# Mensuration Questions for Competitive Exams

Mensuration Questions with solution for Competitive Exams. Latest, new questions from SSC CGL, CHSL previous year exam question paper.

### Mensuration Questions : free online practice for Competitive Exams

Q.1: What is the perimeter (in cm) of an equilateral triangle whose height is 3.46 cm ?Take $\sqrt 3= 1.73$.
(A) 12
(B) 9
(C) 6
(D) 10.4
(SSC CHSL August 2021)

Ans : (A) 12
height of equilateral $\triangle$ $=\frac {\sqrt 3}{2} \times$ side
$=\frac {\sqrt 3}{2} \times$ a = 3.46
$\frac {1.73}{2}\times$ a = 3.46
a = 2 x 2 = 4
Peimeter = 3a = 4 x 3 = 12

Q.2: The perimeters of two similar triangles ABC and PQR are 156 cm and 46.8 cm respectively. If BC = 19.5 cm and QR = x cm, then the value of x is:
(A) 3.76 cm
(B) 5.85 cm
(C) 4.29 cm
(D) 6.75 cm
(SSC CHSL August 2021)

Ans : (B) 5.85 cm
$\frac {(\triangle \text{ABC}) Perimeter}{(\triangle \text{PQR})Perimeter}$= $\frac {BC}{QR}$
$\frac {156}{46.8}=\frac {19.5}{x}$
$x=\frac{46.8\times 19.5}{156}$
5.85 cm

Q.3: The radius of a sphere is 9 cm. It is melted and drawn into a wire of radius 0.3 cm. The length of the wire is:
(A) 112 m
(B) 108 m
(C) 118 m
(D) 106 m
(SSC CHSL August 2021)

Ans : (B) 108 m
Volume of sphere = Wire volume
$\frac {4}{3} \pi r^3$ = $\pi$r2h
$\frac 43 \times \pi \times 9^3$ = $\pi$x 0.3 x 0.3 x h
h =$\frac {4\times9\times9\times9}{3\times0.3\times0.3}$
h = 10800 cm = 108 m

Q.4: One side of a rectangular field is 39 m and its diagonal is 89 m. What is the area of the field ?
(A) 3120 m2
(B) 2100 m2
(C) 2160 m2
(D) 3140 m2
(SSC CHSL August 2021)

Ans : (A) 3120 m2
Side = 39
Diagonal = 89
Second side $= \sqrt{(89)^2 - (39)^2}$
$= \sqrt{7921-1521}$
Second Side $= \sqrt{6400}$ =80
Area of field = First Side x Second side (l x b)
= 80 x 39 = 3120 m2

Q.5: The base of a triangle to the perimeter fo a square whose diagonal is $7\sqrt2$ cm and its height is equal to the side of square whose area is 169 cm2 . The area (in cm2 ) of the triangle is:
(A) 130
(B) 182
(C) 175
(D) 156
(SSC CHSL August 2021)

Ans : (B) 182
Diagonal of square $7\sqrt2$
$a\sqrt2$=$7\sqrt2$
a = 7
Perimeter of Square = 4a = 28
Perimeter of square = Base of Triangle = 28
Area of Square = 169
Side of Square = 13
Height of Triangle = Side of square = 13
Area of Triangle = $\frac12$ x b x h
= $\frac12 \times28\times13$ =182

Q.6: If the adjacent sides of a rectangle whose perimeter is 60 cm are in the ratio 3 : 2, then what will be the area of the rectangle ?
(A) 864 cm2
(B) 216 cm2
(C) 60 cm2
(D) 300 cm2
(SSC CHSL August 2021)

Ans : (B) 216 cm2
Ration of side = 3 : 2
Perimeter = 60
Perimeter of rectangle = 2 ( l + b)
2(3n + 2n) = 60
5n = 30, n = 6
l = 3n = 3 x 6 = 18
b = 2n = 2 x 6 = 12
Area = l x b = 18 x 12 = 216 cm2

Q.7: The perimeter of a right angle triangle whose sides that make right angles are 15 cm and 20 cm is:
(A) 60 cm
(B) 40 cm
(C) 70 cm
(D) 50 cm
(SSC CHSL August 2021)

Ans : (A) 60 cm
Diagonal of Right angle triangle
(Third Side ) $=\sqrt{(20^2 + 15^2)}$
$=\sqrt {400 + 225}$
Diagonal $=\sqrt {625}$ = 25
Perimeter = Sum of all sides = 15 + 20 + 25= 60 cm

Q.8: The difference between the two perpendicular sides of a right-angled triangle is 17 cm and its area is 84 cm2 . What is the perimeter (in cm ) of the triangle ?
(A) 65
(B) 49
(C) 72
(D) 56
(SSC CHSL August 2021)

Ans : (D) 56
b – h = 17
Area = $\frac12$x b x h = 84
b x h = 168 = 24 x 7
b = 24
h = 7
Diagonal2 = b2 + h2 = 242 + 72
=> 576 + 49 = 625
Diagonal= 25
Sides of Triangle= 7, 24, 25
Perimeter = 7 + 24 + 25 = 56

Q.9: A solid metallic sphere of radius 10 cm is melted and recast into spheres of radius 2 cm each. How many such spheres can be made ?
(A) 64
(B) 216
(C) 125
(D) 100
(SSC CHSL August 2021)

Ans : (C) 125
Volume of big sphere = n x Volume of small sphere
V =$\frac 43\pi r^3$
$\frac 43 \pi (10)^3$ = n x$\frac 43\pi (2)^3$
10 x 10 x 10 = n x 2 x 2x 2
1000 = n x 8
n = 125

Q.10: The sum of the squares of the sides of a rhombus is 1600 cm2 . What is the side of the rhombus ?
(A) 10 cm
(B) 15 cm
(C) 20 cm
(D) 25 cm
(SSC CHSL August 2021)

Ans : (C) 20 cm
All the sides of rhombus are equal
The sum of the squares of sides = 1600
a2 + a2 + a2 + a2 = 1600
4a2 = 1600
a2 = 400
a = 20

Mensuration Questions for Competitive Exams

Q.11: The volume of a right circular cone is 462 cm3 . If its height is 12 cm, then the area of its base (in cm2 ) is:
(A) 124.5
(B) 103.5
(C) 115.5
(D) 98.5
(SSC CHSL August 2021)

Ans : (C) 115.5
Volume of Cone =$\frac13 \pi r^2h$
$\frac13 \pi r^2h$ = 462
r2 =$\frac{462\times 3}{\pi \times h}$
=$\frac{462\times 3\times7}{22 \times 12}$
r2 =$\frac{21\times7}{4}$
Area of base =$\pi r^2$
=$\frac {22\times21\times7}{7\times4}$
= 115.5

Q.12: How many bricks each measuring 64 cm x 11.25 cm x 6 cm. will be needed to build a wall measuring 8m x 3m x 22.5m ?
(A) 200000
(B) 250000
(C) 67500
(D) 125000
(SSC CHSL August 2021)

Ans : (D) 125000
Volume of wall = V1
Volume of Brick= V2
Number of Bricks =$\frac{V1}{V2}$
=$\frac{8m\times3m\times22.5}{64cm\times11.25cm\times6cm}$
=$\frac{800\times300\times22.5}{64\times11.25\times6}$
= 125000

Q.13: If the radius of a circle is equal to a diagonal of a square whose area is 12 cm2 , then the area of the circle is:
(A) 28$\pi$ cm2
(B) 36$\pi$ cm2
(C) 32$\pi$ cm2
(D) 24$\pi$ cm2
(SSC CHSL August 2021)

Ans : (D) 24$\pi$ cm2
Area of Square = 12
Side of Square =$2\sqrt3$
Diagonal of square= side x$\sqrt2$
=$2\sqrt6$
Radius of Circle = Diagonal of Square
= $2\sqrt6$
Area of circle =$\pi r^2$
=$\pi(2\sqrt6)^2$
=24$\pi$

Q.14: The sum of three sides of an isosceles triangle is 20 cm, and the ratio of an equal side to the base is 3 : 4, The altitude of the triangle is:
(A) $3\sqrt3$cm
(B) $4\sqrt5$cm
(C) $3\sqrt5$cm
(D) $2\sqrt5$cm
(SSC CHSL August 2021)

Ans : (D) $2\sqrt5$cm
Ration of side and base = 3 : 4
Sides 3x, 3x, 4x
Sum of sides = 20
3x + 3x + 4x = 20
10x = 20
x = 2
Sides = 6, 6, 8
Height =$\sqrt{6^2 - 4^2}$
= 36 – 16
=$\sqrt{20}$
=$2\sqrt5$

Q.15: If the volume of a sphere is $697\frac{4}{21}$cm3 , then its radius is:
$(\text{Take} \pi=\frac{22}{7})$
(A) 5 cm
(B) 6 cm
(C) 4.5 cm
(D) 5.5 cm
(SSC CHSL August 2021)

Ans : (D) 5.5 cm
Volume of Sphere =$\frac43\pi r^3$
$\frac43\pi r^3$ = $697\frac{4}{21}$
r3 =$\frac{14641\times3\times7}{21\times4\times22}$
r3 =$\frac{1331}{8}$
r =$\frac{11}{2}$
= 5.5 cm

Q.16: The area of the largest that can be inscribed in a semi-circle of radius 6 cm is:
(A) 38cm2
(B) 35cm2
(C) 36cm2
(D) 34cm2
(SSC CHSL August 2021)

Ans : (C) 36cm2
Base of Tringle = Diameter of semi-circle
Base of Triangle = 12
Height of Triangle = radius of semi-circle
Height= 6
Area of Triangle=$\frac12$ x base x height
=$\frac12$ x 12 x 6 = 36 cm2

Q.17: A solid metallic sphere of radius 12cm is melted and recast in the form of small spheres of radius 2 cm. How many small spheres are formed ?
(A) 96
(B) 864
(C) 216
(D) 24
(SSC CHSL August 2021)

Ans : (C) 216
Volume of Big sphere= V1
Volume of small sphere= V2
Number of Spheres =$\frac{V1}{V2}$
=$\frac {\frac{4}{3}\pi r^3}{\frac{4}{3}\pi r^3}$
=$\frac{12\times12\times12}{2\times2\times2}$
= 6 x 6 x 6 = 216

Q.18: The square of the diagonal of a cube is 2175 cm2 . What is the total surface area (in cm2 ) of the cube ?
(A) 4272
(B) 4305
(C) 4350
(D) 4530
(SSC CHSL August 2021)

Ans : (C) 4350
Diagonal of square =$a\sqrt3$
$(a\sqrt3)^2$ = 2175
a2 =$\frac{2175}{3}$
a2 = 725
total surface area = 6a2
= 6 x 725 = 4350

Q.19: A rectangle with perimeter 50 cm has its sides in the ratio 1 : 4. What is the perimeter of a square whose area is the same as that of the rectangle ?
(A) 45 cm
(B) 36 cm
(C) 50 cm
(D) 40 cm
(SSC CHSL August 2021)

Ans : (D) 40 cm
Let sides are x , 4x
Perimeter = 2(l + b)
2(x + 4x) = 50
5x = 25, x = 5
Length= 20 cm
Breath= 5 cm
Area= l x b = 20 x 5 = 100
As per question, Area of Rectangle = Area of Square
a2 = 100, a = 10
Perimeter of square= 4 x side = 4 x 10 = 40

Q.20: The inner and outer radius of two concentric are 6.7 cm and 9.5 cm, respectively. What is the difference between their circumferences (in cm ) ?$\text {Take} \pi=\frac{22}{7}$
(A) 10.4
(B) 17.6
(C) 6.5
(D) 20.5
(SSC CHSL August 2021)

Ans : (B) 17.6
Circumference of circle =$2\pi$r
Difference of Circumference
=$2\pi$R – $2\pi$r
=$2\times\frac{22}{7}$ [ 9.5 – 6.7]
=$2\times\frac{22}{7}\times 2.8$
= 17.6

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