ME NSUR ATIO N 5

CHAPTER

1
Concept 1 : Area or area = × (diagonal)2
2
A figure made up of straight line segments is If diagonal = 2.5 cm
called a rectilinear figure. 1 6.25
area = × (2.5)2 cm2 = cm2 = 3.125 cm2.
2 2
 Area of Rectangle and Square
 Rectangle : Ex.2: The area of a square is 42.25 m2. Find the side
Area = length × breadth or A =  × b of the square. If tiles measuring 13 cm × 13 cm
area paved on the square area. find how many
Perimeter = 2 (length + breadth) or
such tiles are used for paving it.
P = 2( + b)
Sol. : The area of the square = 42.25 m2 = 422500 cm2
The side of the square = area = 422500 cm
= 650 cm
The area of 1 tile = 13 cm × 13 cm = 169 cm2
Number of tiles required = 422500 ÷ 169 = 2500

Ex.3: A room is 5 metres long. 4 metres broad and 3

 Square : metres high. Find the area of the four walls. Also
Area = (side)2 or A = s2 find the area of the ceiling and the area of the
Perimeter = 4 × side or P = 4s floor. If it costs Re 0.30 to whitewash 1 dm3 of
wall, find the cost of whitewashing the four walls
and the ceiling.
Sol. : Area of four walls = h + bh + h + bh = 2h(+b)
= 6 × 9 m2 = 54 m2
Area of ceiling = Area of floor = 20 m2

 EXAMPLES 
1
Ex.1: Show that area of a square = × (diagonal)2.
2
Find the area of a square whose diagonal = 2.5 cm. Since 1 m2 = 100 dm2,
Sol. : In right triangle BCD
 54 m2 = 5400 dm2 and
(diagonal)2 = DC2 + CB2 = s2 + s2 = 2s2
20 m2 = 2000 dm2
Cost of whitewashing the four walls at the rate
of Re 0.30 per dm2
= Rs (5400 × 0.30) = Rs 1620
Cost of whitewashing the ceiling at the rate of
Re 0.30 per dm2
= Rs (2000 × 0.30) = Rs 600
But area of square = s2 Total cost of white washing = Rs 1620 + Rs 600
 (diagonal)2 = 2 × area
= Rs 2220

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Ex.4: The length and breadth of a rectangular field is = ½ BD × AC
in the ratio 4 : 3. If the area is 3072 m2, find the cost i.e. Area of rhombus = ½ × product of diagonals
of fencing the field at the rate of Rs 4 per metre.
Sol. : Let the length and breadth of the field be 4x and  Area of a Trapezium :
3x metres respectively. The area of the field
Let ABCD be a trapezium with AB || DC. Draw AE
= 4x × 3x = 12x2 = 3072 m2
2
Hence x = 3072 ÷ 12 = 256 and BF perpendicular to DC.
Then AE = BF = height of trapezium = h
or x = 256 = 16
Area of trapezium ABCD = Area of ADE
Length = 4x = 64 m; Breadth = 3x = 48 m + Area of rectangle ABFE
Length of fencing = Perimeter of the field
+ Area of BCF
= 2 (64 + 48) m = 224 m
Cost of fencing at Rs 4 per meter
= Rs (224 × 4) = Rs 896

 Area of Quadrilaterals
 Area of a Parallelogram :

= ½ × DE × h + EF × h + ½ FC × h
= ½ h (DE + 2EF + FC)
= ½ h (DE + EF + FC + EF)
= ½ h (DC + AB) (since EF = AB)
Consider parallelogram ABCD. i.e. Area of trapezium = ½ × (sum of parallel sides)
Let AC be a diagonal × (distance between parallel sides)
In ADC and CBA
AD = CB, CD = AB  Area of a Quadrilateral :
AC is common Let ABCD be a quadrilateral, and AC be one of
 ADC  CBA its diagonals. Draw perpendiculars BE and DF
 Area of parallelogram ABCD from B and D respectively to AC.
= Area of ADC + Area of ABC
= 2 × Area of ADC
= 2 × (½ CD × AE) (where AE  DC)
= DC × AE
i.e. Area of parallelogram = base × height
 Area of a Rhombus :
Since a rhombus is also a prallelogram, its area
is given by Area of quadrilateral ABCD
Area of rhombus = base × height = Area of ABC + Area of ADC
The area of a rhombus can also be found if the = ½ AC × BE + ½ AC × DF
length of the diagonals are given. Let ABCD be = ½ AC (BE + DF)
a rhombus. We know that its diagonals AC and If AC = d, BE = h1 and DF = h2 then
BD bisect each other at right angles. Area of quadrilateral = ½d (h1 + h2)

 EXAMPLES 
Ex.1: A rectangle and a parallelogram have the same
area of 72 cm2. The breadth of the rectangle is
8 cm. The height of the parallelogram is 9 cm.
Find the base of the parallelogram and the length
of the rectangle.
Area of rhombus ABCD = area of ABD + area Sol. : Area of rectangle =  × b =  × 8 = 72
of CBD   = 9 cm
= ½ (BD × AO) + ½(BD × CO) Area of parallelogram = base × height
= base × 9 = 72
(since AO  BD and CO  BD)
 Base = 8 cm
= ½ BD (AO + CO)

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Ex.2: The area of a parallelogram is 64 cm2. Its sides Ex.5: In the trapezium PQRS, P = S = 90º, PQ = QR
are 16 cm and 5 cm. Find the two heights of the = 13 cm, PS = 12 cm and SR = 18 cm. Find the
parallelogram. area of the trapezium.
Sol. : (i) Area = base × height = 16 × h1 = 64 Sol. : The parallel sides are PQ and SR, and the
 h1 = 4cm distance between them is PS,
since P = S = 90º

(ii) Area = base × height = 5 × h2 = 64

 h2 = 12.8 cm
 Area = ½ × sum of parallel sides × heights
= ½ × (13 + 18) × 12 cm2
= 186 cm2

Ex.6: In trapezium ABCD, AB = AD = BC = 13 cm and

CD = 23 cm. Find the area of the trapezium.
Sol. : From B draw BE || AD, and BF  DC
Ex.3: The diagonals of a rhombus measure 10 cm and Since ABED is a parallelogram, DE, = 13 cm.
24 cm. Find its area. Also find the measure of its  EC = 23 cm – 13 cm = 10 cm
side. Also BE = 13 cm.
Sol. : AC = 10 cm, BD = 24 cm Therefore BEC is an isosceles triangle.
Area = ½ (d1 × d2) = ½ × 10 × 24 cm2
= 120 cm2

In ABO, AOB = 90º, AO = ½ AC = 5 cm,

BO = ½ BD = 12 cm. Since BF  EC, therefore F is the midpoint of EC
 AB2 = AO2 + OB2 = 25 + 144 = 169 = 13 × 13  FC = ½ × 10 cm = 5 cm
 AB = 13 cm In the right triangle BFC
 Measure of  side = 13 cm BF2 = BC2 – FC2 = 132 – 52 = 144
 BF = 12 cm
Ex.4: In rhombus ABCD, AB = 7.5 cm, and AC = 12 cm. Area of trapezium = ½ sum of parallel sides × height
Find the area of the rhombus. = ½ (13 + 23) × 12 cm2
Sol. : In ABO, AOB = 90º, AO = ½ AC = 6 cm, = 216 cm2
AB = 7.5 cm Note: We can also say : Area of ABCD = Area
of ABED + Area of BCE can be found by Hero’s
formula as all its sides are known.

Ex.7: In a quadrilateral ABCD, AC = 15 cm, The

perpendiculars drawn from B and D respectively
to AC measure 8.2 cm and 9.1 cm. Find the area
of the quadrilateral.
Sol. :

 OB2 = AB2 – OA2

= (7.5)2 – 62 = 56.25 – 36 = 20.25
 OB = 20.25 = 4.5 cm
 BD = 2 × OB = 9 cm
Area of rhombus =½ d1 ×d2 = ½ ×9×12 cm2 =54cm2

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 Area of quadrilateral = ½ d (h1 + h2) Area of trapezium PREF
= ½ × 15 × (8.2 + 9.1) cm2 = ½ × PR (PF + RE)
= ½ × 15 × 17.3 cm2 = ½ × 70 × 100 m2 = 3500 m2
= 129.75 cm2 Area of trapezium BQSC
Ex.8: PQRS is a trapezium, in which SR || PQ, and SR = ½ × QS (BQ + SC)
is 5 cm longer than PQ. If the area of the trapezium = ½ × 120 × 80 m2 = 4800 m2
is 186 cm2 and the height is 12 cm, find the Area of SCD = ½ × SD × SC
lengths of the parallel sides. = ½ × 70 × 50 m2 = 1750 m2
Sol. : Let PQ = x cm; then SR = (x + 5) Area of ERD = ½ × RD × ER
Area of PQRS = ½ × 12 × (x + x + 5) cm2
= ½ × 150 × 60 m2 = 4500 m2
= 186 cm2
Total area = (600 + 900 + 3500 + 4800
+ 1750 + 4500) m2
= 16050 m2

Points to Remember :
1. Area of rectangle = length × breadth
 6(2x + 5) = 186 2. Perimeter of rectangle = 2 (length + breadth)
or 2x + 5 = 31  x = 13 3. Area of parallelogram = base × height
 PQ = 13 cm, SR = 13 cm 5 cm = 18 cm 4. Area of rhombus = ½ × product of diagonals.
5. Area of trapezium = ½ × (sum of parallel
 Area of Irregular Rectilinear Figures sides) × (distance between parallel sides)
6. Area of quadrilateral = ½ × d (h1 + h2) where
For field ABCDEF, to find its area, we proceeds d is a diagonal and h1, h2 are the lengths of
as follows : perpendiculars from the remaining vertices
1. Select two farthest corners (A and D) such that on the diagonal.
the line joining them does not intersect any of
the sides. Join the corners. The line joining them Concept 2 : Volume and Surface Area
is called the base line. In this case the base line
is AD.  Introduction :
2. From each corner draw perpendiculars FP, BQ, Triangles, quadrilaterals, circles etc. lie in one
ER and CS to AD. These are called offsets. plane. They have two dimensions only—a length
3. Measure and record the following lengths: AP and a breadth. They are called “two dimensional”
and PF, AQ and QB, AR and RE, AS and SC. figures. Solids do not lie in one plane. They have
4. Record these measurements as shown. three dimensions—length, breadth and height.
D They occupy space. Solids are called “three
Metres dimensional” figures.
To D
S C
250
180 50 to C
60 to E 100
E R 60 30 to B
Q B 40 to F 30
P From A
F

A
The field has been divided into four right triangles
and two trapezia. In the trapezia, the parallel sides
are perpendicular to the base line.
The area of the field is the sum of the areas of
the triangles and trapezia.
Area of APF = ½ × AP × FP = ½ × 30 × 40 m2
= 600 m2
Area of AQB = ½ × AQ × QB = ½ × 60 × 30 m2
= 900 m2

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Formulae to calculate area of some geometrical figures :

S.No. Name Figure Perimeter in Area in

units of length square units

1. Rectangle 2(a + b) ab

a = length
b = breadth

a2
2. Square 4a 1
(diagonal) 2
2
a = side

3. Parallelogram 2(a + b) ah

a = side
b = side adjacent to a
h = distance between
the opp. parallel sides

a
d1
a d2 a 1
4. Rhombus 4a dd
2 1 2
a
a = side of rhombus;
d1d2 are the two
diagonals

1
5. Quadrilateral Sum of its (AC) (h1 + h2)
2
four sides
AC is one of its
diagonals and h1, h2
are the altitudes on
AC from D, B respectively.

1
6. Trapezium Sum of its h (a + b)
2
four sides
a, b, are parallel sides
and h is the distance
between parallel sides

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S.No. Name Figure Perimeter in Area in
units of length square units

1
7. Triangle a + b + c = 2s b ×h or
2
where s is the
b is the base and h is semi perimeter. s(s  a )(s  b)(s  c)
the altitude a, b, c are
three sides of .

1
8. Right triangle b +h+d bh
2

d(hypotenuse)
= b2  h 2

1
9. Equilateral 3a (i) ah
2
triangle
3 2
a = side (ii) a
4
3
h = altitude = a
2

c 4a 2  c 2
10. Isosceles 2a + c
4
triangle c = unequal side
a = equal side

1 2
11. Isosceles 2a + d a
2
right triangle
d(hypotenuse)
= a 2 , a = Each of
equal sides. The angles
are 90º, 45º, 45º.

12. Circle 2r r 2

r = radius of the circle

22
= or 3.1416
7

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 S.No. Name Figure Perimeter in Area in
units of length square units

1 2
13. Semicircle r + 2r r
2
r = radius of the circle

14. Ring (shaded ..... (R2 – r2)

region)
R = outer radius
r = inner radius


15. Sector of  + 2r where × r2
360
a circle

º = central angle of = × 2r
360
the sector, r = radius
of the sector  = length
of the arc

Volume of some solid figures :

S.No. Nature of the Shape of the Lateral/curved Total surface Volume Abbreviations
solid solid surface area area used

l  length
1. Cuboid 2h (l + b) 2(lb + bh + lh) lbh b  breadth
h  height

2. Cube 4a2 6a2 a3 a = length of edge

(perimeter of 2(area of one Area of

3. Right prism base)  end )  lateral base 
Height surface area height

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S.No. Nature of the Shape of the Lateral/curved Total surface Volume Abbreviations
solid solid surface area area used

r  radius of
Right base
4. circular 2rh 2r(r + h) r2h h  height of
cylinder the cylinder

1 1
(Perimeter Area of the (Area
Right 2 3
of the base)  base  lateral of base)
5.
pyramid surface area
(slant height ) height

Right l h  height
1 2
6. circular h r l r(l + r) r h r  radius
3
cone r l  slant
height

r 4 3
7. Sphere – 4r2 r r = radius
3

r
2 3
8. Hemi-sphere 2r2 3r2  r  r = radius
3 

R  outer
R
r radius
4
9. Spherical shell – 4(R2 – r2) (R3 – r3) r  inner
3
radius

R  l arg er
R h radius
3
10. Volume of h r  smaller
Bucket (R 2  r 2  Rr ) radius
r
h  height

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  EXAMPLES  Ex.5: A right circular cylinder has a height of 1 m and
a radius of 35 cm. Find its volume, area of curved
Ex.1: Find the volume and surface area of a cuboid of surface and total area.
 = 10 cm, b = 8 cm and h = 6 cm. Sol. : h = 1m, r = 35 cm = 0.35 m
Sol. : V =  × b × h = 10 cm × 8 cm × 6 cm = 480 cm3 22
Surface area = 2 (b + h + bh) Volume = r2h = × 0.35 × 0.35 × 1 m3
= 2(10 cm × 8 cm + 10 cm × 7
= 0.385 m3
6 cm + 8 cm × 6 cm)
= 2(80 + 60 + 48) cm2 = 376 cm2 22
Area of curved surface = 2rh = 2 × × 0.35 × 1m2
7
Ex.2: How many matchboxes of size 4 cm × 3 cm × 1.5 = 2.2 m2
cm can be packed in a cardbord box of size 30 cm Total surface area = 2r(h + r)
× 30 cm × 20 cm ? 22
Sol. : Volume of cardboard box = 30 cm × 30 cm × 20 cm =2× × 0.35 (1 + 0.35) m2
7
= 18000 cm3
Volume of each matchbox = 4 cm × 3 cm × 1.5 cm 2  22  0.35  1.35 2
= m = 2.97 m2
= 18 cm3 7
 Number of matchboxes that can fit in the
cardboard box Ex.6: An open cylindrical tank is of radius 2.8m and
= 18000 cm3 ÷ 18 cm3 = 1000 height 3.5m. What is the capacity of the tank ?
Sol. : Capacity = volume of cylinder
Ex.3: The dimensions of a cube are doubled. By how 22
many times will its volume and surface area = r2h = × 2.8 × 2.8 × 3.5 m3
7
increase ? = 86.24 m3
Sol. : Let the side of the original cube be s
Then side of the new cube = 2s Ex.7: A metal pipe 154 cm long, has an outer radius
equal to 5.5 cm and an inner radius of 4.5 cm.
what is the volume of metal used to make the
pipe ?
22
s Sol. Outer volume = r2h = × (5.5)2 × 154 cm3
7
(i) Volume of original cube = s × s × s
22
= s3 cubic units Inner volume = × (4.5)2 × 154 cm3
Volume of new cube = 2s × 2s × 2s 7
= 8s3 cubic units  Volume of metal = outer volume – inner volume
 Volume increases eight times if the side is 22 22
doubled. = × 154 × (5.5)2 – × 154 × (4.5)2
7 7
(ii) Surface area of original cube = 6s2 22
Surface area of new cube = 6(2s)2 = 24s2 = × 154 [(5.5)2 – (4.5)2]
= 4(6s2) 7
22
= × 154 (5.5 + 4.5) (5.5 – 4.5)
7
22
= × 154 × 10 × 1 = 4840 cm2
7
2s Ex.8: A cylindrical roller is used to level a rectangular
 Surface area increases four times. playground. The length of the roller is 3.5 m and
its diameter is 2.8 m. if the roller rolls over 200
Ex.4: The outer surface of a cube of edge 5m is painted. times to completely cover the playground, find
if the cost of painting is Re 1 per 100 cm2, find the area of the playground.
the total cost of painting the cube. Sol. : When the roller rolls over the ground once
Sol. : Surface area of cube = 6s2 = 6 × 5m × 5m = 150m2 completely, It covers a ground area equal to its
= 150 × 10000 cm2 curved surface area.
Cost of painting 100 cm2 is Re 1. Area of curved surface = 2rh
 Cost of painting 150 × 10000 cm2 is
22
1 =2× × 1.4 × 3.5 m2
Rs × 150 × 10000 7
100
= Rs 15,000 200  2  22  1.4  3.5 2
 Area of ground = m
7
= 6160 m 2

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Ex.9: A cylindrical pipe has an outer diameter of 1.4m Ex.10: Earth is dug out to a depth of 15 m from a
and an inner diameter of 1.12m. Its length is 10m. circular plot of land of radius 7 m. The earth is
It has to be painted on the outer and inner then spread out evenly on an adjacent
surfaces as well as on the rims at the top and rectangular plot of dimensions 16 m × 7 m. Find
bottom. If the rate of painting is 0.01 per cm2, the height of the earth on the rectangular plot.
find the cost of painting the pipe. 22
22 Sol. : Volumeof dug out earth = r2h = × 7 × 7 ×15m3
Sol. : Outer surface area = 2rh = 2 × × 0.7 × 10m2 7
7 = 2310m3
= 44m2 Let the height of the earth on the rectangular
22 plot be h
Inner surface area = 2rh= 2× ×0.56 ×10m2 Then volume of earth on the plot
7
= 35.2m2 =×b×h
= 16 × 7 × h m3
= 112h m3
Since volume of earth on the plot = volume of
dug out earth
 112 h = 2310
2310
or h= m = 20.625m
112

Ex.11: A rectangular piece of paper of width 20 cm and

length 44 cm is rolled along its width to form a
22 cylinder. What is the volume of the cylinder so
Area of two rims =2× × (0.72 – 0.562) formed ?
7
= 1.1088m2 Sol. : The length of the rectangle becomes the
 Total area to be painted circumference of the base of the cylinder.
= 44m2 + 35.2m2 + 1.1088m2  2r = 44, where r is the radius of the cylinder.
= 80.3088 m2 44  7
 r= = 7 cm
Rate of painting = Re 0.01 per cm2 2  22
The width of the rectangle becomes the height
= Re 0.01 × 10000 per m2
of the cylinder.
= Rs 100 per m2
22
 Total cost = Rs 80.3088 × 100  Volume = r2 h = × 7 × 7 × 20 cm3
= Rs 8030.88 7
= 3080 cm2

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 EXERCISE # 1

Q.1 One side of a rectangular field is 15 m and one Q.14 A room is half as long again as it is broad. The
of its diagonals is 17 m. Find the area of the field. cost of carpeting the room at Rs. 5 per sq. m is
Rs. 270 and the cost of papering the four walls
Q.2 A lawn is in the form of a rectangle having its at Rs. 10 per m2 is Rs. 1720. If a door and
sides in the ratio 2 : 3. the area of the lawn is 2 windows occupy 8 sq. m, find the dimensions
1 of the room.
hectares. Find the length and breadth of the
6
lawn. Q.15 Find the area of a triangle whose sides measure
13 cm, 14 cm and 15 cm.
Q.3 Find the cost of carpeting a room 13 m long and
9 m broad with a carpet 75 cm wide at the rate Q.16 Find the area of a right-angled triangle whose
of Rs. 12.40 per square metre. base is 12 cm and hypotenuse 13 cm.

Q.4 If the diagonal of a rectangle is 17 cm long and Q.17 The base of a triangular field is three times its
its perimeter is 46 cm, find the area of the altitude. If the cost of cultivating the field at
rectangle. Rs. 24.68 per hectare be Rs. 333.18, find its base
and height.
Q.5 The length of a rectangle is twice its breadth. If
its length is decreased by 5 cm and breadth is Q.18 The altitude drawn to the base of an isosceles
increasded by 5 cm, the area of the rectangle is triangle is 8 cm and the perimeter is 32 cm. Find
increased by 75 sq. cm. Find the length of the the area of the triangle.
rectangle.
Q.19 Find the length of the altitude of an equilateral
Q.6 In measuring the sides of a rectangle, one side triangle of side 3 3 cm.
is taken 5% in excess, and the other 4% in deficit.
Find the error percent in the area calculated from Q.20 In two triangles, the ratio of the areas is 4 : 3 and
these measurements. the ratio of their heights is 3 : 4. Find the ratio
of their bases.
Q.7 A rectangular grassy plot 110 m by 65 m has a
gravel path 2.5 m wide all round it on the inside. Q.21 The base of a parallelogram is twice its height.
Find the cost of greavelling the path at 80 paise If the area of the parallelogram is 72 sq. cm, find
per sq. metre. its height.

Q.8 The perimeters of two squares are 40 cm and Q.22 Find the area of a rhombus one side of which
32 cm. Find the perimeter of a third square whose measures 20 cm and one diagonal 24 cm.
area is equal to the difference of the areas of the
two squares. Q.23 The difference between two parallel sides of a
trapezium is 4 cm. The perpendicular distance
Q.9 A room 5 m 55 cm long and 3m 74 cm broad is between them is 19 cm. If the area of the
to be paved with square tiles. Find the least trapezium is 475 cm2, find the lengths of the
number of square tiles required to cover the floor. parallel sides.

Q.10 Find the area of a square, one of whose diagonals Q.24 Find the length of arope by which a cow must be
is 3.8 m long. tethered in order that it may be able to graze an
area of 9856 sq. metres.
Q.11 The diagonals of two squares are in the ratio of
2 : 5. Find the ratio of their areas. Q.25 The area of a circular field is 13.86 hectares. Find
the cost of fencing it at the rate of Rs. 4.40 per
Q.12 If each side of a square is increased by 25%, find metre.
the percentage change in its area.
Q.26 The diameter of the driving wheel of a bus is
Q.13 If the length of a certain rectangle is decreased 140 cm. How many revolutions per minute must
by 4 cm and the width is increased by 3 cm, a the wheel make in order to keep a speed of
square with the same area as the original 66 kmph ?
rectangle would result. Find the perimeter of the
original rectangle.

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Q.27 A wheel makes 1000 revolutions in covering a Q.42 A cube of edge 15 cm is immersed completely in
distance of 88 km. Find the radius of the wheel. a rectangular vessel containing water. If the
dimensions of the base of vessel are 20 cm × 15 cm,
Q.28 The inner circumference of a circular race trak, find the rise in water level.
14 m wide, is 440 m. Find the radius of the outer
circle. Q.43 Three solid cubes of sides 1 cm, 6 cm and 8 cm
are melted to form anew cube. Find the surface
Q.29 Two concentric circles form a ring. The inner area of the cube so formed.
and outer circumferences of the ring are
2 3 Q.44 If each edge of a cube is increased by 50%, find
50 m and 75 m respectively. Find the width
7 7 the percentage increase in its surface area.
of the ring.
Q.45 Two cubes have their volumes in the ratio 1 : 27.
Q.30 A sector of 120º, cut out from a circle, has an Find the ratio of their surface areas.
3
area of 9 sq. cm. Find the radius of the cirlce.
7 Q.46 Find the volume, curved surface area and the
total surface area of a cylinder with diameter of
Q.31 Find the ratio of the areas of the incircle and base 7 cm and height 40 cm.
circumcircle of a square.
Q.47 If the capacity of a cylindrical tank is 1848 m3
Q.32 If the radius of a circle is decreased by 50%, find
and the diameter of its base is 14 m, then find the
the percentage decrease in its area.
depth of the tank.
Q.33 Find the volume and surface area of a cuboid
16 m long, 14 m broad and 7 m high. Q.48 2.2 cubic dm of lead is to be drawn into a
cylindrical wire 0.50 cm in diameter. Find the
Q.34 Find the length of the longest pole that can be length of the wire in metres.
placed in a room 12 m long, 8 m broad and
9 m high. Q.49 How many iron rods, each of length 7 m and
diameter 2 cm can be made out of 0.88 cubic
Q.35 The volume of a wall, 5 times as high as it is metre of iron ?
borad and 8 times as long as it is high, is
12.8 cu. metres. Find the breadth of the wall. Q.50 The radii of two cylinders are in the ratio 3 : 5
and their heights are in the ratio of 2 : 3. Find the
Q.36 Find the number of bricks, each measuring ratio of their curved surface areas.
24 cm × 12 cm × 8 cm, required to construct a
wall 24 m long, 8 m high and 60 cm thick, if
10% of the wall is filled with mortar ? Q.51 If 1 cubic cm of cast iron weighs 21 gms, then
find the weight of a cast iron pipe of length
Q.37 Water flows into a tank 200 m × 150 m through 1 metre with a bore of 3 cm and in which
a rectangular pipe 1.5 m × 1.25 m @ kmph. In thickness of the metal is 1 cm.
what time (in minutes) will the water rise by
2 metres ? Q.52 Find the slant height, volume, curved surface
area and the whole surface area of a cone of
Q.38 The dimensions of an open box are 50 cm, 40 cm radius 21 cm and height 28 cm.
and 23 cm. Its thickness is 3 cm. If 1 cubic cm
of metal used in the box weighs 0.5 gms, find the Q.53 Find the length of canvas 1.25 m wide required
weight of the box. to build a conical tent of base radius 7 metres
and height 24 metres.
Q.39 The diagonal of a cube is 6 3 cm. Find its
volume and surface area. Q.54 The heights of two right circular cones are in the
ratio 1 : 2 and the permieters of their bases are
Q.40 The surface area of a cube is 1734 sq. cm. Find in the ratio 3 : 4. Find the ratio of their volumes.
its volume.
Q.55 The radii of the bases of a cylinder and a cone
Q.41 A rectangular block 6 cm by 12 cm by 15 cm is are in the ratio of 3 : 4 and their heights are in
cut up into an exact number of eqaul cubes. Find the ratio 2 : 3. Find the ratio of their volumes.
the least possible number of cubes.

manishkumarphysics.in
Q.56 A conical vessel, whose internal radius is 12 cm Q.62 Two metallic right circular cones having their
and height 50 cm, is full of liquid. The contents heights 4.1 cm and 4.3 cm and the radii of their
are emptied into a cylindrical vessel with internal bases 2.1 cm each, have been melted together
radius 10 cm. Find the height to which the liquid and recast into a sphere. Find the diameter of the
rises in the cylindrical vessel. sphere.

Q.57 Find the volume and surface area of a sphere of Q.63 A cone and a sphere have equal radii and equal
radius 10.5 cm. volumes. Find the ratio of the diameter of the
sphere to the height of the cone.
Q.58 If the radius of a sphere is increassed by 50%,
find the increase percent in volume and the Q.64 Find the volume, curved surface area and the
increase percent in the surface area. total surface area of a hemisphere of radius 10.5
cm.
Q.59 Find the number of lead balls, each 1 cm in
diameter that can be made from a sphere of Q.65 TA hemispherical bowl of internal radius 9 cm
diameter 12 cm. contains a liquid. This liquid is to be filled into
cylindrical shaped small bottles of diameter 3 cm
Q.60 How many spherical bullets can be made out of and height 4 cm. How may bottles will be needed
a lead cylinder 28 cm high and with base radius to empty the bowl ?
6 cm, each bullet being 1.5 cm in diameter ?
Q.66 A cone, a hemisphere and a cylinder stand on
Q.61 A copper sphere of diameter 18 cm is drawn into equal bases and have the same height. Find the
a wire of diameter 4 mm. Find the length of the ratio of their volumes.
wire.

ANSWERS

1. 120 m2 2. 50 m. 3. 1934.4 4. 120 cm2

5. 20 cm 6. 0.8% 7. 680 8. 24 cm

9. 176 10. 7.22 m2 11. 4 : 5 12. 56.25%

13. 50 cm 14. L = 9, B = 6, H = 6 15. 84 cm2 16. 30 cm2

17. B = 900 m, H = 300 m. 18. 60 cm2 19. 4.5 cm 20. 16 : 9

21. 6 cm 22. 384 cm2 23. 27 cm, 23 cm 24. 56 m

25. 5808 26. 250 27. 14 m 28. 84 m

29. 4 m 30. 3 cm 31. 1 : 2 32. 75%

33. 868 m2 34. 17 m 35. 40 cm 36. 45000

37. 96 min. 38. 8.04 kg. 39. 216 cm3, 216 cm2 40. 4913 cm3

41. 40 42. 11.25 cm 43. 486 cm2 44. 125%

45. 1 : 9 46. 1540 cm3, 880 cm2, 957 cm2 47. 12 m 48. 112 m

49. 400 50. 2 : 5 51. 26.4 kg 52. 12936 cm3, 2310 cm2, 3696 cm2
53. 440 m 54. 9 : 32 55. 9 : 8 56. 24 cm

57. 4581 cm3, 1386 cm2 58. 237.5%, 125% 59. 1728 60. 1792

61. 243 m 62. 4.2 cm 63. 1 : 2

64. 2425.5 cm3, 693 cm2, 1039.5 cm2 65. 54 66. 1 : 2 : 3

manishkumarphysics.in
 EXERCISE # 2

Q.1 The length of a room is 5.5 m and width is 3.75 Q.9 The ratio between the length and the perimeter
m. Find the cost of paving the floor by slabs at of a rectangular plot is 1 : 3. What is the ratio
the rate of Rs 800 per sq. metre. between the length and breadth of the plot ?
(A) Rs. 15,000 (B) Rs. 15,550 (A) 1 : 2 (B) 2 : 1
(C) Rs. 15600 (D) Rs. 16,500 (C) 3 : 2 (D) Data inadequate

Q.2 The length of a rectangle is 18 cm and its breadth Q.10 The ratio between the length and the breadth of
is 10 cm. When the length is increased to 25 cm, a rectangular park is 3 : 2. If a man cycling along
what will be the breadth of the rectangle if the the boundary of the park at the speed of 12 km/
area remains the same? hr completes one round in 8 minutes, then the
(A) 7cm (B) 7.1 cm area of the park (in sq.) is:
(C) 7.2 cm (D) 7.3 cm (A) 15360 (B) 153600
(C) 30720 (D) 307200
Q.3 A rectangular plot measuring 90 meters by 50
meters is to be enclosed by wire fencing. If the Q.11 The length of a rectangular hall is 5m more than
poles of the fence are kept 5 metres apart, how its breadth. The area of the hall is 750 m2. The
many poles will be needed? length of the hall is:
(A) 55 (B) 56 (A) 15 m (B) 22.5 m
(C) 57 (D) 58 (C) 25 m (D) 30 m

Q.4 A length of a rectangular plot is 60% more than Q.12 The area of a rectangle is 460 square metres. If
its breadth. If the difference between the length the length is 15% more than the breadth, what is
and the breadth of that rectangle is 24 cm, what the breadth of the rectangular field?
is the area of that rectangle? (A) 15 metres
(A) 2400 sq. cm (B) 2480 sq. cm (B) 26 metres
(C) 2560 sq. cm (D) Data inadequate (C) Cannot be determined
(E) None of these (D) None of these

Q.5 A rectangular parking space is marked out by Q.13 A rectangular field is to be fenced on three sides
painting three of its sides. If the length of the leaving a side of 20 feet uncovered. If the area
unpainted side is 9 feet, and the sum of the of the field is 680 sq. feet. How many feet of
lengths of the painted sides is 37 feet, then what fencing will be required?
is the area of the parking space in square feet? (A) 34 (B) 40
(A) 46 (B) 81 (C) 68 (D) 88
(C) 126 (D) 252
Q.14 The ratio between the perimeter and the breadth
Q.6 The difference between the length and breadth of a rectangular is 5 : 1. If the area of the rectangle
of a rectangle is 23m. If its perimeter is 206 m is 216 sq. cm, what is the length of the rectangle?
then its area is: (A) 16 cm (B) 18 cm
(A) 1520 m2 (B) 2420 m2 (C) 24 cm (D) Data inadequate
(C) 2480 m 2 (D) 2520 m2 (E) None of these

Q.7 The length of a rectangular plot is 20 metres Q.15 A Farmer wishes to start a 100 sq. m rectangular
more than its breadth. If the cost of fencing the vegetable garden. Since he has only 30 m barbed
plot @ Rs. 26.50 per metre is Rs. 5300, what is wire, he fences three sides of the garden letting
the length of the plot in meters? his house compound wall act as the fourth side
(A) 40 (B) 50 fencing. The dimension of the garden is:
(C) data inadquete (D) none of these (A) 15 m × 6.67 m (B) 20 m × 5 m
(C) 30 m × 3.33 m (D) 40 m × 2.5 m
Q.8 The breadth of a rectangular field is 60% of its
length. If the perimeter of the field is 800 m., what Q.16 The sides of a rectangular field are in the ratio
is the area of the field? 3 : 4. If The area of the field is 7500 sq. m, the
(A) 18750 sq. m (B) 37500sq. m cost of fencing the field @ 25 paise per metre is:
(C) 40000 sq. m (D) 48000 sq. m (A) Rs. 55.50 (B) Rs. 67.50
(C) Rs. 86.50 (D) Rs. 87.50

manishkumarphysics.in
Q.17 A rectangle of certain dimensions is chopped off Q.25 The number of revolutions a wheel of diameter
from one corner of a larger rectangle as shown. 40 cm makes in travelling a distance of 176 m, is:
AB = 8 cm and BC = 4 cm. The perimeter of the (A) 140 (B) 150
figure ABCPQRA (in cm) is: (C) 160 (D) 166

Q.26 The radius of a wheel is 0.25 m. The number of

revolutions it will make to travel a distance of 11
km will be:
(A) 2800 (B) 4000
(C) 5500 (D) 7000

(A) 24 (B) 28
(C) 36 (D) 48 1
Q.27 The wheel of an engine, 7 meters in
2
Q.18 A large field of 700 hectares is divided into two circumference makes 7 revolutions in 9 seconds.
parts. The difference of the areas of the two The speed of the train in km per hour is:
parts is on-fifth of the average of the two areas. (A) 130 (B) 132
What is the area of the smaller part in hectares? (C) 135 (D) 150
(A) 225 (B) 280
(C) 300 (D) 315 Q.28 The wheel of a motorcycle, 70 cm in diameter,
makes 40 revolutions in every 10 seconds. What
Q.19 A rectangular paper, when folded into two is the speed of the motorcycle in km/hr?
congruent parts had a perimeter of 34 cm for (A) 22.32 (B) 27.68
each part folded along one set of sides and the (C) 31.68 (D) 36.24
same is 38 cm when folded along the other set
of sides. What is the area of the paper? Q.29 Wheels of diameters 7 cm and 14 cm start rolling
(A) 140 cm2 (B) 240 cm2 simultaneously from X and Y, which are 1980 cm
(C) 560 cm 2 (D) None of these apart, towards each other in opposite directions.
Both of them make the same number of
Q.20 A rectangular plot is half as long again as it is revolutions per second. If both of them meet
after 10 seconds, the speed of the smaller wheel
2 is:
broad and its area is hectares. Then, its length
3 (A) 22 cm/sec (B) 44 cm/sec
is: (C) 66 cm/sec (D) 132 cm/sec
(A) 100 m (B) 33.33 m
Q.30 A toothed wheel of diameter 50 cm is attached to
100 3
(C) 66.66 m (D) a smaller wheel of diameter 30 cm. How many
3 revolutions will the smaller wheel make when the
larger one makes 15 revolutions?
Q.21 The areas of two circular fields are in the ratio (A) 18 (B) 20
16 : 49. If the radius of the latter is 14 m, then (C) 25 (D) 30
what is the radius of the former:
(A) 4 m (B) 8 m Q.31 Find the diameter of a wheel that makes 113
(C) 18 m (D) 32 m revolutions to go 2 km 26 decametres.
4 4
Q.22 If the ratio of areas of two circles is 4 : 9, then (A) 4 m (B) 6 m
the ratio of their circumferences will be: 13 11
(A) 2 : 3 (B) 3 : 2 4 8
(C) 4 : 9 (D) 9 : 4 (C) 12 m (D) 12 m
11 11
Q.23 The perimeter of a circle is equal to the perimeter
of a square. Then, their areas are in the ratio: Q.32 The font wheels of a wagon are 2 feet in
(A) 4 : 1 (B) 11 : 7 circumference and the rear wheels are 3 feet in
(C) 14 : 11 (D) 22 : 7 circumference. When the front wheels have made
10 more revolutions than the rear wheels, how
Q.24 The diameter of a wheel is 1.26 m. How far will many feet has the wagon travelled?
it travel in 500 revolutions? (A) 30 (B) 60
(A) 1492 m (B) 1980 m (C) 90 (D) 150 
(C) 2530 m (D) 2880 m

manishkumarphysics.in
Q.33 A circular ground whose diameter is 35 meters, Q.43 The area of a sector of a circle of radius 5 cm,
has a 1.4 m broad garden around it. What is the formed by an arc of length 3.5 cm, is :
area of the garden in square metres? (A) 7.5 cm2 (B) 7.75 cm2
(A) 160.16 (B) 176.16 (C) 8.5 cm2 (D) 8.75 cm2
(C) 196.16 (D) Data inadequate
Q.44 In a circle of radius 7 cm, an arc subtends an
Q.34 A circular garden has a circumference of 440 m. angle of 108° at the centre. The area of the sector
These is a 7 m wide border inside the garden is:
along its periphery. The area of the border is: (A) 43.2 cm2 (B) 44.2 cm2
(A) 2918 m2 (B) 2921 m2 (C) 45.2 cm 2 (D) 46.2 cm2
(C) 2924 m 2 (D) 2926 m2
Q.45 The area of the greatest circle which can be
Q.35 The area of two concentric circles forming a ring inscribed in a square whose perimeter is 120 cm,
are 154 sq. cm and 616 sq. cm. The breadth of is:
the ring is: 2 2
(A) 7 cm (B) 14 cm 22  7  22  9 
(A)    cm2 (B)    cm2
(C) 21 cm (D) 28 cm 7 2 7 2
2
Q.36 A circular park has a path of uniform width 22  15 
 152 cm2
22
around it. The difference between outer and inner (C)    cm2 (D)
7  2 7
circumferences of the circular path is 132 m. Its
width is:
(A) 20 m (B) 21 m Q.46 The area of the largest circle, that can be drawn
(C) 22 m (D) 24 m inside a rectangle with sides 18 cm by 14 cm, is:
(A) 49 cm2 (B) 154 cm2
(C) 378 cm 2 (D) 1078 cm2
Q.37 A circular swimming pool is surrounded by a
concrete wall 4 ft. wide. If the area of the concrete
Q.47 The area of a circle is 220 sq. cm. The area of a
11 square inscribed in this circle will be:
wall surrounding the pool is that of the pool,
25 (A) 49 cm2 (B) 70 cm2
then the radius of the pool is: (C) 140 cm2 (D) 150 cm2
(A) 8 ft (B) 16 ft
(C) 20 ft (D) 30 ft Q.48 A square is inscribed in a circle whose radius is
4 cm. The area of the portion between the circle
Q.38 The ratio of the outer and the inner perimeters of and the square is:
a circular path is 23 : 22. If the path is 5 meters (A) (8 – 16) (B) (8 – 32)
wide, The diameter of the inner circle is: (C) (16 – 16) (D) (16 – 32)
(A) 55 m (B) 110 m
(C) 220 m (D) 230 m Q.49 The circumference of a circle is 100cm. The side
of a square inscribed in the circle is:
Q.39 What will be the area of a semi-circle of 14 m 100
diameter? (A) 50 2 cm (B) cm
(A) 22 m2 (B) 77 m2 
(C) 154 m 2 (D) 308 m2 50 2 100 2
(C) cm (D) cm
 
Q.40 A semi-circular shaped window has diameter of
63 cm. Its perimeter equals:
Q.50 Four equal sized maximum circular plates are cut
(A) 126 cm (B) 162 cm
off from a square paper sheet of area 784 cm2.
(C) 198 cm (D) 251 cm
The circumference of each plate is:
Q.41 What will be the area of a semi-circle whose (A) 22 cm (B) 44 cm
perimeter is 36 cm? (C) 66 cm (D) 88 cm
(A) 154 cm2 (B) 168 cm2
(C) 308 cm 2 (D) None of these Q.51 There are 4 semi-circular gardens on each side of
a square-shaped pond with each side 21m. The
Q.42 If a wire is bent into the shape of a square, then cost of fencing the entire plot at the rate of Rs
the area of the square is 81 sq. cm. When the 12.50 per metre is:
wire is bent into a semi-circular shape, then the (A) Rs. 1560 (B) Rs. 1650
area of the semi-circle will be: (C) Rs. 3120 (D) Rs. 3300
(A) 22 cm2 (B) 44cm2
(C) 77cm2 (D) 154 cm2

manishkumarphysics.in
Q.52 The ratio of the area of the incircle and Q.61 The area of the largest triangle that can be
circumcircle of an equilateral triangle is: inscribed in a semi-circle of radius r, is:
(A) 1 : 2 (B) 1 : 3 (A) r2 (B) 2r2
(C) 1 : 4 (D) 1 : 9 (C) r3 (D) 2r3

Q.53 The radius of the circumcircle of an equilateral Q.62 ABC is a right-angled triangle with right at B. If
triangle of side 12 cm is: the semi-circle on AB with AB as diameter
4 2 encloses an area of 81 sq. cm and the semicircle
(A) cm (B) 4 2 cm on BC with BC as diameter encloses an area of
3 36 sq. cm, then the area of the semi-circle on AC
4 3 with AC as diameter will be:
(C) cm (D) 4 3 cm (A) 117 cm2 (B) 121 cm2
3 2
(C) 217 cm (D) 221 cm2
Q.54 The area of the incircle of an equilateral triangle
of side 42 cm is: Q.63 If the radius of a circle is increased by 75%, then
its circumference will increase by:
(A) 22 3 cm2 (B) 231 cm2 (A) 25% (B) 50%
(C) 462 cm2 (D) 924 cm2 (C) 75% (D) 100%

Q.55 The area of a circle inscribed in an equilateral Q.64 A can go round a circular path 8 times in 40
triangle is 154 cm2 . Find the perimeter of the minutes. If the diameter of the circle is increased
triangle. to 10 times the original diameter, then the time
(A) 71.5 cm (B) 71.7 cm required by A to go round the new path once,
(C) 72.3 cm (D) 72.7 cm travelling at the same speed as before, is:
(A) 20 min. (B) 25 min.
Q.56 The sides of a triangle are 6 cm, 11 cm and 15 cm. (C) 50 min. (D) 100 min.
The radius of its incircle is:
Q.65 If the radius of a circle is increased by 6%, Then
4 2
(A) 3 2 cm (B) cm the area is increased by:
5 (A) 6% (B) 12%
5 2 (C) 12.36% (D) 16.64%
(C) cm (D) 6 2 cm
4
Q.66 The capacity of a tank of dimensions
Q.57 The perimeter of a triangle is 30 cm and the (8m × 6m × 2.5m) is:
circumference of its incircle is 88cm. The area of (A) 120 litres (B) 1200 litres
the triangle is: (C) 12000 litres (D) 120000 litres
(A) 70 cm2 (B) 140 cm2
(C) 210 cm2 (D) 420 cm2 Q.67 Find the surface area of a 10cm × 4cm × 3cm
brick.
Q.58 If in a triangle, the area is numerically equal to (A) 84 sq. cm (B) 124 sq. cm
the perimeter, then the radius of the inscribed (C) 164 sq. cm (D) 180 sq. cm
circle of the triangle is:
(A) 1 (B) 1.5 Q.68 A cistern 6m long and 4 m wide contains water
(C) 2 (D) 3 up to a depth of 1m 25 cm. The total area of the
wet surface is:
Q.59 An equilateral triangle, a square and a circle have (A) 49 m2 (B) 50 m2
(C) 53.5 m2 (D) 55 m2
equal perimeters. If T denotes the are of the
triangle , S the area of the square and C, the area
of the circle, then: Q.69 A boat having a length 3 m and breadth 2 m is
(A) S < T < C (B) T < C < S floating on a lake. The boat sinks by 1cm when
(C) T < S < C (D) C < S < T a man gets on it. The mass of man is:
(A) 12 kg (B) 60 kg
Q.60 If an area enclosed by a circle or a square or an (C) 72 kg (D) 96 kg
equilateral triangle is the same, then the maximum
perimeter is possessed by: Q.70 The area of the base of a rectangular tank is
(A) circle 6500 cm2 and the volume of water contained in it
(B) square is 2.6 cubic metres. The depth of water in the
(C) equilateral triangle tank is:
(D) triangle and square have equal perimeters (A) 3.5 m (B) 4 m
greater than of circle (C) 5 m (D) 6 m

manishkumarphysics.in
Q.71 Given that I cu. cm of marble weighs 25 gms, the Q.80 The number of bricks, each measuring
weight of a marble block 28 cm in width and 5 cm 25cm × 12.5cm × 7.5cm, required to construct a
thick is 112 kg. The length of the block is: wall 6 m long, 5m high and 0.5 m thick, while the
(A) 26.5 cm (B) 32 cm mortar occupies 5% of the volume of the wall, is:
(C) 36 cm (D) 37.5 cm (A) 3040 (B) 5740
(C) 6080 (D) 8120
Q.72 Half cubic metre of gold sheet is extended by
hammering so as to cover an area of 1 hectare. Q.81 50 men took a dip in a water tank 40 m long and
The thickness of the sheet is: 20 m broad on a religious day. If the average
(A) 0.0005 cm (B) 0.005 cm displacement of water by a man is 4 m3 , then the
(C) 0.05 cm (D) 0.5 cm rise in the water level in the tank will be:
(A) 20 cm (B) 25 cm
(C) 35 cm (D) 50 cm
Q.73 In a shower, 5 cm of rain falls. The volume of
water that falls on 1.5 hectares of ground is:
Q.82 A tank 4 m long, 2.5 m wide and 1.5 m deep is
(A) 75 cu. m (B) 750 cu. m dug in a field 31 m long and 10 m wide. If the
(C) 7500 cu. m (D) 75000 cu. m earth dug out is evenly spread out over the field,
the rise in level of the field is:
Q.74 The height of a wall is six times its width and the (A) 3.1 cm (B) 4.8 cm
length of the wall is seven times its height. If (C) 5 cm (D) 6.2 cm
volume of the wall be 16128 cu. m, its width is:
(A) 4 m (B) 4.5 m Q.83 A river 1.5 m deep and 36 m wide is flowing at
(C) 5 m (D) 6 m the rate of 3.5 km per hour. The amount of water
that runs into the sea per minute (in cubic metres)
Q.75 The volume of a rectangular block of stone is is:
10368 dm3. Its dimensions are in the ratio of (A) 3150 (B) 31500
3 : 2 : 1. If its entire surface is polished at 2 paise (C) 6300 (D) 63000
per dm2, then the total cost will be:
(A) Rs. 31.50 (B) Rs. 31.68 Q.84 A rectangular water tank is 80m × 40m. Water
(C) Rs. 63 (D) Rs. 63.36 flows into it through a pipe 40 sq. cm at the
opening at a speed of 10 km/hr. By how much,
Q.76 The edges of a cuboid are in the ratio 1 : 2 : 3 the water level will rise in the tank in half an
and its surface area in 88 cm2. The volume of the hour?
cuboid is: 3 4
(A) 24 cm3 (B) 48 cm3 (A) cm (B) cm
2 9
(C) 64 cm3 (D) 120 cm3 5
(C) cm (D) none of these
8
Q.77 The maximum length of a pencil that can be kept
in a rectangular box of dimensions 8 cm × 6 cm Q.85 A hall is 15 m long and 12 m broad. If the sum
× 2 cm, is: of the areas of the floor and the ceiling is equal
(A) 2 3 cm (B) 2 14 cm to the sum of areas of four wall the volume of the
hall is:
(C) 2 26 cm (D) 10 2 cm (A) 720 (B) 900
(C) 1200 (D) 1800
Q.78 Find the length of the longest rod that can be
Q.86 A closed metallic cylindrical box is 1.25 m high
2 and its base radius is 35 cm. If the sheet metal
placed in a room 16 m long, 12 m broad and 10
3 costs Rs 80 per m2, the cost of the material used
m high. in the box is:
1 2 (A) Rs. 281.60 (B) Rs. 290
(A) 22 m (B) 22 m (C) Rs. 340.50 (D) Rs. 500
3 3
(C) 23 m (D) 68 m
Q.87 The curved surface area of a right circular
cylinder of base radius r is obtained by
Q.79 How many bricks, each measuring
multiplying its volume by:
25cm × 11.25cm × 6cm, will be needed to build a
wall 8m × 6m × 22.5cm? 2
(A) 2r (B)
(A) 5600 (B) 6000 r
(C) 6400 (D) 7200 2
(C) 2r2 (D) 2
r

manishkumarphysics.in
Q.88 The ratio of total surface area to lateral surface Q.96 Two cones have their height in the ratio of 1 : 3
area of a cylinder whose radius is 20cm and and radii 3 : 1 . The ratio of their volumes is:
height 60cm, is: (A) 1 : 1 (B) 1 : 3
(A) 2 : 1 (B) 3 : 2 (C) 3 : 1 (D) 2 : 3
(C) 4 : 3 (D) 5 : 3
Q.97 The radii of two cones are in the ratio 2 : 1, their
Q.89 A powder tin has a square base with side 8 cm volumes are equal. Find the ratio of their heights.
and height 14 cm another tin has circular base of (A) 1 : 8 (B) 1 : 4
with diameter 8 cm and height 14 cm. The (C) 2 : 1 (D) 4 : 1
difference in their capacities is:
Q.98 If the volumes of two cones are in the ratio of
(A) 0 (B) 132 cm3 1 : 4 and their diameters are in the ratio of 4 :
(C) 137.1 cm3 (D) 192 cm3 5, then the ratio of their heights is:
(A) 1 : 5 (B) 5 : 4
Q.90 The ratio between the radius of the base and the (C) 5 : 16 (D) 25 : 64
height of a cylinder is 2 : 3. If its volume is 12936
cu. cm, the total surface area of the cylinder is:
Q.99 The volume of the largest right circular cone that
(A) 2587. cm2 (B) 3080 cm2 can be cut out of a cube of edge 7 cm is:
(C) 25872 cm 2 (D) 38808 cm2 (A) 13.6 cm3 (B) 89.8 cm3
(C) 121 cm3 (D) 147.68 cm3
Q.91 The radius of the cylinder is half its height and
area of the inner part is 616 sq. cms.
Approximately how many litres of milk can it Q.100 A cone of height 7 cm and base radius 3 cm is
contain? carved from a rectangular block of wood 10cm ×
5cm × 2cm. The percentage of wood wasted is:
(A) 1.4 (B) 1.5
(A) 34% (B) 46%
(C) 1.7 (D) 1.9
(C) 54% (D) 66%
Q.92 The sum of the radius of the base and the height
of a solid cylinder is 37 metres. If the total surface Q.101 A right circular cone and a right circular cylinder
area of the cylinder be 1628 sq. metres, its volume have equal base and equal height. If the radius
is: of the base and the height are in the ratio 5 : 20,
then the ratio of the total surface area of the
(A) 3180 m3 (B) 4620 m3 cylinder to that of the cone is:
(C) 5240 m 3 (D) none of these (A) 3 : 1 (B) 13 : 9
(C) 17 : 9 (D) 34 : 9
Q.93 The curved surface area of a cylinder pillar is 264
m2 and its volume is 924 m3. Find the ratio of its
diameter to its height. Q.102 A cylinder with base radius of 8 cm and height
of 2 cm is melted to form a cone of height 6 cm.
(A) 3 : 7 (B) 7 : 3 The radius of the cone will be:
(C) 6 : 7 (D) 7 : 6 (A) 4 cm (B) 5 cm
(C) 6 cm (D) 8 cm
Q.94 The height of a closed cylinder of given volume
and the minimum surface area is:
Q.103 A right cylindrical vessel is full of water. How
(A) equal to its diameter many right cones having the same radius and
(B) half of its diameter height as those of the right cylinder will be
(C) double of its diameter needed to store that water ?
(D) None of these (A) 2 (B) 3
(C) 4 (D) 8
Q.95 If the radius of the base of a right circular cylinder
is halved, keeping the height same, what is the Q.104 A solid metallic cylinder of base radius 3 cm and
ratio of the volume of the reduced cylinder to height 5 cm is melted to form cones, each of
that of the original one? height 1 cm and base radius 1 mm. The number
(A) 1 : 2 (B) 1 : 4 of cone is:
(C) 1 : 8 (D) 8 : 1 (A) 450 (B) 1350
(C) 4500 (D) 13500

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Q.105 Water flows at the rate of 10 meters per minute Q.113 Surface area of a sphere is 2464 cm2. If its radius
from a cylindrical pipe 5 mm in diameter. How be doubled, then the surface area of the new
long will it take to fill up a conical vessel whose sphere will be:
diameter at the base is 40 cm and depth 24 cm? . (A) 4928 cm2 (B) 9856 cm2
(A) 48 min. 15 sec. (B) 51 min. 12 sec. (C) 19712 cm 2 (D) Data insufficient
(C) 52 min. 1sec. (D) 55 min.
Q.114 If the radius of a sphere is doubled, how many
Q.106 A solid cylindrical block of radius 12 cm and times does its volume become?
height 18 cm is mounted with a conical block of (A) 2 times (B) 4 times
radius 12 cm and height 5 cm. The total lateral (C) 6 times (D) 8 times
surface of the solid thus formed is:
5 Q.115 If the radius of a sphere is increased by 2 cm,
(A) 528 cm2 (B) 1357 cm2
7 then its surface area increases by 352 cm2. The
(C) 1848 cm2 (D) None of these radius of the sphere before the increase was:
(A) 3 cm (B) 4 cm
Q.107 Consider the volume of the following: (C) 5 cm (D) 6 cm
1. A parallelopiped of length 5 cm, breadth 3 cm
and height 4 cm Q.116 How many lead shots each 3 mm in diameter can
2. A cube of each side 4 cm be made from a cuboid of dimensions 9 cm × 11
cm × 12 cm?
3. A cylinder of radius 3 cm and length 3 cm
(A) 7200 (B) 8400
4. A sphere of radius 3 cm
(C) 72000 (D) 84000
The volume of these in the decreasing order is:
(A) 1, 2, 3, 4 (B) 1, 3, 2, 4
Q.117 A sphere and a cube have equal surface areas.
(C) 4, 2, 3, 1 (D) 4, 3, 2, 1 The ratio of the volume of the sphere to that of
the cube is:
Q.108 The volume of a sphere is 4851 cu. cm. Its curved
surface area is: (A) : 6 (B) 2: 
(A) 1386 cm2 (B) 1625 cm2 (C) : 3 (D) 6: 
(C) 1716 cm2 (D) 3087 cm2

Q.118 The ratio of the volume of a cube to that of a

Q.109 The curved surface area of a sphere is 5544 sq.
sphere which will fit inside the cube is:
cm. Its volume is:
(A) 4 :  (B) 4 : 3
(A) 22176 cm3 (B) 33951 cm3
3 (C) 6 :  (D) 2 : 
(C) 38808 cm (D) 42304 cm3

Q.119 The surface area of a sphere is same as the

Q.110 The volume of a sphere if radius r is obtained by
curved surface area of a right circular cylinder
multiplying its surface area by:
whose height and diameter are 12 cm each. The
4 r 4r radius of the sphere is :
(A) (B) (C) (D) 3r (A) 3 cm (B) 4 cm
3 3 3
(C) 6 cm (D) 12 cm
Q.111 If the volume of a sphere is divided by its surface
area, the result is 27 cm. The radius of the sphere Q.120 The diameter of the iron ball used for the shot-
is: put game is 14 cm. It is melted and then a solid
(A) 9 cm (B) 36 cm 1
(C) 54 cm (D) 81 cm cylinder of height 2 cm is made. What will be
3
the diameter of the base of the cylinder?
Q.112 Spheres A and B have their radii 40 cm and 10 cm 14
respectively. The ratio of the surface area of A to (A) 14 cm (B) cm
the surface area of B is: 3
(A) 1 : 4 (B) 1 : 16 28
(C) 28 cm (D) cm
(C) 4 : 1 (D) 16 : 1 3

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Q.121 The volume of the greatest sphere that can be Q.124 The diameter of a sphere is 8 cm. It is melted
cut off from a cylindrical log of wood of base and drawn into a wire of diameter 3 mm. The
radius 1 cm and height 5 cm is: length of the wire is:
4 10 (A) 36.9 m (B) 37.9 m
(A)  (B)  (C) 38.9 m (D) 39.9 m
3 3
20
(C) 5 (D)  Q.125 A cylindrical vessel of radius 4 cm contains
3
water. A solid sphere of radius 3 cm is lowered
into the water until it is completely immersed.
Q.122 How many spherical bullets can be made out of The water level in the vessel will rise by:
a lead cylinder 15 cm high and with base radius
3 cm, each bullet being 5 mm in diameter? 2 4
(A) cm (B) cm
(A) 6000 (B) 6480 9 9
(C) 7260 (D) 7800 9 9
(C) cm (D) cm
4 2
Q.123 A cylindrical rod of iron whose height is eight
times its radius is melted and cast into spherical
balls each of half the radius of the cylinder. The
number of spherical balls is:
(A) 12 (B) 16 (C) 24 (D) 48

ANSWER KEY

Ques . 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Ans . D C B C C D D B B B D D D B B D A D A A
Ques . 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40
Ans . B A C B A D B C C C B B A D A B C C B B
Ques . 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60
Ans . D C D D D B C D C B B C D C D C C C C C
Ques . 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80
Ans . A A C C C D C A B B B B B A D B C B C C
Ques . 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 10 0
Ans . B C A C C A B C D B B B B A B C B D B A
Ques . 10 1 10 2 1 03 1 04 1 05 1 06 1 07 1 08 1 09 1 10 1 11 1 12 1 13 1 14 1 15 1 16 1 17 11 8 11 9 12 0
Ans . C D B D B D D A C B D D B D D D D C C C
Ques . 12 1 12 2 1 23 1 24 1 25
Ans . A B D B C

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 Hints & Solution # 1

Sol.1 Other side = (17) 2  (15) 2 = 289  225 Sol.7 Area of the plot = (110 × 65)m2 = 7150 m2
Area of the plot excluding the path
= 64 = 8 m. = [110 – 5) × 65 – 5)] m2 = 6300 m2
Area = (15 × 8) m2 = 120 m2.  Area of the path = (7150 – 6300) m2 = 850 m2.
Cost of gravelling the path
Sol.2 Let lenth = 2x metres and breadth = 3x metres.  80 
= Rs.  850   = Rs. 680.
1   5000  2  100 
Now, area =  1000  m2 =  m .
 6   3 
5000 2500  50   40 
So, 2x × 3x =  x2 =  x =  Sol.8 Side of first square =   cm = 10 cm;
3 9  3   4 
100 1  32 
 length = 2x = m = 33 and side of second square =   cm = 8 cm.
3 3  4 
Area of third square = [(10)2 – (8)] cm2
 50 
Breadth = 3x =  3   m = 50 m. = (100 – 64) cm2 = 36 cm2.
 3 
Side of third square = 36 cm = 6 cm.
Sol.3 Area of the carpet = Area of the room  Required perimeter = (6 × 4) cm = 24 cm.
= (13 × 9)m2 = 117 m2.
Sol.9 Area of the room = (544 × 374) cm2.
 Area  
 = 117   m
4
Length of the carpet =  Size of largest square tile = H.C.F. of 544 cm and
 Width   3 374 cm = 34 cm.
= 156 m. Area of 1 tile = (34 × 34) cm2.
 Cost of carpeting = Rs. (156 × 12.40)
 544  374 
= Rs. 1934.40  Number of tiles required =   = 176.
 34  34 
Sol.4 Let length = x and breadth = y. Then, 1
Sol.10 Area of the square = × (diagonal)2
2 (x + y) = 46 or x + y = 23 and 2
x2 + y2 = (17)2 = 289. 1 
Now, (x + y)2 = (23)2  (x2 + y2) + 2xy = 259 =   3.8  3.8  m2 = 7.22 m2.
2 
 289 + 2xy = 529  xy = 120.
 Area = xy = 120 cm2. Sol.11 Let the diagonals of the squares be 2x and
5x respectively.
Sol.5 Let breadth = x. Then, length = 2x. Then, 1 1
 Ratio of their areas = × (2x)2 : × (5x)2
(2x – 5) (x + 5) – 2x × x = 75  5x – 25 = 75 2 2
 x = 20. = 4x2 : 25x2 = 4 : 25.
 Length of the rectangle = 20 cm. Sol.12 Let each side of the square be a. Then, area = a2.
125a 5a
Sol.6 Let x and y be the sides of the rectangle. Then, Now side = = .
Correct area = xy. 100 4
2
 105   96  504  5a  25a 2
Calculated area =  x y  xy.. New area =   = .
 100   100  500  4  16
 25a 2 2  9a
2
 504 
Error in measurement =  xy  – xy =
4
xy.. Increase in area =  16  a  = .
 500  500   16
 4 1  4  9a 2 1 
 Error % =  xy  100 % = % = 0.8%.  Increase   2  100  % = 56.25%.

 500 xy  5  16 a 

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Sol.13 Let x and y be the length and breadth of the = (13.5 × 10000) m2 = 135000 m2.
rectangle respectively. Let altitude = x metres and base = 3x metres.
Then, x – 4 = y + 3 or x – y = 7 ... (i) 1
Area of the rectangle = xy; Then, × 3x × x = 135000  x2 = 90000
2
Area of the square = (x – 4) (y + 3)  x = 300.
 (x – 4) (y + 3) = xy  3x – 4y = 12 ... (ii)  Base = 900 m and Altitude = 300 m.
Solving (i) and (ii), we get x = 16 and y = 9.
 Perimeter of the rectangle Sol.18 Let ABC be the isosceles triangle and AD be the
= 2(x + y) = [2(16 + 9)] cm = 50 cm. altitude.
Let AB = AC = x. Then, BC = (32 – 2x).
3x Since, in an isosceles triangle, the altitude bisects
Sol.14 Let breadth = x metres, length = metres,
2 the base,
height = H metres. so BD = DC = (16 – x).
 Total cos t of carpeting  2 In ADC, AC2 = AD2 + DC2
Area of the floor =  m  x2 = (8)2 + (16 – x)2
 Rate / m 2 
 32x = 320  x 10.
 270  2
=  m = 54 m2.  BC = (32 – 2x) = (32 – 20) cm = 12 cm.
 5 
A
3x  2
 x× = 54  x2 =  54   = 36  x = 6.
2  3
3  x x
So, breadth = 6 m and length =   6  m = 9 m.
2 
 1720  2 B D C
Now, papered area =   m = 172 m2.
 10 
Area of 1 door and 2 windows = 8 m2. 1 
Hence, required area =   BC  AD 
Total area of 4 walls = (172 + 8) m2 = 180 m2. 2 

 180  1 
 2(9 + 6) × H = 180  H =   = 6 m. =   12  10  cm2 = 60 cm2.
 30  2 

Sol.15 Let a = 13, b = 14 and c = 15. Then,

3 27 3
1 Sol.19 Area of the triangle = × ( 3 3 )2 = .
s = (a + b + c) = 21. 4 4
2 Let the height be h.
 (s – a) = 8, (s – b) = 7 and (s – c) = 6.
1 27 3 27 3 2
 Area = s(s  a )(s  b)(s  c) Then, × 3 3×h= h= ×
2 4 4 3 3
= 21 8  7  6 = 84 cm2. 9
= = 4.5 cm.
2
Sol.16 Height of the triangle = (13) 2  (12) 2 cm
Sol.20 Let the bases of the two triangles be x and y and
= 25 cm = 5 cm. their heights be 3h and 4h respectively. Then,
1 1
 Its area = × Base × Height  x  3h 4
2 2 x  4 4  16
=  =  = .
1  1 3 y 3 3 9
=   12  5  cm2 = 30 cm2.  y  4h
2  2
 Required ratio = 16 : 9.
Total cost
Sol.17 Area of the field =
Rate Sol.21 Let the height of the parallelogram be x cm. Then,
base = (2x) cm.
 333.18 
=  hectares = 13.5 hectares  2x × x = 72  2x2 = 72  x2 = 36  x = 6.
 24.68 
Hence, height of the parallelogram = 6 cm.

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Sol.22 Let other diagonal = 2x cm.
 1100 
Since diagonals of a rhombus bisect each other  Number of revolutions per min. =  
at right angles, we have :  4.4 
= 250.
(20)2 = (12)2 + x2  x = (20) 2  (12) 2
 88  1000 
= 256 = 16 cm. Sol.27 Distance covered in one revolution =  m
 1000 
So, other diagonal = 32 cm.
= 88 m.
1 22
 Area of rhombus = × (Product of diagonals)  2R = 88  2 × × R = 88 m.
2 7
1   7 
=   24  32  cm2 = 384 cm2.  R =  88   = 14 m.
 2   44 
Sol.23 Let the two parallel sides of the trapezium be a Sol.28 Let inner radius be r metres. Then 2r = 440
cm and b cm.
 7 
Then, a – b = 4 ... (i)  r =  440   = 70 m.
 44 
And,  Radius of outer circle = (70 + 14) m = 84 m.

1 Sol.29 Let the inner and outer radii be r and R metres.

× (a + b) × 19 = 475 
2 352  352 7 1 
Then, 2r =  r =    = 8 m.
 475  2  7  7 22 2 
(a + b) =    a + b = 50 ... (ii)
 19  528  528 7 1 
2r =  R =    = 12
Solving (i) and (ii), we get : a = 27, b = 23. 7  7 22 2 
So, the two parallel sides are 27 cm and 23 cm. m.
 Width of the ring = (R – r) = (12 – 8) m = 4 m.
Sol.24 Clearly, the cow will graze a circular field of area
9856 sq. metres and radius equal to the length of Sol.30 Let the radius of the circle be r cm. Then,
the rope. r 2  66 22 120
Let the length of the rope be R metres. =  × r2 ×
360 7 7 360
 7  66  66 7 
Then, R2 = 9856  R2 =  9856   =  r2 =    3  = 9  r = 3.
 22  7  7 22 
= 3136  R = 56. Hence, radius = 3 cm.
 Length of the rope = 56 m. Sol.31 Let the side of the square be x. Then, its diagonal
Sol.25 Area = (13.86 × 10000) m2 = 138600 m2. = 2 x.
x
 7  Radius of incircle = and
R2 = 138600  R2 = 138600   2
 22 
2x x
 R = 210 m. radius of circumcircle = = .
2 2
 22 
Circumference = 2R =  2   210  m =1320 m.
 7 
 Cost of fencing = Rs. (1320 × 4.40) x/
2
= Rs. 5808 m. x/2

 66  1000 
Sol.26 Distance to be covered in 1 min. =  m
 60  x
= 1100 m.
 22   x 2 x 2  1 1
Circumference of the wheel =  2   0.70  m  Required ratio  :  = : = 1 : 2.
 7   4 2  4 2
= 4.4 m.

manishkumarphysics.in
 50 R Sol.38 Volume of the metal used in the box
Sol.32 Let original radius = R. New radius = R= . = External Volume – Internal Volume
100 2
Original area = R2 and New area = [(150 × 40 × 23) – (44 × 34 × 20)] cm3
2 = 16080 cm3.
R R 2
=   = .  16080  0.5 
2 4  Weight of the metal =   kg
 1000 
 3R 2 1 
 Decrease in area =   100  %

= 8.04 kg.
 4 R 2

= 75%. Sol.39 let the edge of the cube be a
3 a = 6 3  a = 6.
Sol.33 Volume = (16 × 14 × 7) m3 = 1568 m3.
So, Volume = a3 = (6 × 6 × 6) cm3 = 216 cm3.
Surface area = [2 (16 × 14 + 14 × 7 + 16 × 7)] m2
Surface area = 6a2 = (6 × 6 × 6) cm2 = 216 cm2.
= (2 × 434) m2 = 868 m2.
Sol.40 Let the edge of the cube be a. Then,
Sol.34 Length of longest pole = Length of the diagonal 6a2 = 1734  a2 = 289  a = 17 cm.
of the room  Volume = a3 = (17)3 cm3 = 4913 cm3.
= (12) 2  8 2  9 2 = 289 = 17 m.
Sol.41 Volume of the block = (6 × 12 × 15) cm3
= 1080 cm3.
Sol.35 Let the breadth of the wall be x metres. Side of the largest cube
Then, Height = 5x metres and = H.C.F. of 6 cm, 12 cm, 15 cm = 3 cm.
Length = 40 x metres. volume of this cube = (3 × 3 × 3) cm3 = 27 cm3.
 x × 5x × 40x = 12.8
 1080 
12.8 128 64 Number of cubes =   = 40.
 x3 = = = .  27 
200 2000 1000
Sol.42 Increase in volume
4  4 
So, x = m =  100  cm = 40 cm. = Volume of the cube = (15 × 15 × 15) cm3.
10  10 
 Volume 
 Rise in water level =  
Sol.36 Volume of the wall = (2400 × 800 × 60) cu. cm.  Area 
Volume of bricks = 90% of the volume of the wall  15  15  15 
 90  =  cm = 11.25 cm.
 2400  800  60  cu. cm.  20  15 
=
 100 
Sol.43 Volume of new cube = (13 + 63 + 83) cm3
Volume of 1 brick = (24 × 12 × 8) cu. cm.
= 729 cm3.
 90 2400  800  60 
 Number of bricks =    Edge of new cube = 3 729 cm = 9 cm.
 100 24  12  8 
 Surface area of the new cube
= 45000.
= (6 × 9 × 9) cm2 = 486 cm2.
Sol.37 Volume required in the tank = (200 × 150 × 2) m3
= 60000 m3 Sol.44 Let original length of each edge = a.
Volume of water column flown in 1 min. Then, original surface area = 6a2.

 20  1000   150  3a
= m=
1000
m. New edge = (150% of a) =  a = .
 60  3  100  2
2
Volume flown per minute  3a  27 2
New surface area = 6 ×   = a.
 1000  3  2  2
= 1.5  1.25   m = 625 m3.
 3  Increase percent in surface area
 60000   15 2 1 
 Required time =   min. = 96 min. =  a  2  100  % = 125%.
 625   2 6a 

manishkumarphysics.in
Sol.45 Let their edges be a and b. Then, Outer radius = (1.5 + 1) = 2.5 cm.
3
 a  1
3  Volume of iron
a3 1 a 1
= or   =   or = . = [ × (2.5)2 × 100 –  × (1.5)2 × 100] cm3
b 3 27    
b 3 b 3
22
6a 2 a2 = × 100 × [(2.5)2 – (1.5)2] cm3
 Ratio of their surface areas = = 7
6b 2 b2
 8800  3
a 1
2
=  cm .
=   = , i.e., 1 : 9.  7 
 b 9
 8800 21 
 Weight of the pipe =    kg
 7 1000 
 22 7 7  = 26.4 kg.
Sol.46 Volume = r2h =     40  cm3 = 1540 cm3.
 7 2 2 
Curved surface area = 2rh Sol.52 Here, r = 21 cm and h = 28 cm.
 22 7   Slant height, l = r 2  h 2
=  2    40  cm2 = 880 cm2.
 7 2 
Total surface area = 2rh + 2r2 = 2r (h+r) = (21) 2  (28) 2 = 1225 = 35 cm.

 22 7  1  1 22 
= 2    ( 40  3.5)  cm2 = 957 cm2. Volume = r2h =    21 21  28  cm3
 7 2  3 3 7 
= 12936 cm3.
Sol.47 Let the depth of the tank be h metres. Then,  22 
Curved surface area =   21 35  cm2
 7 1   7 
 × (7)2 × h = 1848  h = 1848    2
 22 7  7 = 2310 cm .
= 12 m. Total surface area = (rl + r2)
 22 
Sol.48 Let the length of the wire be h metres. Then, =  2310   21 21 cm2 = 3696 cm2.
 7 
2
 0.50  2.2
 ×  ×h=
 2  100  1000 Sol.53 Here, r = 7m and h = 24 m.

 2.2 100  100 7  So, l = r 2  h 2 = 7 2  (24) 2 = 625 = 25 m.

h =    = 112 m.
 1000 0 .25  0. 25 22   22 
Area of canvas = rl =   7  25  m2 = 550 m2.
 7 
 22 1 1   Length of canvas
Sol.49 Volume of 1 rod =     7  cu. m
 7 100 100   Area   550 
11 =  =  m = 440 m.
= cu. m.  Width   1.25 
5000
Volume of iron = 0.88 cu. m
 5000  Sol.54 Let the radii of their bases be r and R and their
Number of rods =  0.88   = 400. heights be h and 2h respectively.
 11 
2r 3 r 3 4
Then, =  =  R = r..
Sol.50 Let the radii of the cylinders be 3x, 5x and their 2R 4 R 4 3
heights be 2y, 3y respectively. Then, 1 2
Ratio of their curved surface areas r h
3
 Ratio of volumes =
2   3x  2 y 2 1 4 
2
= = = 2 : 5.  r  (2h )
2  5x  3 y 5 3 3 
9
3 = = 9 : 32.
Sol.51 Inner radius =   cm = 1.5 cm, 32
2

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Sol.55 Let the radii of the cylinder and the cone be 3r
4 
and 4r and their heights be 2h and 3h Sol.59 Volume of larger sphere =    6  6  6  cm3
respectively. 3 
= 288  cm3.
Volume of cylinder   (3r ) 2  2h
 =1 4 1 1 1
Volume of cone Volume of 1 small lead ball =       cm3
  (4r 2 )  3h 3 2 2 2
3
9
= = 9 : 8. 
8 = cm3.
6
Sol.56 Volume of the liquid in the cylindrical vessel  6
= Volume of the conical vessel  Number of lead balls =  288   = 1728.
  
 1 22 
=    12  12  50  cm3 Sol.60 Volume of cylinder = ( × 6 × 6 × 28)cm3
3 7 
= (36 × 28)  cm3.
 22  4  12  50  3
=   cm .
 7  4 3 3 3
Volume of each bullet =       cm3.
Let the height of the liquid in the vessel be h.  3 4 4 4
22 22  4  12  50 9 3
Then,  10  10  h = = cm .
7 7 16
 4  12  50  Volume of cylinder
or h =  = 24 cm. Number of bullets =
 10  10  Volume of each bullet
 16 
4  4 22 21 21 21  = (36  28)    = 1792.
Sol.57 Volume = r3 =       cm3  9 
3 3 7 2 2 2 
= 4851 cm3.
4 
 22 21 21  Sol.61 Volume of sphere =    9  9  9  cm3 = 972 cm3.
Surface area = 4r2=  4     cm2 3 
 7 2 2
= 1386 cm2. olume of wire ( × 0.2 × 0.2 × h) cm3.
2 2
Sol.58 Let original radius = R.  972 =  × × ×h
10 10
50 3R
Then, new radius = R= .  972  5  5 
100 2  h = (972 × 5 × 5) cm =  m
 100 
4
Original volume = R3. = 243 m.
3
3
4  3R  9R 3
New volume =   = . Sol.62 Volume of sphere = Volume of 2 cones
3  2  2
1 1 
 19 3  =    ( 2.1)  4.1    ( 2.1)  4.3 cm3
2 2
Increase % in volume =  R   100  %
3
4R  3 3 
 6 3

= 237.5%. 1
=  × (2.1)2 (8.4) cm3.
Original surface area = 4R2. 3
2 Let the radius of the sphere be R.
 3R 
New surface are = 4   = 9R2. 4 1
 2   R3 =  × (2.1)2 × 4 or R = 2.1 cm.
3 3
 5R 2  Hence, diameter of the sphere = 4.2 cm.
Increase % in surface area =   100  %

 4R
2

= 125%.

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Sol.63 Let radius of the each be R and height of the 2 
cone be H. Sol.65 Volume of bowl =    9  9  9  cm3
3 
4 1
Then, R3 = R2H = 486 cm3.
3 3
 3 3 
R 1 2R 2 1 Volume of 1 bottles =      4  cm3 = 9
or = or = = .  2 2 
H 4 H 4 2 cm3.
 Required ratio = 1 : 2.
 486 
Number of bottle =   54.
 9 
2 3  2 22 21 21 21  3
Sol.64 Volume = r =       cm
3 3 7 2 2 2 
3 Sol.66 Let R be the radius of each.
= 2425.5 cm .
Height of hemisphere = Its radius = R.
Curved surface area = 2r2
 Height of each = R.
 22 21 21 
=  2     cm3 = 693 cm2. 1 2 2
 7 2 2 Ratio of volumes = R × R : R3 : R2 × R
3 3
Total surface area = 3r2
= 1 : 2 : 3.
 22 21 21 
=  3     = cm2
 7 2 2
= 1039.5 cm2.

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