 2 = 152 − 92 = 12 cm.
3
1 2 Therefore height of the cylinder = EF = 2 EG = 24 cm
4 π × (r)2 × 14 = π × (7 )3 ⇒ r = 7 cm
3 3 Now, Lateral surface area of cylinder
4 = 2π × 9 × 24 = 432 π sq cm
5 Volume of the spherical shell = π (R 3 − r3 )
3 Total surface area of original sphere
4 872
= π (7 3 − 53 ) = π = 4π (radius of sphere)2 = 4π × 152 = 900 π sq cm
3 3
6 Since volume is constant Lateral surface area of a spherical cap = perimeter of the
4 4 sphere × height of the spherical cap
∴ n × π (1)3 = π × (8)3 ⇒ n = 512
3 3 = 2π × (radius of the sphere) × (MG − EG )
2 = 2π × 15 × (15 − 12) = 90π
7 Volume of hemisphere = πr3
3 Therefore, required surface area
2 22 = 432 π + [ 900 π − 2(90 π )] = 1152 π sq cm
= × × (21)3 = 19404 m3
3 7
11 Look at the following digaram and recall the definition of a
8 Change in height (or level) of water zone.
Volume of sphere
=
Base area of cylinder Zone
Hemisphere
4 Spherical cap
π × (9)3
3 27
= = cm
π × (12)2
4
Hence choice (d) is correct.
9 Volume of cone = Volume of sphere Solutions (for Q. Nos. 12 to14) Given that AB and CD are
1 2 7 4 parallel, if we draw a line passing from the centre and
πr × = π (7 )3 ⇒ r = 14 2 cm
3 2 3 intersecting AB and CD, this line EF will be perpendicular to both
10 Required surface area = lateral surface area of cylinder + the lines AB and CD. If you have any doubt please refer the
[Total surface Area of Sphere − 2 (lateral surface area of Geometry chapter, especially, Lines and Angles section to
spherical cap)] understand this concept and even you can refer the Circles
section also.
504 QUANTUM CAT
But, OF = OE − EF = 15 − 7 = 8 cm
2 cm
C 4 E D Now, in ∆CFO, we have

5
3 CF = CO 2 − OF 2 = 17 2 − 82 = 15 cm
O 7 cm
The height of larger spherical cap
5 4 = radius of sphere + length of OF
A 3 F B 1 cm = 17 + 8 = 25 cm
And the height of the smaller segment = radius of sphere
Now, we know that ∆AFO and ∆CEO are right angle − length of OE = 17 − 15 = 2 cm.
triangles. We know that EF = 7 cm and AF = 3 cm.
15 Volume of the zone
Now in ∆OAF, we can use Pythagoras theorem and find OF πh 2
= (h + 3r12 + 3r22 )
= OA 2 − AF 2 = 52 − 32 = 4 cm. 6
7π 2 3206
Therefore, OE = EF − OF = 3 cm. = (7 + 3 (8)2 + 3(15)2 ) = π
6 3
Now, in ∆CEO, we can use Pythagoras theorem and find
Hence, choice (b) is correct.
CE = CO 2 − OE 2 = 52 − 32 = 4 cm. 16 Volume of the larger spherical cap
Therefore, AB = 2 ( AF ) = 6 cm and CD = 2 (CE ) = 8 cm. πh 2 25 π 16250
= (h + 3r2 ) = (252 + 3 (15)2 ) = π
Now, we know that the segment with larger base radius is 6 6 3
the larger segment, so the height of larger segment Hence, choice (c) is correct.
= 5 − 3 = 2 cm. Similarly, the hight of the smaller segment
17 Total surface area of the smaller spherical cap
= 5 − 4 = 1 cm
= 2π Rh + π r2 = 2 π (17 × 2) + π (2)2 = 72π
12 Volume of the zone
Hence, choice (a) is correct.
πh 2 7π 2
= (h + 3r12 + 3r22 ) = (7 + 3 (3)2 + 3 (4)2 ) 18 Volume of the spherical sector = Volume of the spherical
6 6
434 cap + Volume of the conical section
= π
3 h
Hence, choice (a) is correct.
13 The smaller spherical cap has 3 cm radius and 1 cm height. R
Therefore the volume of the smaller spherical cap
πh 2 π (5 − 4) 14
= (h + 3r2 ) = ((5 − 4)2 + 3(3)2 ) = π.
6 6 3
Hence, choice (c) is correct. But the height of the conical section = 252 − 7 2 = 24 cm
14 The larger spherical cap has 4 cm radius and 2 cm height. Therefore, the height of the smaller spherical cap
Therefore, the lateral surface area of the larger spherical = 25 − 24 = 1 cm
cap perimeter of the sphere × height of the spherical cap
Thus, the total surface area of the sector of the sphere
= 2π (5) × (2) = 20 π
πh 2 1
Hence, choice (d) is correct. = (h + 3r2 ) + (base area of cone × height of cone)
6 3
Solutions (for Q. Nos. 15 to 17) Given that OA = OC = 17cm. π 1 1250
= (1 + 3 (7 ) ) + (π × 7 2 ) × 24 =
2
π
Now drop the perpendicular OF from the centre on the line CD 6 3 3
and another perpendicular OE on the line AB. These Hence, choice (d) is correct.
perpendiculars will bisect the lines CD and AB respectively.
1
19 Volume of pyramid = × base area × height
3
1
= × 25 × 12 = 100 cm3
3
O
17 Hence, choice (b) is correct.
F
C
17
D 20 Volume of prism = Base area × height
E
A B Since, base area is constant and height is being halved
therefore volume will also be halved. Hence, its volume
Now, in ∆AEO, we have will be reduced by 50%.
OE = OA 2 − AE 2 = 17 2 − 82 = 15 cm Hence, choice (a) is correct.
Mensuration 505

Level 01 Basic Level Exercise

1 With the given perimeter the area of parallelogram will be ⇒ a + b = 17 …(i)
maximum when it will be a rhombus. A
Hence, option (a) is correct.
Alternatively 13 cm
b
D C
B
d h = d/2 C a
Again (a − b)2 = (a + b)2 − 4ab
30°
A b B (a − b)2 = 289 − 240
Perimeter = 2d + 2b = 4h + 2b = 64 cm ⇒ (a − b) = 7 …(ii)
64 − 4h From eq. (i) and (ii), we get
⇒ b=
2 a = 12 cm and b = 5 cm
64h − 4h 2 (Altitude+ Base − Hypotenuse)
∴ Area = b × h = = f ( A) Now, Inradius =
2 2
1 (12 + 5 − 13)
∴ f ′ ( A ) = (64 − 8h) (Refer the differentiation) = = 2 cm
2 2
For the maximum area = f ′ ( A ) = 0 Area
Alternatively Inradius (r) =
1 Semiperimeter
∴ (64 − 8h) = 0
2 30
= = 2 cm
⇒ h = 8 cm and d = 16 cm 15
∴ b = 16 cm  (12 + 5 + 13) 
Semiperimeter = = 15
 2 
2 Volume of water displaced = volume of sphere
NOTE If you are well aware with Pythagorian triplets then
4
π × (40)2 × h =
π × (30)3 it is very easy to find the sides ( i . e. , altitude and base) of the right
3 angle triangle.
90
⇒ h= = 22.5 cm 5 General phenomenon: A convex polygon in which
4 there is maximum number of sides, it has the greatest
Thus, the level of water rises by 22.5 cm. enclosed area when the perimeter of the polygon is
constant.
NOTE The volume of water will be calculated by
considering it in the cylindrical shape since the water takes the Remember that in a circle there are infinite sides of
shape of vessel in which it is filled. minimum possible length. So, the area of circle will be
maximum.
3 O is the starting point and L is the end point. Alternatively Perimeters of hexagon and
24 m circumference of circle are same i.e.
L
N
10 m 6a = 2πr ⇒ 21a = 22 r
where a is the side of hexagon and r is the radius of circle.
28 m

O 12 m W E then Area of circle = πr2

18 m 3
and Area of hexagon = 6 × × a2
S 4
2
12 m 3 3  22r
= × 
4  21 
(OL )2 = (24)2 + (10)2
OL2 = 676 3 3 22 22
= × × × r2
⇒ OL = 26 m North-East 4 7×3 7×3
3 22 2
4 a + b2 = 132 = 169
2
= × πr = 0.45πr2
4 21
a×b
and = 30 Therefore, area of circle πr 2 is greater than area of hexagon.
2
Similarly you can compare other figures.
Now (a + b)2 = a2 + b2 + 2ab
Alternatively Consider some suitable values. Let us
(a + b)2 = 169 + 120 assume perimeter of square and hexagon is 24 cm, then
each side of square = 6 cm.
506 QUANTUM CAT
and Area = 36 cm2 a 3
⇒ h=
Similarly, each side of hexagon = 4 cm 2
3 3 ∴ Area of rhombus a
h a
and area of hexagon = × (4)2 = 24 3 cm2
2 a × 3a
=a×h= 60°
2 a
Thus, we can say that for a particular (or constant) value of
perimeter, area of hexagon is greater than that of square. and Area of square = a2
 24 a2
Similarly each side of octagon = 3 cm  =  ∴ Required ratio = ×2
 8
3a2
∴ Area of octagon = 2a2 (1 + 2) 2  3 2 3
= ×  =
= 2 × 9 (1 + 2) ≈ 43.5 cm2 3  3 3
Thus, the area of octagon is greater than the area of 10 D R C
hexagon.
12
Similarly, 2πr = 24 ⇒ r = cm S Q
π
12 12
∴ Area of circle = π × r2 = π × × = 45.81 cm2 A B
π π P
Thus, we can say that ultimately the area of the circle is ABCD is a square, AB = BC = CD = AD = 20 cm
greatest. and P , Q , R and S are the mid-points.
D C D C ∴ PS = SR = RQ = PQ = 10 2cm (Q AP = BP etc = 10 cm)
6
90° ∴ Area of square PQRS = (10 2)2 = 200 cm2
or D C
Alternatively We can see that there
90° are total 8 equal parts (or triangles) out
A B A B
of 8 we have taken only 4 parts i . e. ,
Since sum of all the angles of a quadrilateral is 360°. 50% area has been cut out.
Therefore, ∠ A + ∠ B + ∠ C + ∠ D = 360° So, the remaining area = 50% of 400 B
A
⇒ ∠ B + ∠ C = 180° = 200 cm2.
⇒ ∠ B = 180° − ∠ C A
11 Area of right angle triangle
∠ ABG = 60° ( x + 2) × (2x + 3)

(x + 2)
7 = = 60
∴ ∠ GBC = 90° − 60° = 30° 2
Again ∠ BGC + ∠BCG = 180° − 30° = 150° ⇒ 2x 2 + 7 x + 6 = 120
⇒ 2x 2 + 7 x − 114 = 0 C B
Now, since ∠ BGC = ∠ BCG (2x + 3)
D C Solving the above quadratic equation, we get
75° x=6
75°
G ∴ x + 2 = 8 cm
and 2x + 3 = 15 cm
30°

∴ Hypotenuse AB = (8)2 + (15)2 = 17 cm

60° 60°
A B (Using Pythagorus theorem)
U T
∴ ∠ BGC = 75° 12 Let us assume PQRSTUVW is a regular D C
octagon with each side a unit. V S
NOTE In a triangle when two sides are equal, then the two
angles opposite to these sides are also equal.
Again, if we produce all the sides of
octagon in both sides we get a square. W R
r 3x
8 = Since each of the side of octagon is ‘a’ A B
R 4x P Q
then
Q 4 x − 3x = 2 r
UV = WP = QR = ST = a
⇒ x=2 a
R ∴ DU = DV (etc. ) =
∴ Outer radius = 8 cm 2
∴ Area of outer circle a a
∴ DU + UT + TC = + a+
= π × (8)2 = 64π cm2 2 2
9
h
= sin 60° = a (1 + 2)
a ∴ Area of square = a2 (1 + 2)2
h 3 = a2 (3 + 2 2) sq unit
=
a 2
Mensuration 507

Again, area of each shaded portion (i . e. , an isosceles right ∴ Area of the shaded region B
angle triangle)  2 
= 102 − 49 3 cm2 A 60° 60°
1 a a a2  3 
= × × = 60°
2 2 2 4 18 l1 × b1 = l2 × b2
a2 C
∴ Total area of all the shaded region = 4 × = a2 New area of first rectangle
4 5 3 15
∴ Area of octagon = Area of square ABCD = l1 × b1 = l1b1
4 4 16
− Area of total shaded region 3 5 15
= a2 (3 + 2 2) − a2 New area of the second rectangle = l2 × b2 = l2b2
4 4 16
= 2a2 (1 + 2) sq. unit Hence, areas of both the new rectangles are same.
13 Let the each side of the square be DM D C
19 = sin 60°
2 cm, then D C AD
area of square = 4 cm2 and 6 3 3
a
63 a
area of 4 quadrants of the four =
AD 2 60°
circles (i . e. , unshaded part inside A M B
⇒ AD = 12 cm
the square) A B
1 4 3 1 2
= 4 × π × (1)2 = π cm2 20 πr = πr h
4 3 3
∴ Area of shaded region = (4 − π ) cm2 (Since radii of sphere and cone are same)
4−π 6/7 3 ⇒ 4r = h
Therefore, the required ratio = = = h 4
π 22/ 7 11 ∴ = ⇒ h: r = 4:1
r 1
14 From the concept of mid point theorem, it is the average of 1 1 
the length of the parallel sides. 21 Area of the shaded region =  π(22)2 − 2  π × (11)2 
2  2 
OP 1 PC
15 From the concept of similarity of triangles = = 1
OQ 5 BQ = π × (11)2 [ 4 − 2] = 121π cm2
O 2
x 22 Area of path = (l + b − w ) w
D C = (60 + 40 − 5) 5 = 475 m2
P h
Cost = Area × Rate
4x
= 475 × 0.8 = ` 380
Q 23 Go through options. Let the length and breadth of the
A B
second rectangle is l and b respectively, then the area of
Since, the ratio in radii of the two cones is 1 : 5. Therefore,
21 second field = l × b.
the radius of smaller cone ODC is = 4.2 cm.
5 NOTE Now consider option (d) i. e. , x = 80, then
Alternatively Solve it in the proper way, which is
1
actually a very tedious process. Still you have to apply the Area of first field = l × 5b = lb
concept of similarity of triangles, which you will study in 5
Geometry. Hence, area of first field is same as that of second field,
16 l × b = 2 (l + b) hence the presumed option is correct.
2b 2 (b − 2) + 4 4 1
⇒ l= = = 2+ 24 Area of quadrant = πa2
(b − 2) b−2 (b − 2) 4
a2
Area of triangle ACB =
Since, l is an integer, so 4 must be divisible by (b − 2). 2
Thus, b can be 4 or 6 or 3.
Therefore, if b = 4, l = 4, it will be a square. If b = 6, l = 3and
if b = 3, l = 6. → Area of required region
A B
Hence, l = 6 and b = 3
∴ l−b=3 → Area of segment
3
17 Area of ∆ ABC = × (14)2 = 49 3 cm2
4
C
Area of sector
πa2 a2 a2
ABC = π × (14)2 ×
60 2
= 102 cm2 ∴ Area of segment = − = [ π − 2]
360 3 4 2 4
508 QUANTUM CAT
2 2
1  a 2 πa2 ∴ Area of equilateral triangle =
3  4a
Area of semi-circle = π  =  
2  2  4 4  3

πa2 a2 a2 4 3 2
∴ Area of required region = − [ π − 2] = sq unit a =
4 4 2 9
a ×9
2
25 Wrongly calculated area = 1.05 × 0.92 = 0.966 = 96.6% ∴ Required ratio of area =
4 3 × a2
∴ % error = 100 − 96.6 = 3.4%
Alternatively Actual area = l × b
= 9: 4 3
Wrongly calculated area = 1.05l × 0.92b = 0.966lb 31 ABCD is a square O is the point of intersection of diagonals.
P , Q , R and S are the mid-points on the sides AB, BC , CD,
Deficit in area = lb − 0.966lb = 0.034lb
DA respectively.
0.034lb
% error in area = × 100 = 3.4%
lb D R C
26 Volume of water = Area of cross-section × Length of pool
(10 + 6) S Q
= × 6 × 120 = 5760 m3 O
2
22176
27 Area of path = = 22176 m2 A
P
B
1 Fig. (i)
Fig. (ii)
π (R 2 − r2 ) = 22176
22176 r In the above figure you can see that there are total 16
(R 2 − r2 ) = ×7 congruent isosceles right angled triangles.
22
In figure (iii) you can see that OMPN is a square of
R 2 − (112)2 = 7056 R
maximum possible area which is made up of 2 isosceles
⇒ R 2 = 19600 ⇒ R = 140 m right angled triangles OMP and ONP. Thus, there are 4
∴ Width of the path = 140 − 112 = 28 m smaller squares around O and thus the total area of these 4
squares is half of the larger square. Hence, the required
28 From the figure itself it is clear that there are total area = 800 cm2.
9 equilateral (congruent) triangles. Out of 9 triangles,
O
3 triangles are cut out.
A A
M 90° 90° N

T S T S
A B
U R U R P
Fig. (iii)

B C B C Alternatively ABCD is a square

P Q P Q R
Fig. (i) Fig. (ii) AB = 40 cm D C
1 OA = OB = 20 2 cm
It means (i . e. , 33.33%) area has been removed.
3 and OP = 20 cm O Q
M N
SP 3 C Square OMPN has maximum area inside
29 = sin 60° = the triangle and OP is the diagonal of A B
AS 2 P
square OMPN.
3 S R
⇒ SP = AS (20)2
2 ∴ Area of OMPN = = 200 cm 2
Also, SP = SR = CS = CR 2
A B
P Q Since, there are total 4 similar squares in each of the
(since, SR || AB ∴ ∠ CSR = ∠ SRC = ∠ RCS = 60°) 4 triangles AOB, BOC , COD, DOA inside the larger square.
So, the total required area of 4 smaller squares
3
∴ CS = AS = 4 × 200 = 800 cm 2
2 C
CS 3 CR 32 Let the each side of equilateral triangle be
∴ = = a
AS 2 RB ‘a’, then circumradius of the circle = a
3
2 O
30 Let the each side of a square be a then its area will be a . ∴ Area of circumcircle
Therefore, area of circle will also be a2. 2
 a π A
D
B
Again since the perimeter of square and equilateral triangle = πr2 = π ×   = a2
4a  3 3
is same then, the each side of equilateral triangle is ⋅
3
Mensuration 509

Area of square whose side is equal to that of equilateral 38 Area of large cube = 6 × (5)2 = 150 (unit) 2
triangle = a2. Area of cuboid = 2 (1 × 1 + 1 × 125 + 125 × 1)
π 2 = 502 (unit) 2
a
π
∴ Required ratio = 3 2 = (502 − 150)
a 3 Therefore, increase in surface area = × 100
150
NOTE A circumcircle always passes through the vertices of 2
= 234 %
the inscribed figure (say triangle). 3
33 Best way is to go through option. Given that height of room 39 AO = BO = 13 cm
= 10 m. ⇒ AC = BD = 26 cm (Q O is the point of bisector)
Volume of room = 25 × 400 = 10000 m3 Now, since the diagonals are equal, it D C
and Surface area of walls = 2h (l + b) = 1300 m2 means the given figure is actually a
Now, consider option (c) and verify it. rectangle.
where l = 40 and b = 25 A B
∴ BC 2 = AC 2 − AB 2
34 Let the each side of equilateral triangle be a, then its
perimeter = 3a BC 2 = (26)2 − (24)2
C BC 2 = 100
Again, 2 (l + b) = 3a
3 3a ⇒ BC = 10 cm
⇒ l + b = a⇒ a + b = Q P
2 2 40 AB = 16 cm
AB = l = a
(for the rectangle) A B AQ = 10 cm
a AB
⇒ b= and AO = = 8 cm
2 2
a a2
∴ Area of rectangle = a × = ∴ OQ = AQ − AO
2 2
3 = 10 − 8 = 2 cm
and Area of triangle = × a2
4
a2/ 2 2
∴ Required ratio = 2
=
3a /4 3
A P O Q B
35 Since, b, l and 2 (l + b) are in GP, therefore
l 2(l + b)
= OQ (= OP ) is the radius of smaller enclosed circle between
b l
l two arcs.
Suppose = x,
b ∴ Area of circle with centre O is π × (2)2 = 4π cm2.
 1 41 Area of the shaded region
then x = 2 1 +  ⇒ x 2 − 2x − 2 = 0
 x
D C
l
⇒ x = 3+1= S R
b
36 Non-polished area = 4π (r)2
 πr2  P Q
Polished area = 4 ×  2 ×  = 4πr2
 2  A B

In the adjoining figure, one of the four = (Area of square PQRS − 4 × area of each quadrant of circles)
parts of the sphere is shown (To understand it properly,  1 
= (2)2 − 4 × π × (1)2 = (4 − π ) cm2
take an apple and cut it in the four parts one across  4 
horizontal and another cut make vertical to it then you will
notice that in a piece there are 2 semicircles.) Therefore, 42 Let each side of the square be a, then
required ratio = 1 : 1. D C
85 × 17 × 5.1
37 Number of cubes = = 1500
1.7 × 1.7 × 1.7
(1.7 is the HCF of 85, 17 and 5.1)
Area of each cube = 6 × (1.7 )2 A
E
B

∴ Area of all the 1500 cubes = 1500 × 6 × (1.7 ) 2

AE = AB − BE = (a − 17 )
= 26010 cm2
510 QUANTUM CAT
∴ Area of triangle =
1
× AD × AE ⇒ b × b = 10000
2 ⇒ b = 100 m
1 ∴ Perimeter of Kaurav’s land = 4 × 100 = 400 m
= a × (a − 17 ) = 84
2 ∴ Expenses = 2 × 400 = ` 800
⇒ a2 − 17 a − 168 = 0 48 Ratio of height
⇒ a = 24 cm h1 9 3 9 cm
= =
∴ Area of square = (24)2 = 576 cm2 h2 12 4
12 cm
43 The solid with the least number of sides will have (3)3 27
∴ Ratio of volumes = =
maximum surface area. So, tetrahedron will have (4)3 64
maximum surface area. Notice that in a sphere there are
Hence, the volume of smaller cone
infinite number of sides with the least possible length. So,
the surface area of sphere will be the least. = 27 x
l (l + b) and the volume of larger (actual) cone = 64 x
44 =
b l ∴ the volume of frustum = 64 x − 27 x = 37 x
⇒ l = b (l + b) = lb + b2
2 64 x
…(A) ∴ Required ratio = = 64 : 37
37 x
⇒ l2 − b2 = lb …(i)
49 AB = 45 km, AC = 53 km
⇒ (l + b)(l − b) = lb …(ii)
(Since distance = speed × time)
(l + b) b
and = …(iii) C
l (l − b)
Therefore, statements (ii) and (iii) are true from
equation (A) A B
l2 bl + b2 ∴ BC = AC − AB 2
2
=
b2 b2
l2
l BC = (53)3 − (45)2
⇒ =1 +
b2 b or BC = 28 km
l l2 ∴ Area of field = AB × BC
⇒ = −1
b b2 = 45 × 28 = 1260 km2
Hence, statement (i) is wrong. 50
3
45 Area of triangle = × (2)2 = 3 cm2
4 A B 30 cm
Area of 3 circles enclosed by the triangle 60°
60 π
= 3 × π × (1)2 × = cm2 O
360 2 Fig. (i) Fig. (ii)
C D
∴ Area of shaded region ∠ AOB = 60°
π (2 3 − π ) AB = 8 3 cm
= 3− = cm2 M
2 2 OC = 30 cm 30 cm 60° B
46 Diameter of cube = 3 cm OM 30°
= tan 60° = 3
9×9×9 MB
∴ Number of cubes = O
3× 3× 3 OM = 3MB Fig. (iii)
= 27cubes OM = 12 cm
47 OM OC
b Now, =
MB CD
(Triangles OMB and OCD are similar)
l
∴ CD = 10 3 cm(which is the radius of cone)
l × b = 20000 1
l ∴ Volume of cone = πr2h
⇒ × b = 10000, Area of Kaurav’s land 3
2 1
For the given area, a square gives the minimum perimeter. = π × (10 3)2 × 30
3
l
∴ =b = 3000π cm3
2
Mensuration 511
P
51 Surface area of one cube = 6a2 56 Consider the following Figure.
When 6 cubes are fixed on the 6 faces of a cube then only 5 Let RS = b and QS = h.
faces of a cube are visible of each cube. Since, the central Then, RQ 2 = 512 = b2 + h 2 …(i)
cube is completely covered. So, the only 6 cubes are visible Q
each with 5 faces. Hence, the total surface area of this solid and RP 2 = RS 2 + (PQ + QS )2
= 5a2 × 6. ⇒ 742 = b2 + (25 + h)2 …(ii)
h
5a × 6 5
2
Solving the above two equations we
∴ Required ratio = = ⇒ 5: 1 get h = 45 and b = 24.
6a2 1
Now, volume of cone (V1 ) when ∆ PRS R
52 Ratio of areas = (Ratio of diagonals) 2 b S
is rotated about PS
2
 3 9 1
=  = = × π × 242 × 70 = 13440 π cubic cm
 2 4 3
53 You should know that And volume of cone (V2 ) when ∆ QRS is rotated about QS
1
= × π × 242 × 45
E D 3
= 8640 π cubic cm
F C Therefore the required volume of the solid (V3 )
= V1 − V2 = 4800 π cubic cm
Hence, choice (b) is correct.
A B
57 Let 4a be the side of each square.
Area of ∆ BDF 1 Then the radius of the circle inscribed in the squares
= ABCD = 2a
Area of hexagon 2
So the area of this circle = π (2a)2 = 4 πa2
Actually there is a perfect symmetricity.
3 3
∴ Area of hexagon = × (6)2 = 54 3 cm2
2
∴ Area of ∆ BDF = 27 3 cm2
Alternatively Subtract the areas of three triangles
DEF , BAF , BCD from the area of hexagon. And the radius of each circle inscribed in the square PQRS
1 =a
Area of ∆ DEF = × 6 × 6 sin 120° = 9 3 cm2
2 So the total area of four circles = 4 × π (a)2 = 4 πa2
∴ Area of all the three triangles = 27 3 cm
Therefore, required area of ∆BDF = 54 3 − 27 3 It shows that the area occupied by the circle(s) in each
square is same. Therefore the required ratio = 1 : 1
= 27 3 cm2
Hence, choice (d) is correct.
Solutions (for Q. Nos. 54 and 55)
58 If 2a be the side of any square, the area of square would be
4x 4x 4x 4x x 4a2.
x
x And, the radius of the inscribed circle would be a, so the
4x x 4x area of the circle would be πa2.
x
Therefore the ratio of area of each circle to the area of
Total length of plot = 17x concerned square = π : 4
and total breadth of the plot = 4x Since the total area of all the smaller squares = total area of
∴ Area of plot = 17x × 4 x = 9792 the Original Square.
⇒ x = 12 Area of all the circles π
∴ =
∴ l = 17 × 12 = 204 m Area of original square 4
and b = 4 × 12 = 48 m Hence, choice (c) is correct.
4 x × 4 x 16 59 After you cut out any cube or cuboid from the corner of a
54 =
x×x 1 solid cube or cuboid, the surface area remains unchanged,
⇒ 16 : 1 which you can easily understand by looking at the diagram.
55 l = 204 m and b = 48 m So, the required area is 216 cubic ft.
Hence, choice (a) is correct.
512 QUANTUM CAT
60 Since ∠DPC = 90°,
1
Area of ∆DPC = × DP × PC = 189 A P B
2
Now drop a perpendicular PQ from P on DC.
Then you can see that the area of
∆APD = ∆PQD and area of ∆PBC = ∆PQC. It means each side of the square inscribed by the circle is
Therefore, area of rectangle ABCD D Q C
2 5x
= = 10 x
= 2 × area of ∆ DPC = 378 mm2 2
Hence, choice (a) is the correct one. (2x )2 2
Therefore, the required ratio = =
Alternatively Area of rectangle ABCD = DC × BC ( 10 x )2
5
= DC × PQ Hence, choice (a) is the correct one.
= DP × PC Alternatively Look carefully at the following diagrams.
= 18 × 21 = 378 In the first diagram, the pertinent square covers 4 units of
smaller squares. In the second diagram, the pertinent
Hint In right angle triangle DPC,
square covers 10 units of smaller squares.
DC × PQ = DP × PC = area of ∆ DPC.
Hence, choice (a) is the correct one.
61 The shortest path that the ant will
follow to reach the farthest corner from
one particular corner is shown below in 12
the diagram.
Thus the required length
= (12)2 + (12)2 = 12 2 6 6 2
Therefore, the required ratio = 4 / 10 = .
Hence, choice (c) is the correct one. 5

62 Let the longer side be x and the shorter side be y, the 64 In ∆ABO , AB = OA 2 − OB 2 = 2
semiperimeter of the parallelogram would be x + y. Since AB = OB, so ∆ABO is an
x+ y isosceles triangle.
x x+ y x x Therefore, ∠BAO = ∠BOA = 45° B C
∴ = ⇒ = √2 O 1
y x y x √2 √3
In ∆OCD, 2 2
x
 y CD = OD 2 − OC 2 = 3 A D
1+  
x  x x y Since, the three sides of the triangle are in the ratio
⇒ = ⇒ =1 +
y 1 y x 1: 3 : 2, it implies that ∠CDO = 30° and ∠COD = 60°
x y Therefore, ∠AOD = 180 − (∠BOA + ∠COD ) = 75°
⇒ − −1 = 0
y x Now, area of the shaded region = area of ∆ABO + area of
1 x sector AOD + area of ∆OCD
⇒ k − −1 = 0 (assuming = k) 1
k y Area of ∆ABO = × 2 × 2 = 1sq cm
2
⇒ k2 − k − 1 = 0 75 5π
Area of sector AOD = π (2)2 × = sq cm
1+ 5 x 1+ 5 360 6
⇒ k= ⇒ =
2 y 2 1 3
Area of ∆OCD = × 1 × 3 = sq cm
∴ y = 2 cm ∴ x = 1 + 5 cm. 2 2
But you know that the maximum area can be obtained only 5π 3
Therefore, the required shaded area = 1 + + sq cm.
when parallelogram is a rectangle. Therefore, area of the 6 2
rectangle = x × y = (1 + 5) × 2 = 2 (1 + 5) Hence, choice (d) is the correct one.
Hence, choice (b) is the correct one. 65 Since area of the circle = 1
63 Let each side of the square inscribed by the semicircle is 2x, ∴ πr2 = 1 ⇒ r = 1 / π
then the radius of semicircle would be 5x.
It means the diameter of the circle would be 2 5x. ∴ Each side of hexagon =2/ π
2
It means the diagonal of the square inscribed by the circle is 3 3 2 6 3
∴ Area of the hexagon =   = sq cm.
2 5x. 2  π π
Hence, choice (c) is the correct one.
Mensuration 513

πa2 1 (a − 2)
66 Area of the semicircle with a as its diameter = So, we have EP = FP = (EF ) =
4 2 2
πb2 Therefore, total area of roads (shaded region)
Area of the semicircle with b as its diameter =
4 = 4 (area of EPSG) + 4 (area of ∆AEG) + (area of PQRS)
πc2
Area of the semicircle with c as its diameter =  (a − 2)
= 4 G
4  + (1) + (1) = 2( 2 a − 1) A H D
ab  2 
Area of right triangle = E
S
2 And, the area of the square = a2
Therefore area of the shaded region P R
But, it is given that a2 = 2[ 2( 2 a − 1)]
 πa2 πb2 ab πc2 π 2 ab F
= + +  − = (a + b2 − c2 ) + ⇒ a2 − 4 2 a + 4 = 0 Q
 4 4 2 4 4 2 B C
⇒ a= 2 2+ 2
π 2 ab ab Please note that the other value of a = (2 2 − 2) < 1. Since
= (c − c2 ) + =
4 2 2 the length of the square cannot be less than the breadth of
Hence, choice (a) is the correct one. the road, so it is an inadmissible value.
67 Let r be the circumradius of the circumcircle and O be the Hence choice (c) is the correct one.
centre of the circle, where CD is the perpendicular bisector 69 Point F is the centre of the semicircle. Now connect the
of AB, as ∆ABC is an isosceles triangle. points as shown in the diagram. Now, you see that there
Therefore, AD = BD = 8 are four equilateral triangles of same height.
Now, since AO = OC = r C
Therefore, OD = 16 − r
C
D E

A B A B
D F

Now, we have AO 2 = AD 2 + OD 2 Therefore, AF = DF = EF = BF = 1cm = radius of circle.

2
⇒ r2 = 82 + (16 − r)2 ⇒ r = 10 cm Area of shaded region = Area of semicircle – area of
3
∴Area of circle = 100 π cm2 quadrilateral ADEB
Hence, choice (c) is the correct one. 2  1   3 
=  π × 1 −  3 × × 1 
68 Let the length of each side of the square be a unit and the 
3 2   4 
breadth of the road 1 unit. π 3
1 1 = −  cm2
Then, we have AE = AG = (EG ) = 3 2
2 2
There fore, EF = AB − ( AE + BF ) = a − 2 Hence, choice (b) is correct.

Level 02 Higher Level Exercise

1 2πr = a  4π 2 − 1 
h 2 = a2  2 
a  4π 
a
∴ h= 4π 2 − 1
a a a a 2π

1 2 1 a2 a
a ∴ Volume = πr h = π × × 4π 2 − 1
Also, slant height (l) = a ∴ r= 3 3 4π 2 2π
2π a3
= 4π 2 − 1
2 Q l2 = h 2 + r2 24π 2
2
 a
⇒ h 2 = l2 − r2 = a2 −  
 2π 
514 QUANTUM CAT
3 It will be in the form of a right angled triangle. and area of sector COD (figure ii)
πr 120
πr = πr2
90° 90° 360
2
 a 1 πa2
=π×  × =
a a  3 3 9

∴ Area of segment = (Area of sector − Area of triangle)

4 2πr (r + h) = 1540 cm2 πa2 a2

= −
9 4 3
and (r + h) = 35 cm
∴ 2πr =
1540
= 44 cm  πa2 a2 
35 Total area of all the four segments = 4  − 
 9 4 3
1
5 Total volume = πr2h1 + πr2h2  πa2 a2 
3 16 cm and the total area of whole figure = a2 + 4  − 
2  h   9 4 3
= πr h1 + 2
 3 
80 cm
7 2 (l + b) = 26 ⇒ l + b = 13
22  16 
= × (21)2 80 + 12 + 1 = 13 11 + 2 = 13
7  3 
10 + 3 = 13 9 + 4 = 13
22 256 21 cm
=
× 441 × 8 + 5 = 13 7 + 6 = 13
7 3
22 256 8.45 Since, l > b, therefore, there are only 6 integral values of the
Weight = × 441 × × = 999.39 kg length viz., 7, 8, 9, 10, 11 and 12.
7 3 1000
6 ABCD is a square, each side of square is ‘a’. 8 Total surface area = 2πR 2 + 2πr2 + (πR 2 − πr2 )
= 3πR 2 + πr2 r
D C R
= π (3R 2 + r2 )
2
⇒ 1436 = π [ 3 × (12)2 + r2]
7
A B
10054 1
⇒ × = 432 + r2 ⇒ r = 5 cm
Fig. (i) 7 π
In figure (ii), ∠ DOC = 120° 2
∴ Internal volume of hemisphere = π (R 3 − r3 )
and ∠ ODC = ∠ OCD = 30° 3
2
= π [(12)3 − (5)3]
D P C 3
D a C
90° 30° 2
= 3358 cm3
3
60° 9 Since, there are 3 faces which are visible in a corner
120°
cube. When the cube of a corner is removed then the 3
O
O faces of other cubes will be visible from outside. So,
Fig. (ii) Fig. (iii) there will not be any change in the surface area of this solid
In figure (iii), figure.
PC
= sin 60° 1
OC
a/ 2 3 a 2
= ⇒ OC = 3
OC 2 3
⇒ radius of the arc ‘CD ’. 3
4  15
1 π 
∴ Area of triangle OCD = × CD × OP 3  2
2 10 Number of spheres = 2
= 125 spheres
4  3
2 π 
=
1
×a×
a
=
a 3  2
2 2 3 4 3 2
 15
Surface area of a large sphere = 4π ×  
 OP 1  2
Q = tan 30° and tan 30° = 
 PC 3
Mensuration 515

2 and area of lawn = 10 × 8 = 80 m2

 3
and surface area of a small sphere = 4π   Reduced area of lawn = 8 × 8 = 64 m2
 2
∴ New area of path = 88 + (80 − 64) = 104 m2
and total surface area of all the smaller spheres 104 13
∴ Ratio of areas of path = =
2 88 11
 3
= 125 × 4π   Hence, option (c) is correct.
 2
Alternatively Let the breadth of the lawn be ‘ b’ then
% change in area the length will be (b + 2).
  3
2
 15
2  ∴ Area of path = (l + 4)(b + 4) − lb
 500π   − 4π  
 2  2 
= (b + 6)(b + 4) − (b + 2) b = 8b + 24
= × 100 = 400%
  15
2
 and the new area of path = (l + 4)(b + 4) − b × b
 4π   
  2  = (b + 6)(b + 4) − b2
Alternatively Surface area of larger sphere = 25x
= 10b + 24
(10b + 24) 13
Surface area of smaller sphere = x Q = ⇒ b = 8 m ∴ l = 10 m
(8b + 24) 11
Q The ratio of radii is 5 : 1.
14 From the figure you can see that just half of the liquid has
Therefore, ratio in surface areas = 25 : 1
been flown off and half the liquid is remained in the
Now, since there are 125 smaller spheres. cylindrical jar.
∴ Total surface area of smaller spheres = 125x
125x − 25x
∴ % change in area = × 100 = 400%
25x
11 Let the radius of cone be R and radius of sector = r, K K

2πR Thus it is clear that the capacity (or volume) of the cylinder
= 2 × 2.1 = 4.2 L
15 When the height and base of the cone are same as that of
1
cylinder, then the volume of cone is that of the cylinder.
3
l r ⇒l=r 1
60° Thus the capacity of cone = × 4.2 = 1.4 l
3
Thus the remaining volume = 2.1 − 1.4 = 0.7 l
0.7 1
∴ The required ratio = =
then the slant height of cone (l) = r 4.2 6
60 D C
and 2πR = 2πr × 16 AC = (30)2 + (16)2
360
r 14 7 AC = 34 m 16 m
⇒ R= = = cm
6 6 3 But since elephant is itself 4 m
A B
long. So he has to travel only 30 m
∴ Total surface area = πr (l + r)
(34 − 4) = 30 m.
22 7  7
= × 14 +  = 119.78 cm2 30
7 3 3 ∴ the speed of elephant = = 2 m/s
15
12 Between 26 poles, total length is (26 − 1) × 4 = 100 m 2πr 60 2πr
17 Arc of sector = =
It means the length of each side of a square field is 100 m. 360 6
∴ Area of field = (100)2 = 10, 000 m2 = 1 hectare
13 It is clear that length of the lawn is 2 m more than the
breadth of lawn.
To solve this problem quickly, go through options. Let us r l
take option (c). 60°
l = 10 m ⇒ b = 8 m
Area of path = (l + b + 2w ) 2w This arc of sector will be equal to the perimeter of cone. Let
= (10 + 8 + 4) 4 = 88 m2 2πr r
the radius of cone be R, then 2πR = ⇒ R=
6 6
516 QUANTUM CAT
Further the radius of sector will be equal to the slant height ∴ (l − b)2 = l2 + b2 − 2lb = 20 − 16 or (l − b)2 = 4
of cone. ⇒ l−b=2
∴ l=r Q l + b = 6 and l − b = 2
Now, since l2 = h 2 + R 2 ⇒ h = l2 − R 2 ∴ l = 4 and b = 2
2
 r 35 ∴ Area of rectangle = 4 × 2 = 8 cm2
h = r2 −   = r
 6 6 ∴ Total area of the figure = 8 + 10 3
18 The diagonal of cube will be equal to the diameter of = 2 (4 + 5 3) cm2
sphere.
4  d πd 3
3 21 Area of each square = 16 cm2
∴ Volume of sphere = π  = 1
3  2 6 Area of quadrant ADMB = π × (4)2 = 4π
d 4
and each side of cube = a = and radius of smaller quadrant
3
d3 CPMQ = CM = AC − MA
∴ Volume of cube = a3 =
3 3 = 4 2 − 4 = 4 ( 2 − 1)
1
πd 3 d3 d3  π 1 ∴ Area of smaller quadrant = π [ 4 ( 2 − 1)]2
∴ Remaining volume = − =  −  4
6 3 3 3 2 3
= 4π (3 − 2 2)
19 Let AP = x, then AM = x and MS = x
Area of shaded region inside the square ABCD
S R
= 16 − [ 4π + 4π (3 − 2 2)]

M N = 16 − [ 4π (1 + 3 − 2 2)]
= 16 − [ 4π (4 − 2 2)]
A P Q B = 8 [ 2 − 2π + 2π ]
Now, area of quadrants = AEG + EFG = 2AEG
∴ AS = AM + MS 1
AS = 2x
S = 2 × π (4)2 = 8π
4
∴ PS = AS − AP
2 2
∴ Area of shaded region inside the square EAGF
PS = 3x 2x = 8π − 16 = 8 (π − 2)
M
∴ Area of square 8 (2 − 2π + 2π )
∴ Required ratio =
PQRS = ( 3x )2 = 3x 2 8 (π − 2)
Area of circle = πr2 = π × x 2 = πx 2 A x P [ 2 + π ( 2 − 2)]
=
Area of both the circles = 2πx 2 (π − 2)
2πx 2 2π
∴ Required ratio = = 22 Given that
AB AD
=
3x 2 3 BC DF
20 Let the length of rectangle be ‘ l’ and Also BE = BC
breadth be ‘ b’, then D C
Q
Let AD = 1 and AE = x
2 (l + b) = 12 S
AE AE AE
⇒ l + b = 6 cm
A B ∴ = = =x
EF AD BC
and area of larger equilateral triangle
AE AD Q AD = BC = BE 
3 2 ∴ =  
= l EF AB  and AB = AE − BE 
4
x 1
3 2 =
Similarly area of smaller equilateral triangle = b 1 x −1
4
∴ Total area of all the 4 triangles (1 ± 5)
⇒ x2 − x − 1 = 0 ⇒ x =
3 2 2
= 2× (l + b2 ) = 10 3
4 (1 + 5)
∴ x=
⇒ l2 + b2 = 20 2
∴ (l + b)2 = l2 + b2 + 2lb Since ratio of two sides can never be negative.
⇒ 36 = 20 + 2lb Alternatively Since ratio of two side can never be
negative therefore only option (c) is correct.
⇒ lb = 8
Mensuration 517

Solutions (for Q. Nos. 23 to 25) Let DP = a, then

DC = DP + PQ + QC  Q DP = QC 
AB = 4 D C  and DP = PG = GQ = QC 
=a+ a 2+ a  
4 2 ∠ G = 90°
∴ AO = AC = =2 2 DC = a (2 + 2)  
2 42
2 O 2
∴ Area of  ∴ PQ = 2a 
∴ Area of circle  
4 1 a 2
 
ABCD = π × (2 × 2)2 = 8π A B ∆PGQ = × a × a =
2 2
Area of region 2 (only left part)
∴ Area of all the triangles outside the square ABCD
Area of circle – Area of square
= a2
4 =4× = 2a2
8π − 16 2
D
= = (2π − 4) C
But DC = a (2 + 2) = 4 cm
4 3
4
Area of region 3 = Area of square ⇒ a=
(2 + 2)
− 2 (Area of semicircle) 3 2
1  A B  4 
= 16 − 2  × π × 4 ∴ 2 (a)2 = 2 ×  
2   2 + 2
= 16 − 4π = 4 (4 − π ) cm2 16 (3 − 2 2)
= × = 16 (3 − 2 2)
Area of region 1 = Area of semicircle AD D (3 + 2 2) (3 − 2 2)
− Area of region 2 and Area of square = 16 cm
1 ∴ Total area of the figure = 16 + 16 (3 − 2 2)
= π × (2) − (2π − 4) = 4 cm2
2
2 A = 16 (4 − 2 2)
23 Total area of region 1 = 2 × 4 = 8 cm2 = 32 (2 − 2) cm2
24 Total area of region 2 = 2 × (2π − 4) = 4 (π − 2) cm2 28 When l = CD, then the volume of cone will be maximum,
25 Total area of region 3 = 4 (4 − π ) cm2 where l is the slant height of the cone and the largest
D C possible angle at the vertex of cone is 90°.
26 Total area of square = 64 cm 2

y x A A
∴ 4 ( x + y ) = 64 x
⇒ x + y = 16 …(i) y y 20 cm P
Again in a semicircle O
xx y xx 16 cm D D
1
AOB = x + y + x = π × (4) 2
A B 9.6
2
2x + y = 8π …(ii) B B
C 12 cm C 9.6 Q
From eqs. (i) and (ii), we get
x = 8π − 16 12 × 16
CD = = 9.6 cm,
Total area of shaded region = 4 (8π − 16) 20
= 32 (π − 2) cm2 which is the radius of the sector.
Alternatively Area of square − 2 (Area of semicircles) Therefore, arc of the sector
= 2y = 2π × 9.6 ×
90
= 4.8π
(64 − 16π ) = 2y 360
∴ 4 y = (128 − 32π ) Let the radius of the cone be r, then
∴ Required area of shaded region 2πr = arc of the sector
(4 x ) = (Area of square − 4 y) 2πr = 4.8π
= 64 − (128 − 32π ) r = 2.4
= 32π − 64 = 32 (π − 2) cm2 ∴ Height of the cone (h) = l2 − r2
27 You can see in the figure that the sides of one square is = (9.6)2 − (2.4)2 = 2.4 15 cm
parallel to the diagonals of the other square. 1 2
G ∴ Volume of the cone = πr h
3
D C
P Q 1 22
= × × (2.4)2 × 2.4 15
H F 3 7
= 56.1 cm3
A B
E
518 QUANTUM CAT
29 To increase the value (or price of diamond) they should cut 36 Volume of the whole body
(divide) the diamond in such a way that the surface area 1 2 h1
V1 = πr1 h1 + πr12h2
will be maximum. 3 r1
h1 2
but =
h2 3 h2
11h1 h3
∴ V1 = πr12
a 6
2 5h 2r2
a and h3 = (h1 + h2 ) = 1
3 3
a Hence, volume of the hole (V2 ) = πr22h3
Thus, when four parts are parallel to each other. 5
= πr22h1
In this way total surface area 3
= 6a2 + 2a2 + 2a2 + 2a2 = 12a2 V1 − V2
But it is given that V2 =
3
Actual surface area of cubical diamond = 6a2
∴ V1 = 4V2
Therefore, percentage increase in area
5 2 11
12a2 − 6a2 ⇒ 4× πr2 h1 = πr12 × h1
= × 100 = 100% 3 6
6a2
Remember that for the given volume, minimum surface 55
⇒ r2 = cm
area is possessed by a cube. So to maximize the area we 8
have to increase the maximum possible difference between
the edges of cuboid. 37 19 × 19 = 361
Thus, we make equal 19 measurements each of 19°, then
30 Side of square I = a
we get (361 − 360) = 1° angle at the centre. Thus, moving
a
Side of square II = continuously in the similar fashion, we can get all the 360°
2
angle i . e. , 360 equal sectors of 1°.
a
Side of square III = 38 When we open the paper after cutting it,
2
a we will find it as shown in the following
Side of square IV =
2 2 figure.
a Radius of the larger circle = 5 cm
Side of square V =
4 ∴ Area of larger circle = 25π
Therefore, sum of perimeters of all the squares and the radius of each smaller circle is 1 cm.
 a a a a
= 4 a + + + +  Therefore, total area of all the 9 circles = 9 × π × (1)2 = 9π
 2 2 2 2 4
∴ Remaining area = (25 − 9) π = 16π
 1 1 1 1
= 4a 1 + + + +  Hence, the required ratio = 25 : 16
 2 2 2 2 4
4 + 2 2 + 2+ 39 In the top layer we can see that total
2 + 1 D C
= 4a   = a (7 + 3 2) 13 cubes get a cut. So, in 7 layers total
 4  13 × 7 = 91 cubes will get a cut and
the remaining (7 3 − 91) = 252 cubes
31 Total area of the five squares
2 2 2 2 are without any cut.
 a  a  a   a Total number of pieces which are not A B
= a2 +   +   +   + 
 2  2  2 2  4 a cube
 1 1 1 1 = 12 × 2 × 7 + 4 × 7 = 196
= a2 1 + + + +
 2 4 8 16 
16 + 8 + 4 + 2 + 1  31 31a2 (Since 84 cubes are diagonally cut into two parts and 7
= a2 = a2 × =
 16  16 16 cubes which are in the centre are divided into 4 parts.)
Thus, total 196 children and teenagers will get one-one
32 (n − 2)3 piece and 252 adults get one-one piece.
33 6 (n − 2)2 Thus total 252 + 196 = 448 people can get a piece of cake.
34 12 (n − 2) NOTE It is clear that everyone get equal number of pieces
35 There are 8 cubes at the corners, which are always fix. but not according to the volume of pieces.
Mensuration 519
2
Solutions (for Q. Nos. 40 to 42) Diameter ( 2R ) of the   3 
outermost circle is equal to the diagonal of larger square.   R 
Area of inner circle π  2   3
2R 44 = =
Hence, the side of square = = 2R Area of outer circle π (R )2 4
2
Again the side of larger square is equal to the diameter of 45 Each side of the first hexagon = R
middle most circle. 3
R Each side of the second hexagon = R
Hence, the radius of mid-circle is . 2
2 3
Each side of the third hexagon = R
Once again the diameter of the mid-circle is equal to 4
the diagonal of smaller square. Hence, side of the smaller 3 3
Each side of the fourth hexagon = R
square = R. Similarly the diameter of innermost circle is equal 8
to the side of the smaller square. R 8
∴ Required ratio = =
R 3 3 3 3
Hence, radius of the innermost circle = . R
2 8
R
40 46 From the concept of similarity of triangles. All the five
2 quadrilaterals viz., AOA′, BOB′ , COC ′ , DOD′ and EOE′ are
41 Area of larger square = ( 2R )2 = 2R 2 similar.
and area of smaller square = R 2 From the figure (ii),
r2 − r1 r3 − r2
∴ Total area of both squares = 3R 2 =
r2 + r1 r3 + r2
 R R
r −r r −r
42 Sum of all the circumferences = 2π  R + + 
 2 2 = 4 3 = 5 4 =K
r4 + r3 r5 + r4
 2 + 2 + 1
= 2πR  
 2  90°
E 90° E'
= (3 + 2) πR D'
D
Sum of perimeters of all the squares = 4 ( 2R + R )
C C'
= 4R ( 2 + 1)
B B'
(3 + 2) πR (3 + 2) π
∴ Required ratio = = A A'
( 2 + 1) 4R ( 2 + 1) 4
Alternatively Since circumference and perimeter both
has R. So R will be cancelled in the ratio of circumference O
Fig. (i)
to the perimeter. Thus, neither of the choices a, b and c are
admissible. Hence, the only correct choice is (d). r2 r3 r4 r5
⇒ = = = =K
r1 r2 r3 r4
Solutions (for Q. Nos. 43 to 45) Each side of outer (larger)
(By componendo and dividendo)
hexagon is equal to the radius of circle which is R.
Now, OC = ON = OD O
C
radii of the inner (smaller) circle B
ON 3 A
But = sin 60° = M
OA 2 C D r1 r2 r3
3 3 O
⇒ ON = OA = R , radius of the A B P Q R
2 2 N Fig. (ii)
inner circle
It means all the radii are in GP.
and this is also equal to the side of the inner hexagon. 4
r 81  3
3 Therefore, 5 = (K )4 = = 
43 Sum of perimeters of both the hexagons = 6R + 6 × R r1 16  2
2 3
 3 ⇒ K= ∴ r3 = r1 (K )2
= 6R 1 +  2
 2 9 9r 9
r3 = r1 × = 1 = × 16 = 36 cm
= 3 (2 + 3)R 4 4 4
520 QUANTUM CAT
47 Q r1 = 16, r2 = 24, r3 = 36, … etc. Area of region y = Area of square − 4 (area of quadrant)
OP OQ 1 
∴ = = 4 − 4  π × (1)2 = (4 − π )
AP BQ 4 
∴ Required area (of shaded region)
h + r1 h + 2r1 + r2 h + 16 h + 56
= ⇒ = = Area of square − [Area of region x + Area of region y]
r1 r2 16 24
= 4 − [ 4 − π + 4 − π ] = 2π − 4
⇒ h = 64 cm
51 Let the volume of solid block be V and radius of the spheres
48 60 = 1 × 1 × 60
formed from the first block be r1, then the volume of each
= 1 × 2 × 30
sphere be V1.
= 1 × 3 × 20
Similarly, let the radius of each sphere obtained from
= 1 × 4 × 15
second block be r2 (= 2r1 ), then the volume of each sphere
= 1 × 5 × 12
= 1 × 6 × 10 be
= 2 × 2 × 15 V2 = (8V1 )
= 2 × 3 × 10 ∴ V = kV1 + 14 …(i)
… … … and V = lV2 + 36
… … … or V = 8lV1 + 36 …(ii)
= 3× 4 × 5
Out of the given different combinations the first From eqs. (i) and (ii),
combination (1 × 1 × 60) gives maximum length of kV1 + 14 = 8lV1 + 36
diagonal of cuboid, but in this case two of the edges are
same. So, the second combination gives the proper value ⇒ V1 (k − 8l) = 22
i . e. , which gives the maximum length of diagonal whose all The possible value of V1 = 22, 11, 2 or 1
sides are different. Hence, the length of such a pencil is
But V1 can never be equal to or less than 14 (since
equal to the diagonal of cuboid = 12 + 22 + 302 = 905
remainder is always less than divisior) So, the only possible
49 value of V1 = 22.
∴ V2 = 8 × V1 = 176 cm3
O O
52 The length of tether of the horse is 80 m.
A B A P B S
Q
C D C D 40
Fig. (i) Fig. (ii)
R 40
D
In figure (ii) C
80
3 Q
OP = OA = 4 3 cm 40
2
30
OP OA 4 3 8 90°
Again = , = (OQ = OP + PQ = 4 3 + 2 3) P
30 50
OQ OC 6 3 OC A B
80
⇒ OC = 12 cm
∴ Each side of the outer hexagon is 12 cm.
∴ Required area = (Area of outer hexagon
− Area of inner hexagon)
3 3
= [12 − 8 ] = 120 3 cm2
2 2
2
50 Area of region x = Area of square − Area of inscribed circle
Area grazed by horse
= (4 − π )
 270 90 90 
x x x x = π × (80)2 × + π × (30)2 × + π × (40)2 ×
 360 360 360 
 3 1 1
y y = π  6400 × + 900 × + 1600 × 
 4 4 4
x x x x  21700 
=π = 5425π m2
 4 
Fig. (i) Fig. (ii) Fig. (iii)
Mensuration 521

Hint When horse is tethered at B, then he can move freely 270° i . e. , 4

from P to S with the full length of his tether as the radius of the arc = π (7 )3 [125 − 8]
3
PS. Again when horse reaches P and tries to move further in the
4
right direction i . e. , towards D, his tether gets fixed at A and now = π × 343 × 117
only 30 m tether is free which works as a radius of the quadrant 3
PAQ. Similarly when horse reaches S and tries to move left further in 4
the direction of D, due to wall BC his tether gets fixed at C and now π × 343 × 117
only 40 m tether is free to move further. At this moment point C Required time to explode = 3 = 1078 s
156
behaves like a fixed point of tether (as centre of quadrant RSC).
NOTE You can tether horse at any of the four corners 57 Let the each side of cube be a, then
A , B , C and D area grazed by horse remains same. CD = 2a
a
53 Here each side is broken up into 6 parts i . e. , n = 6 ∴ CQ =
2
Now, N 0 = (n − 2) = (4) = 64
3 3
Let the radius of cone be r and height be h, then
N1 = 6 (n − 2)2 = 6 × (4)2 = 96 r=h 2
N 2 = 12 (n − 2) = 12 (4) = 48 ∴ In ∆ APO and ∆ CQO (Similar triangles)
∴ N 0 : N1 : N 2 = 64 : 96 : 98 = 4 : 6 : 3 a
AP CQ r 2
54 Let the radius of seed be r and radius of = = =
PO OQ h (h − a)
the whole fruit (pulp + seed) be R, then
thickness of the pulp = (R − r) a
r
4 2 = 2
Volume of mango fruit = πR 3 ⇒ A B
3 (h − a) P
4 R
and Volume of pulp = π (R 3 − r3 ) ⇒ a = 2 (h − a)
C D
3 3a Q
⇒ h=
4  2  
3
but = π R 3 −  R   2
3  7  3a 3a
 ∴ r= × 2 and h =
 r 2 2  2 2 O
Q R − r = 5 ⇒ r = 7 R  1  3a 2 3a
2
  ∴ Volume of cone = π ×   ×
∴ Percentage of volume of pulp to the total volume of fruit 3  2  2
4  8  9 3
πR 3 1 − = aπ
3  343 4
=
4
πR 3 and Volume of cube = a3
3 9 3
335 πa
= × 100 = 97.66% 9
∴ Required ratio = 4 3 = π = 2.25π
343 a 4
55 Let the radius of each smaller circle is D C 58 For the given volume, cube has minimum possible length of
r and radius of the larger circle is R, R diagonal.
then Therefore each side of cube = 4 cm
O
πR 2 = 4πr2 ⇒ R = 2r and its diagonal = 4 3 cm.
P
OR = OP = R + r = 3r l
59 l = 2πr ⇒ r =
Also PM = r A M B 2π
(PM is the perpendicular on AB)
∴ AP = 2r
b
∴ AO = AP + PO
= r 2 + 3r = r (3 + 2) l
∴ AC = 2AO = 2r (3 + 2), which is the diagonal of square
where r is the base radius of cylinder and l is the length of
2r (3 + 2) paper and h = b, where h is the height of cylinder and b is
∴ Required ratio = = (2 + 3 2)
2r the breadth of the paper.
2
56 Initial radius = 14 cm  l 
∴ Volume of cylinder = πr2h = π ×   × b
 2π 
Radius at a time when the balloon explodes = 35 cm
4 π × l2b 385
Change in volume = π [(35)3 − (14)3] ⇒ = 48.125 =
3 4π 2
8
⇒ l2b = 11 × 11 × 5
522 QUANTUM CAT
⇒ l = 11 and b = 5 (Q l > b) ∴ Width of the sheet = AK + MC + CT
∴ Volume of the box = l × b × 4 = 1 + 3 + 1 = (2 + 3) ft
= 10 × 4 × 0.5 = 20 cm3
w or

Fig (i) Fig (ii)

K L
4 4 cm x M x B
A

cm
A M B

5
10 10 cm

0.
2x
C
Height of cone C
60 Vertical spacing between any two turns = T
Number of turns Fig. (iii) Fig. (iv)
h
= 64 Recall that for a given perimeter the polygon of minimum
n number of sides has minimum area and the polygon of
h maximum number of sides has maximum area. So, the
61 Number of turns =
x correct relation is h > s > r.
Length of string in each turn Thus, hexagon (6 sides) has maximum area.
4 Now, between square and rhombus, square has greater area
= 2πr = 2π ×= 8 cm
π than rhombus. For easier understanding consider some
∴ Total length of string in all the values.
x
n turns x 5
h 8h x
= ×8= cm x h
x x x 5 5 4 5
62 Total length of string = 8n cm x
Since, total length of string 5 3 2
= number of turns × perimeter of cylinder Area = 25 cm2 Area = base × height
= 8 × n = 8n cm
= 5 × 4 = 20 cm2
Alternatively
D a D a C a
D C

a a a a
h
C a
a
θ
A a B A B
8n a
Length of string required for 1 turn (or round) = = 2n Square Rhombus
4
 a
2 In rhombus ABCD: Area of rhombus = a × h
but 2n =   + (4a)2 h 
 4 = a × a sin θ  = sin θ
a 
a/4 = a sin θ
2
1 4 4 2 4 4 3

a/4 As you know the maximum value of sin θ is 1 at θ = 90° but

a a at θ = 90° rhombus will become a square. So except θ = 90°
a/4
for all the rest values the area of rhombus will be less than
a/4 the area of square.
a a a a
65 C
4a C
8n
⇒ a= 60°
m
1c

257 P a Q
where a is the side of cube.
a
10
63 From the sheet of 10 ft long, maximum = 5 circular discs 60° 60°
2 A B A a B
M N
can be cut along the length of the iron sheet. Fig. (i) Fig. (ii)
CM = AC 2 − AM 2 = 4 x 2 − x 2 PCQ is also an equilateral triangle
CM = x 3 = 3 ft (Since x = 1 ft) ∴ PC = PQ = PM = a
Mensuration 523

∴
a
=
3
⇒ PA =
2a 8r2 − 2πr2 2r2 (4 − π ) 1
∴ Required ratio = = =
PA 2 3 6 πr 2 6r2 π 11
2a
∴ AC = AP + PC = + a = 1 cm 67 Short cut: Very quickly check the options. If all the
3
options have values.
3
⇒ a= = 3 (2 − 3) Alternatively The required capacity of box is 864 m3 .
(2 + 3)
C Let the length of the base be l and height of the box is ‘ h’,
Now, in figure (iii) then
PM = MT = a 864
Let the each side of square RSYX 864 = l2h ⇒ h = 2
l
be K, then RT = K also (since P
T
Q
RTS is an equilateral triangle) S
Now, surface area of the box A = l2 + 4lh
R
4l × 864
A B = l2 +
∴
K
=
3 M X Y N l2
RM 2 Fig. (iii) 3456
= l2 +
2K l
∴ RM =
3 Differentiating w.r. to l, we get
2K dA 3456
∴ MT = RT + RM = K + = 2l − 2
3 dl l
( 3 + 2) dA
MT = K For the minimum area = =0
3 dl
But MT = a 3456
∴ 2l = 2
 3 + 2 l
∴ a=  K
 3  ⇒ l3 = 1728
3a ⇒ l = 12
∴ K=
( 3 + 2) ∴ Base area = (12)2 = 144 m2
But a = 3 (2 − 3) 864
and Height = =6m
3 l2
∴ K= [ 3 (2 − 3)]
( 3 + 2) 68 Let the initial radius be r and volume be V, then
3 (2 − 3) (2 − 3) V = πr 2 × 4
K= ×
(2 + 3) (2 − 3) Ist case: V1 = π (r + 12)2 × 4
3 (2 − 3) 2
IInd case: V2 = πr2 × (4 + 12)
⇒ K= = 3 (7 − 4 3)
1 But V1 = V + K
∴ Area of square RSYX = K 2 = [ 3 (7 − 4 3)]2 and V2 = V + K
K 2 = 9 (49 + 48 − 56 3) ∴ V1 = V2
⇒ π (r + 12)2 × 4 = πr2 (16)
K 2 = (873 − 504 3) cm2
⇒ r = 12 ft
66 For the minimum wastage of sheet he has to cut the sheet
∴ Increased volume = V1 = V2
in the given manner.
= π × (24)2 × 4 = 2304π cubic ft
2r 2πr 2r
69 Let us denote the distinct regions of the square by x , y and p
2r as shown in the following figure.
r r
r y
x x
Total area of sheet required
y p y
(2πr + 4r) × 2r = 4r2 (π + 2)
Area of sheet utilised = (2πr × 2r) + 2 (πr2 ) = 6πr2
x x
Area of wastage sheet = 4r2 (π + 2) − 6πr2 y

= 8r2 − 2πr2
524 QUANTUM CAT
Now, from this figure we have It implies that ∠ MDC = 30° and ∠ MCD = 60°, since
4x + 4y + p = 1 (i) ∠ CMD = 90°
π Therefore ∠ NCD = 120°.
And, x + 2y =1−
4 So the length of arc ND
⇒ 4 ( x + 2y ) = 4 − π (ii) = 2 π × 6370 ×
120
= 13346. 67 km
π 360
But, it is given that p = − 3 + 1 (iii) Hence, choice (c) is correct.
3
Now, substituting the values from eqs. (i) and (iii) in the 72 Case 1: When the radius of each circle is same, then in
equation (ii), we get this case radius of each circle would be 49/4 and the total
4x + 8y = 4 − π area of both the circles would be 2π (49 / 4 )2 ≈ 300 π
⇒ (4 x + 4 y ) + 4 y = 4 − π Case 2: When there is one circle of the largest possible
2π
⇒ 4y = 4 − 3 − size and another circle is fitted in the remaining area.
3
Hence, choice (a) is correct. A B

70 Let each side of the square at the corner be x cm and each

side of the base of the cuboid be b cm, then O
x + b + x = 24 Q P
⇒ b = 24 − 2x
C M N D
Therefore, each side of the base of the cuboid, which is
b = (24 − 2x ) cm Let the radius of larger circle be R and that of smaller circle
be r.
Consequently, the height of the cuboid is h = x cm
Then, OM = R = 16
Therefore, volume of the cuboid = ( x )(24 − 2x )(24 − 2x )
and QM = PN = r.
= 2 (2x )(12 − x )(12 − x )
Therefore, OQ = OM − QM = 16 − r
x And OP = R + r = 16 + r
Similarly,
PQ = MN = 49 − (CM + ND ) = 49 − (16 + r) = 33 − r
b
Now, OP 2 = OQ 2 + PQ 2
⇒ (16 + r)2 = (16 − r)2 + (33 − r)2
x ⇒ r2 − 130r + 1089 = 0
⇒ r = 9 or 121
We know that b + 2x = 24, is constant.
Since, 121 is greater than 16, so it’s unacceptable.
We also know that if p1 + p2 + p3 + ..... + pn = some Therefore r = 9. Thus, the total area of the two circles
constant, then p1 ⋅ p2 ⋅ p3..... pn will be maximum when = π (162 + 92 ) = 337 π
p1 = p2 = p3 = . . . . = pn. As in case 2, the total arc is more than that in case 1, so
Accordingly, 2x + (12 − x ) + (12 − x ) = 24 is a constant, 337 π is max. possible area.
therefore, the volume of cuboid will be maximum when Hence, choice (a) is the correct one.
2x = (12 − x ) = (12 − x )
It implies that x = 4 73 Since, area of square field = 54 m2,

So the volume of the cuboid will be maximum when so each side of the square A B
M
x = 4 cm. = 3 6 m and
Hence, choice (d) is correct. diagonal of the square
71 The man has to reach D from N. Let N
=6 3m Consider the
M is the centre of the latitude where following diagram, now.
P
the man is supposed to reach and MD MB = BN = CN = CM = 6 m
is the radius of the pertinent latitude. CP = 1 / 2 (BC ) = 3 3 m
Now, extend the radius NC to M so C
Therefore, using Pythagoras N
that MD becomes perpendicular to theorem, you can find PM C 6 D
CM. M D
PM = CM 2 − CP 2 = 3 3√6
Given that CD = 6370 km and MD =
∴ MN = 2 (PM ) = 6.
3185 3 km. That is MD/CD = 3/2.
Mensuration 525

It implies that ∆MCN and ∆BMN are equilateral triangles. tangent to the circle, as shown in the diagram. And,
therefore, the diameter of the circle must make a 45° angle
Area of the common region = Area
with the base in order to conform to the symmetry.
encapsulated between two arcs
Now, the radii GO = EO = HO = r
  60  3 2
= 2 π 62   − 6  Therefore, in ∆HOF either by using Pythagoras theorem or
  360  4  HO r
= 2 (6π − 9 3) Trigonometric ratio, we have OF = HF = =
2 2
Area grazed by sheep and goat = 2(Area Therefore, CD = EF
of a quadrant) – area of the common ⇒ CD = EO + OF
region r
⇒ a=r+
π  2
= 2 (6)2 − 2 (6 π − 9 3)
4   2 
⇒ r = a 
= 6(π + 3 3)  1 + 2
Area that cannot be grazed by sheep and goat = (total area r = a (2 − 2)
of square – area grazed by sheep and square)
Therefore, the required area of the semicircle
= 54 − 6 (π + 3 3) = 3. 97 m2 1 1
= πr2 = π (a(2 − 2))2
Hence, choice (b) is the correct one. 2 2
1 2
Alternatively From the given information, you can find = πa (6 − 4 2)
that side of the square is 3 6 = 7.4 m but the length of rope 2
= πa2 (3 − 2 2) ≈ 0.172 πa2
=6m
If you look at the following figure you see Case II gives the highest possible value.
that at the corner of the field there are Hence choice (a) is the correct one.
two white (or un-grazed) regions.
Roughly, each such region looks like a 75 Consider the following diagram, in which O is the centre of
square. And the side of each such square the circle and OG is perpendicular to BC and EF is parallel
= 7. 35 − 6 = 1. 35 m. to BC.
Therefore area of both the un-grazed regions A E B

= 2 (1. 35) = 3. 65 ≈ 4
2 Q
2
So, the total un-grazed area is approx. 4 m
O G
NOTE The actual area at the corners would be slightly
more than the total area of the assumed squares at the two
corners. That’s why choices (a), (c) and (d) are invalid. 60°
D P F C
74 Given that each side of the square is a unit. Let r be the
radius of such semicircle. Then, we can proceed as follows. Since, PQ makes a 60° angle with DC, therefore
Case I: If the diameter of the semicircle ∠ QPC = ∠QOG = 60° and ∠PQC = ∠POF = 30°
coincides with one of the sides of the Since OP = OQ = OE = r therefore, using 30-60-90 degree
square, the maximum area of semicircle theorem (or trigonometric ratios) in ∆ OPF, we can find
2
1 1  a
= πr 2 = π   3
2 2  2 that PF = r / 2 and OF = r.
2
1
= (πa ) = 0.125πa2
2
3 r (2 + 3)
8 Therefore, AB = BC = EF = OE + OF = r + r=
2 2
Case II: If the diameter of the circle makes some angle
So, the area of the smallest possible square
with one of the sides of the square. 2
A  r (2 + 3)  7 + 4 3 2
B =  =  r
H  2   4 

r
Hence, choice (a) is the correct one.
E F 76 Look at the following figure (i), you will have AB = RB, as
r O r
r √2 both of them are radii of the same circle centered at B.
45° Similarly, AB = AR .
C G D Therefore, AR = BR = AB . That is
Let O is the centre of the semicircle. Since square is a ∠RAB = ∠RBA = ∠ARB = 60°
symmetric diagram, so both the sides of the square must be
526 QUANTUM CAT
Now, look at the figure (ii) and find all the important  10 2  100
⇒ a2 =   = = 100 (2 − 3) sq cm
angles.  3 + 1 2 + 3
D C D C
15° 150° 15°
Hence, choice (b) is the correct one.
R 75° 75°
75° 75° 77 Look at the following figure (i), you will have AB = RB, as
R
60° both of them are radii of the same circle centered at B
Similarly, AB = AR
Therefore, AR = BR = AB.

30°
That is ∠RAB = ∠RBA = ∠ARB = 60°
30°
D C D C
60° 60° 15° 150° 15°
R 75° 75°
A B A B 75° 75°
Fig (i) Fig (ii) R
60°
Now, with the help of fig (ii) find all the important angles
in fig (iii).
∠CDR = 15°, similarly ∠ DSA = 150°
30°
∴ ∠RDS = 90 − (15 + 15) = 60° 30°

But, since DR = DS 60° 60°

∴ ∠DSR = ∠ DRS = 60° A B A B
Fig (i) Fig (ii)
That is triangle DRS is an equilateral triangle.
That’s why all the four triangles ∆ DSR , ∆PBQ , ∆RCQ , ∆SAP Now, with the help of fig (ii) find all important angles in
in fig (iv) are equilateral ones. fig (iii).
Since, RS = RQ = PQ = SP so the quadrilateral SRQP is a ∠CDR = 15, similarly ∠ DSA = 15°
square. ∴ ∠ RDS = 90° − (15° + 15° ) = 60°
D C D C But, since DR = DS, ∴∠DSR = ∠ DRS = 60°
R R
That is triangle DRS is an equilateral triangle.
That’s why all the four triangles
S Q S Q ∆DSR , ∆ PBQ , ∆RCQ , ∆SAP in fig (iv) are equilateral ones.
Since, RS = RQ = PQ = SP , so the quadrilateral SRQP is a
square.
P P D C D C
A B A B
Fig (iii) Fig (iv) R R

Now, look at the fig (v) and fig (vi).

S Q S Q
Let us assume that each side of the square SRQP is a, then
SR = SD = DR = a and PQ = PB = QB = a
3 P P
∴ DM = BN = a A B A B
2 Fig. (iii) Fig. (iv)
D D C
R R
Now, look at the fig (v) and fig (vi).
M Let us assume that each side of the square SRQP is a, then
Q SR = SD = DR = a and PQ = PB = QB = a
S S Q
N 3
∴ DM = BN = a
2
P P But, DB = DM + MN + NB
B
A B
Fig (v) Fig (vi) 3 3
⇒ 10 2 = a+ a+ a
But, DB = DM + MN + NB 2 2
10 2 = ( 3 + 1) a
3 3
⇒ 10 2 = a + a+ a 10 2
2 2 ⇒ a=
( 3 + 1)
⇒ 10 2 = ( 3 + 1)a
10 2  10 2  100
⇒ a= ⇒ a2 =   = = 100 (2 − 3) sq cm
( 3 + 1)  3 + 1 2 + 3
Mensuration 527

D C D 25π
R R Therefore, area of the segment = − 25 sq cm
3
M
π 
= 25 − 1 sq cm
S Q S
Q 3 
N
Thus the total area of all the four segments around the
π  π 
P P inscribed square = 4 × 25  − 1 = 100  − 1 sq cm
A B
B 3  3 
Fig (v) Fig (vi)
The area of the common region by two quadrants in fig D C D C
R R
(vii) is determined as follows. Area of common region
= 2 × area of each quadrant – area of the square
1 
= 2 ×  π × 102 − 102 = 50π − 100 sq cm S Q S
4 
30°
Now, look at the fig (viii). The total area of the 6 shaded 30°
segments = Total area of the common region between the P 30°
two quadrants – Total area of the two equilateral triangles A B A B
Fig (i) Fig (ii)
and one square
 3  10 2 
2
 10 2  
2
= (50 π − 100) −  2 × ×  +    sq cm Now, using cosine rule you can obtain RS, the side of the
 4  3 + 1  3 + 1  inscribed square, as follows.
 
 3  BS 2 + BR 2 − RS 2
= (50 π − 100) −  × 100 (2 − 3) + 100 (2 − 3) sq cm cos 30° =
 2 
2 (BS )(BR )
 3  3 102 + 102 − RS 2
= (50 π − 100) − 100 (2 − 3)  + 1 sq cm =
2 2 2 × 10 × 10
 
= (50 π − 100) − (50) sq cm = 50 (π − 3) sq cm 200 − RS 2
=
200
D C D C
R 200 − RS 2
3=
100
Q RS 2 = 100 (2 − 3)
S
D C
Therefore, area of the square R
SRQP = SR 2 = 100 (2 − 3) sq cm
P
A B A B Thus, the required area
Fig (vii) Fig (viii) S Q
π 
4
Therefore, the area of 4 shaded segments = × 50 (π − 3) = 100  − 1 + 100 (2 − 3)
3 
6 P
sq cm π  A B
100 (π − 3) = 100  + 1 − 3 sq cm Fig (iii)
= sq cm 3 
3
D C Alternatively First of all, let us denote the different
Thus, the required area = area of the R regions of the square as shown below.
4 shaded segments around the
square SRQP + area of the square Now, consider the fig (ii).
SRQP inscribed in that region S Q The area of the unshaded region = Area of square – area of
100 (π − 3) the shaded quadrant x + 2y = 100 − 25π (i)
= + 100 (2 − 3) sq cm
3 Now, consider the fig (iv).
P
π  A
= 100 + 1 − 3 sq cm B D C D C
3  Fig (ix)
y y
x x x
Hence, choice (b) is the correct one.
Alternatively Look at the following figures: fig (i) and y z y y
fig (ii).
30° 25π
Area of the sector BSR = π × 102 × = sq cm x x
360° 3 y
1
Area of isosceles triangle BSR = × 102 sin 30° A B A B
2 Fig (i) Fig (ii)
= 25 sq cm
528 QUANTUM CAT
Area enclosed by the two overlapping In the fig (ii), we have GB = 30 − x and HA = GA = x,
quadrants = total areaof the two D C ∴ AB = GB − GA ⇒ 16 = 30 − 2x ⇒ x = 7 cm
quadrants – area of the square x
2x + z = 2 × 25π − 100 Now, from fig (i), we have EB = GC = FD = HA = 7 cm and
⇒ 2x + z = 50 π − 100 (ii) Z AE = BG = CF = DH = 23 cm
Now, consider the fig (iv). Now, in fig (i), if we draw a perpendicular line FJ on AB,
x
such that FJ = AD = 30
Area of the shaded region = area of A B
the equilateral triangle + 2(area of Fig. (iii) Since, AJ = DF = 7, therefore JE = AE − AJ = 23 − 7
the segments outside the equilateral = 16 cm
3
triangle) 2x + y + z = × 100 + 2 (area of the segments Therefore, in the right angle triangle FJE, we have
4
outside the equilateral triangle) FE = 302 + 162 = 34 cm
⇒ 2x + y + z = 25 3 + 2 (area of the segments outside the FE
But, since OF = OE = = 17 cm
equilateral triangle) 2
But the area of each segment = area of each sector with Similarly, OH = OG = 17 cm
central angle 60°− area of equilateral triangle Therefore, the side of the new larger square = 34 cm.
100π Hence, choice (a) is correct.
= − 25 3
6 79 Since, the ratio of area of polygon to that of rectangle is 16/25,
th
D C D C it means 9/25 or 36% area is lost due to overlapping.
P P B′

D E C

A B
A Fig. (iv) B A Fig. (v) B

Therefore, area of the shade region in fig (iv) is Therefore the lost area is equal to the area of triangle AEC.
 100 π  It implies that area of triangle AEC = 36% of area of
2x + y + z = 25 3 + 2  − 25 3
 6  rectangle ABCD.
100π That is area of ∆AED + ∆AEC + ∆CEB′ = 64% of the area of
⇒ 2x + y + z = − 25 3 …(iii)
3 rectangle ABCD
Now, solving the eqs. (ii) and (iii), we get Since area of ∆AED = ∆CEB′ therefore,
50 π
x = 100 − 25 3 − ∆ AED = CEB′ = 14%( ABCD )
3
Now, see there are two triangles AEC and ECB′ of same
Then, from the eq. (i), we get height.
25π
y= + 50 3 − 100 1
( AE × CB′ )
3 2 ∆AEC
∴ =
Again, from eq. (ii), we get 1
(EB′ × CB′ ) ∆ ECB′
 π  2
z = 100 + 1 − 3
 3  AE 36 18k
⇒ = =
78 Given that AB = BC = CD = DA = 30 cm. Due to EB′ 14 7 k
But, EC = EA = 18k and EB′ = 7 k
symmetrical cutting, EF and GH are perpendicular. Also,
AE = BG = CF = DH and EB = GC = FD = HA. Therefore, by applying Pythagoras theorem in ∆ ECB′, we
D F C get CB′ = (18k )2 − (7 k )2 = 5 11 k
O H G O
C And, AB′ = AE + EB′ = 18k + 7 k = 25k
G
Therefore, the required percentage
O F F
B 25k 5
E D = =
E
H 5 11k 11
A Hence, choice (a) is correct.
A E B O H G O
30 – x x
Fig. (i) Fig. (ii)