Areas and Mensuration

Areas and mensuration is a topic depends entirely on application of formulas.

If you remember the formula, solving problems in this area is a cakewalk.
For easy learning and to remember the formulas we provided them into a
simple table.  Before you move on to have a look at the solved examples,
study the tables and try to understand the relevant formula

2 Dimensional Figures: 

Two dimensional figures have only any two of length, breadth, and height.

They have only areas but not volumes.  Perimeter is a uni-dimensional
measure. If you observe carefully, the power of the terms in the formulas of
perimeter is 1, and of the terms in the areas is 2.
 Solved Examples

1. If sides of a triangle are 8 cm, 15 cm and 17 cm respectively. Find its area.

Area of the triangle when all the three sides are given = s(s−a)(s−b)(s−c)−−−
−−−−−−−−−−−−−−√ where s=a+b+c2

s = 8+15+172 = 20 cm

Therefore, Area = 20×(20−8)×(20−15)×(20−17)−−−−−−−−−−−−−−−−−−−−−

−−−−−−−−√

= 20×12×5×3−−−−−−−−−−−−√
= 4×5×3×4×5×3−−−−−−−−−−−−−−−−−√ = 4 x 5 x 3 = 60 cm2
Trick:
The triangle is right angle triangle as 172=152+82
Therefore, Area of right angle triangle = 12 x 8 x 15 = 60 cm2

2. Two parallel sides of a trapezium are 4 cm and 5 cm respectively. The

perpendicular distance between the parallel sides is 6 cm. Find the area of the
trapezium.
Area of trapezium when height and two parallel sides are given
= 12×h×(a+b) = 12 x 6 x (4 + 5) = 27cm2

3. If perimeter and area of a square are equal. Side of the square (in cm) is:
Given, (Side)2 = 4 x (Side)
Therefore, Side = 4 cm

4. The diameter of the wheel of a vehicle is 5 metre. It makes 7 revolutions

per 9 seconds. What is speed of the vehicle in km/h?
Radius of the wheel = 52 metre
Distance covered in 1 revolution = Circumference of the wheel = 2πr = 2 x
227×52 metre

Therefore, Distance covered in one second = 2 x 227×52×79 metre

Therefore, Speed per hour = 2 x 227×52×79×185 = 44 km/h

5. The perimeter of a rhombus is 60 cm and one of its diagonal is 24 cm. Find

the other diagonal of the rhombus.
Let ABCD is the rhombus whose diagonals BD and AC intersecting at point
'O'.
Side of rhombus = 14 x Perimeter = 14 x 60 = 15 cm.
Let BD = 24 cm
Then, BO = 12BD = 12 cm
Now, AO = AB2−BO2−−−−−−−−−√=152−122−−−−−−−√ = 9
Therefore, AC = 2 AO = 2 x 9 = 18 cm

6. A field is 40 metre long and 35 metre wide. The field is surrounded by a

path of uniform width of 2.5 metre runs round it on the outside. Find the area
of the path.
Remember the formula for the Area of path
= 2 x Width x [Length + Breadth + (2 x Width)]
= 2 x 25 x (40 + 35 + 2 x 2.5)
= 5 x (75 + 5) = 400 m2

7. Find area of uniform path of width 2 metre running from centre of each
side of the opposite side of a rectangle field measuring 17 metre by 12 metre.
Remember the formula for the Area of path
= Width of path x (Length of field + Breadth of field) - (Width of path)2
= 2 x (17 + 12) - (2)2
= 58 - 4 = 54 m2

8. A square and a rectangle each has perimeter 60 metre. Difference between

the areas of the two figures is 16 square metre. Find length of the rectangle.
Side of the square = 604 = 15 metre
Difference in areas of square and rectangle = 16m2
Therefore, Increase and decrease in dimensions = 16−−√ = 4 metre
Therefore, Sides of rectangle are (15 + 4) and (15 - 4) metre
Therefore, Length of rectangle = 19 metre

9. Find the radius of a circle whose area is equal to the sum of areas of three
circles with radii 8 cm, 9 cm, 12 cm respectively.
Let radius of new circle is 'R' cm.
Then πR2 = π82+π92+π122=64π+81π+144π=289π
Therefore, R2 = 289 => R = 17 cm.

10. The are of the ring between two concentric circles, whose circumferences
are 88 cm and 132 cm is:
Area of Ring = 14π×(1322−882)=14π x (132 + 88) x (132 - 88)
= 14×722 x 220 x 44 = 770 cm2

11. Two poles whose heights are 11 metre and 5 metre stand upright in a
field. If the distance between their feet is 8 metre, what is the distance
between their tops?
Distance between feet of big pole and that of small pole = 8 metre
Difference between heights of two poles = 11 - 5 = 6 metre
Distance between the tops of poles = 62+82−−−−−−√ = 10 metre

12. Find the are of the largest circle that can be inscribed in a square of 14
cm, a side.
Side of the square = Diameter of the circle = 14 cm
Therefore, Radius of the circle = 7 cm.
Area of the circle = πr2=227 x 7 x 7 = 154 cm2.

13. A horse is tied to one of the corners of a square field whose side is 20
metre. If length of the rope is 14 metre, find the area of ungrazed field.
Area of square field = (20m)2 = 400 m2
Area of the field grazed by the horse = 14πr2
= 14×227 x 14 x 14 = 154 m2
Therefore, Field ungrazed = (400 - 154) = 246 m2

14. A horses are tied one to each corner of a square with 14 m a side, and
length of the rope is 7 m. Find the area of ungrazed field.
Ungrazed area = Area of square - Area of circle
= 142−227 x 7 x 7 = 196 - 154 = 42 m2
Short-Cut Method:
Radius of circle = 7 m.
Therefore, Ungraged field = 67×72 = 42 m2
MCQ's

1. A rectangular carpet has an area of 60 sq.m. Its diagonal and longer side
together equal 5 times the shorter side. The length of the carpet is :
a. 5 m
b. 12 m
c. 13 m
d. 14.5 m
Correct Option: B
Explanation:
Let the length = p metres and breadth = q metres
Then pq = 60
p2+q2−−−−−−√+p=5q
⇒p2+q2=(5q−p)2 =25q2+p2−10pq
As pq = 60,
⇒25q2−10×60=0
⇒q2=25 or q = 5
p = 60/5 = 12m
Length of the carpet = 12m

2.  A rectangular carpet has an area of 120 sq.m and perimeter of 46 m. The

length of its diagonal is :
a. 15 m
b. 16 m
c. 17 m
d. 20 m
Correct Option: C
Explanation:
Let length = a metres and breadth = b metres
Then, 2(a+b)=46 or (a+b) = 23
and ab = 120
Diagonal = a2+b2−−−−−−√ = (a+b)2−ab−−−−−−−−−−√=(23)2−2×120−−−−
−−−−−−−−√ = 289−−−√=17m

3.  A parallelogram has sides 60 m and 40 m and one of its diagonals is 80 m

long. Then, its area is :
a. 480 sq.m
b. 320 sq.m
c. 60015−−√ sq.m
d. 450 15−−√ sq.m
Correct Option: C
Explanation:
AB = 60m, BC=40m or AC = 80m
s=12(60+40+80)m=90m
(s-a) = 30m, (s-b) = 50m and (s-c)=10m
Area of ΔABC = s(s−a)(s−b)(s−c)−−−−−−−−−−−−−−−−−√
=90×30×50×10−−−−−−−−−−−−−−√m2=30015−−√m2
Area of //gm ABCD = 60015−−√m2

4. If a square and a parallelogram stand on the same base, then the ratio of the
area of the square and the parallelogram is :
a. greater than 1
b. equal to 1
c. equal to 12
d. equal to 14
Correct Option: B
Explanation:

Let ABCD be the square and ABEF be the parallelogram.

Then, in right triangles ADF and BCE, we have AD = BC (sides of a square )
and AF = BE (sides of parallelogram). Therefore, DF = CE
[DF2=AF2−AD2=BE2−BC2=CE2]
Thus, ΔADF = ΔBCE
 ⇒ΔADF+□ABCF = △BCE+□ABCF
⇒ Area of parallelogram ABEF = Area of ABCD
5. In a rhombus, whose area is 144 sq.cim, one of its diagonals is twice as
long as the other. The length of its diagonals are :
a. 24 cm, 48 cm
b. 12 cm, 24 cm
c. 62√ cm, 122√cm
d. 6 cm, 12 cm
Correct Option: B
Explanation:
12×x×2x=144⇒x2=144 or x = 12
Length of diagonals = 12cm, 24cm

6. The length of a rope by which a cow must be tethered in order that she
may be able to graze an area of 9856 sq.m.is:
a. 56 m
b. 64 m
c. 88 m
d. 168 m
Correct Option: A

Explanation:
Grazing area is equal to the area of a circle with radius r.
227×r2=9856
Then r2=(9856×722)
r = 56 m

7. The circumferences of two concentric circles are 176 m and 132 m

respectively. What is the difference between their radii?
a. 5 metres
b. 7 metres
c. 8 metres
d. 44 metres
Correct Option: B
Explanation:

2πR−2πr=(176−132)
⇒2π(R−r)=44
⇒(R−r)=44×72×22=7m

8. The diameter of a circle is 105 cm less than the circumference. What is the
diameter of the circle ?
a. 44 cm
b. 46 cm
c. 48 cm
d. 49 cm
Correct Option: D
Explanation:
πd−d=105⇒(π−1)d=105
⇒(227−1)d = 105
d = (715×105)cm=49cm
9. A circle and a square have same area.  The ratio of the side of the square
and the radious of the circle is :
a. π√:1
b. 1 : r√
c. 1 : r
d. r : 1
Correct Option: B
Explanation:
x2=πr2⇒xr=π√=π√:1

10. The number of rounds that a wheel of diameter 711m will make in going
4 km, is :
a. 1000
b. 1500
c. 1700
d. 2000
Correct Option: D
Explanation:
Number of rounds = 4×1000227×711=2000

11. A circular wire of radius 42 cm is cut and bent in the form of a rectangle
whose sides are in the ratio of 6:5. The smaller side of the rectangle is :
a. 30 cm
b. 60 cm
c. 72 cm
d. 132 cm
Correct Option: B
Explanation:
Circumference = (2×227×42) cm = 264 cm
Let the rectangle sides are 6x, 5x.  Then circumference is
2×(6x+5x)=264 or x = 12
Smaller side of recatngle = 5x = 60 cm

12. If the diameter of a circle is increased by 100% , its area is increased by :

a. 100% 
b. 200%
c. 300%
d. 400%
Correct Option: C
Explanation:
Original area = π×(d2)2=πd24
New area = π×(2d2)2=πd2
Increase in area = (πd2−πd24)=3πd24
Increase percent = (3πd24×4πd2×100)% = 300%
Shortcut: 
You can use (A+B+AB100)% formula. Substitute A = B = 100

13.If the radius of a circle is reduced by 50%, its area is reduced by :

a. 25%
b. 50%
c. 75%
d. 100%
Correct Option: C
Explanation:
Original area = π×r2
New are = π×(r2)2=πr24
Reduction in area = (πr2−πr24)=3πr24
Reduction percent = (3πr24×1πr2×100)%=75%
Shortcut: 
You can use (A+B+AB100)% formula. Substitute A = B = - 50.

14. If the circumference of a circle is 352 metres, then its area in m2 is :

a. 9856
b. 8956
c. 6589
d. 5986
Correct Option: A
Explanation:
2×227×r=352⇒r=(352×722×12)=56m
Area = (227×56×56)m2=9856m2

15. Area of square with side x is equal to the area of a triangle with base x .
The altitude of the triangle is :
a. x2 
b. x
c. 2x
d. 4x
Correct Option: C
Explanation:
x2=12×x×h or h = 2x2x=2x

16.  The perimeter of an isosceles triangle is equal to 14 cm; the lateral side is
to the base in the ratio 5:4. The area of the triangle is :
a. 1221−−√cm2
b. 3221−−√cm2
c. 21−−√cm2
d. 221−−√cm2
Correct Option: D
Explanation:
Let lateral side = (5x)cm and base = (4x) cm
5x+5x+4x=14 or x = 1
So, the sides are 5 cm, 5 cm and 4 cm
Semi perimeter, S = 12(5+5+4)cm = 7cm.
(s-a)=2 cm, (s-b)=2cm and (s-c)=3 cm
Area = 7×2×2×3−−−−−−−−−−−√cm2=221−−√cm2

17.  If the diagonal of a square is doubled, how does the area of the square
change ?
a. Becomes four fold
b. Becomes three fold
c. Becomes two fold
d. None of these
Correct Option: A
Explanation:
12×d212×(2d)2=14
New area becomes 4 fold.

18.  If the base of a rectangle is increased by 10% and the area is unchanged,
then the corresponding altitude must be decreased by :
a. 9111%
b. 10%
c. 11%
d. 1119%
Correct Option: A
Explanation:
Let base = b and altitude = h.
then area = (bh)
New base = (110100b)=(1110b)
Let new altitude  = H
Then, 1110b×H=bh or H = (1011h)
Decrease  = (h−1011h)=111h
Decrease percent = (111h×1h×100)% = 9111%
Short cut: 
Assume Length is 10 units and Altitude is 11 Units. Now Area = 110
New length is 11 units and New altitude is h units. Now Area = 11h
But given 110 = 11h ⇒ h = 10
So change in altitude = 1/11 x 100 = 9111%

19.  The length of a rectangle is twice its breadth. If its length is decreased by
5 cm and the breadth is increased by 5 cm, the area of the rectangle is
increased by 75 cm2.  Therefore, the length of the rectangle is :
a. 20 cm
b. 30 cm
c. 40 cm
d. 50 cm
Correct Option: C
Explanation:
Let breadth = x cm and length = (2x) cm
Then. (2x-5)(x+5) - x×2x=75
2x2+5x−25−2x2=75 or 5x = 100
or x = 20
Length = (2x)cm = 40 cm.

20. A rectangle has 15 cm as its length and 150 cm2 as its area. Its area is
increased to 113 times the original area by increasing only its length of its
new perimeter is :
a. 50 cm
b. 60 cm
c. 70 cm
d. 80 cm
Correct Option: B
Explanation:
Breadth of the rectangle = (15015) cm=10 cm
New area = (43×150)cm2=200cm2
New length = (20010)cm=20cm
New perimeter = 2(20+10)cm = 60 cm

21. The length of a rectangular room is 4 metres. If it can be partitioned into

two equal square rooms, what is the length of each partition in metres?
a. 1
b. 2
c. 4
d. Data inadequate.
Correct Option: B
Explanation:
Let the side of each new room = y metres.
Then, y2=2x
Clearly, 2x is a complete square when x = 2
y2=4 or y = 2 m

22. The length and breadth of a square are increased by 40% and 30%
respectively. The area of the resulting rectangle exceeds the area of the
square by :
a. 42%
b. 62%
c. 82%
d. None of these
Correct Option: C
Explanation:
Let the side of the square = 100 m
New length = 140 m. new breadth = 130 m
Increase in area =[(140×130)−(100×100)]m2 = 8200 m2
Increase percent = (8200100×100×100)% = 82%
23. A hall 20m long and 15m broad is surrounded by a verandah of uniform
width of 2.5m. The cost of flooring the verandah at the rate of Rs.3.50 per
sq.metre is :
a. Rs.500
b. Rs.600
c. Rs.700
d. Rs.800
Correct Option: C
Explanation:
Area of verandah = [(25×20)−(20×15)]m2=200 m2
Cost of flooring = Rs.(200×3.50) = Rs.700

24. If the ratio of the areas of two squares is 9:1, the ratio of their perimeters
is :
a. 9:1
b. 3:1
c. 3:4
d. 1:3
Correct Option: B
Explanation:
Let the areas of squares be : (9x2)m2and(x2)m2
Then, their sides are 9x2−−−√,x2−−√ or  (3x) metres & x metres respectively.

25. The length of a rectangle is increased by 60% . By what percent would

the width have to be decreased to maintain the same area ?
a. 3712
b. 60%
c. 75%
d. 120%
Correct Option: A
Explanation:
Initially, let length = x and breadth = y
Let, new breadth = z. Then new length = (160100x)=85x
85x×z =xy or z = 5y8
Decrease in breadth = (y−5y8)=38y
Decrease percent = (38y×1y×100)% = 3712%
Mensuration II (Important results)

The following results are very important to solve various mensuration

problems.

1.  The largest possible sphere that can be chiseled out from a cube of side
"a" cm. 

 Diagonal of the sphere is a, so radius = a/2.

Remaining empty space in the cube = a3−πa36

2. The largest possible cube that can be chiseled out from a sphere of radius
"a" cm

Here OA = radius of the sphere. So diagonal of the sphere = 2a.

Therefore side of the square = 2a3√ [if side of the square is x cm, diagonal
= 3√x]
Remaining empty space in the sphere = 43π(a)3−(2a3√)3=4a33[π−23√3]

3.  The largest possible sphere that can be chiselled out from a cylinder of
radius 'a' cm and height 'h' cm. then 
Case 1: for h>A
Radius of the sphere is equal to radius of the cylinder. 

Case 2: a>h
Radius of the sphere = h2

 4. The largest possible sphere that can be inscribed in a cone of radius 'a' cm
and slant height equal to the diameter of the base (L = 2a)
 The radius of the sphere = a3√

 5.  The largest possible cube that can be chiseled out from a hemisphere of
radius 'a' cm. 

 The edge of the cube = a23−−√

6.  The largest square that can be inscribed in a right angled triangle ABC
when one of its vertices coincide with the vertex of right of the triangle.
 Side of the square = aba+b and area of the square = (aba+b)2

7.  The largest square that can be inscribed in a right angled triangle ABC
when one of its vertices lies on the hypotenuse of the triangle

 Side of the square = abca2+b2+ab

Area of the square = (abca2+b2+ab)2

8.  The largest square that can be inscribed in a semi circle of radius 'r' units
Area of the square = 35r2

9. The largest cube that can be chiseled out from a cone of height 'h' cm and
radius of 'r' cm

Square side = 2√.h.rh+2√.r

10.  The largest square that can be inscribed in a quadrant of radius 'r' cm. 
Side of the square = r2√, and area of the square = r22

11.  The largest circle that can be inscribed in the semi circle of radius 'r' cm
is 

Inscribed circle area = πr24

(Rememeber: Inscribed circle area is half of the semi circle area)

12.  The largest circle that can be inscribed in a quadrant of radius 'r' cm is 
Area of the circle = πr23+22√

13.  The ratio of the volumes of the cylinder and the largest cube chiseled out
from it are in the ratio = 11:7 (here cube side is equal to height of the
cylinder)

14.  The ratio of the volumes of the cylinder and the largest cone chiseled out
from it are in the ratio : 3:1
(here cone and cylinder have same base radius and heights)