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Today, we are sharing Mensuration Maths Formulas for 2D and 3D Shapes (With PDF). It can prove very useful for upcoming competitive exams like SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, and many more. So mensuration formula pdf download in hindi is very important for any competitive exam. This free PDF will be very helpful for your exam.

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Our mensuration formulas for competitive exams pdf are very simple and easy. We cover basic topics like Maths, Geography, History, Polity, etc for the upcoming SSC CGL, BANK, RAILWAYS, RRB NTPC, LIC AAO, Exams including previous year Question Papers, Current Affairs, Important Formulas, etc. Our PDF will help you to upgrade your marks in any competitive exam. ### Important Terms Related to Mensuration Formula

Before understanding the mensuration formulas, we need to understand certain terms. These are –

● Perimeter: This is measured in units such as m, cm, etc and it is the measure of or sum of the continuous length of the boundary of a figure.

● Area: This is measured in square units such as m², cm², etc and it is the surface enclosed in a figure.  (Mensuration Formula PDF Download)

● Volume: This is measured in cubic units such as m³, cm³, etc, and is nothing but the space occupied by an object.

● Curved/Lateral Surface area: This is measured in square units such as m², cm², etc and it is the area of the curved surface in a figure.

● Total Surface area: This is measured in square units such as m², cm², etc and it is the area of the total surface in a figure including the top and bottom portions. #### Mensuration Questions with Solutions –

Q1. 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?

Solution: (△ABC) Perimeter/ (△PQR) Perimeter) = BC/QR

=> 15.6/46.8 = 19.5/x

=> x = 46.8 x 19.5 / 156

Q2. 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?

Solution: As, V = 4/3 π r³

Volume of big sphere = n x volume of small sphere

=> 4/3 π (10)³ = n x 4/3 π (2)³

=> n = 125

Q3. A circle has a radius of 21 cm. Find its circumference and area. (Use π = 22/7)

Solution: We know, Circumference of circle = 2πr = 2 x (22/7) x 21 = 2 x 22 x 3 = 132 cm

Area of circle = πr² = (22/7) x (21)² = 22/7 x 21 x 21 = 22 x 3 x 21

Q4. The length, width, height of a cuboidal box are 30 cm, 25 cm and 20 cm, respectively. Find its area.

Solution: Here, l- 30 cm, b= 25 cm, and h= 20 cm

Total surface area = 2 (lb +bh +hl)

= 2 (30 × 25 + 25 × 20 + 20 × 35)

TSA = 2( 750 + 500 + 700) = 3900 cm²

Q5. A rectangular piece of paper 11 cm × 4 cm is folded without overlapping to make a cylinder of the height of 4 cm. Find the volume of the cylinder.

Solution: The length of the paper will be the perimeter of the base of the cylinder and the width will be its height.

Circumference of base of cylinder = 2πr = 11 cm

2 x 22/7 x r = 11 cm (Mensuration Formula PDF Download)

r = 7/4 cm

Volume of cylinder = πr²h = (22/7) x (7/4)² x 4 = 38.5 cm³

Q7. The total surface area of a hemisphere is 166.32 cm², find its curved surface area?

Solution:  Let radio of hemisphere = r cm

Tital surface area of hemisphere = 3 π r² = 166.32 ……………….. (i)

Multiplying equation (i) by ⅔

=> ⅔ x 3 π r² = ⅔ x 166.32

=> Curved surface area of hemisphere = 2 π r² = 2 x 55.44 = 110.88 cm²

Q8. If G is the centroid and AD, BE, and CF are three medians of the triangle with 72 cm², then the area of triangle BDG is:

Solution: The area of the triangle formed by any two vertices and centroid is ⅓) times the area of △ABC. (Mensuration Formula PDF Download)

Also, the median divides the triangle into two equal areas.

So, the area of △BDG = 1/6 times of △ABC

=> 1/6 x 72

=> 12

Q9. Find the area of a rhombus whose diagonals are of lengths 9 cm and 7.2 cm.

Solution: Now, d1=9 cm, d2=7.2 cm where d1 and d2 are the lengths of diagonals of a rhombus.

Therefore, area of the given rhombus = ½ x 9 x 7.2 cm² = 32.4 cm²

Note: Here, the value of π is either considered as 22/7 or 3.14, r means radius, and h means height.

Q10. A rectangular paper of a width of 10 cm is rolled along its width and a cylinder of a radius of 18 cm is formed. Find the volume of the cylinder. (Take π=22/7)

Solution: A cylinder is formed by rolling a rectangle of 10 cm about its width. The radius of the cylinder is 18 cm and the width of the paper becomes the height.

Height of the cylinder, h = 10 cm and Radius of the cylinder, r = 18 cm

Now, volume of the cylinder, V = πr²h = (22/7) x (18)² x 10

= 10,183 cm³

Hence, the volume of the cylinder is 10183 cm³. (Mensuration Formula PDF Download)

### Also Check:-

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

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

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

Ans : (B) 108 m
Volume of sphere = Wire volume
\frac {4}{3} \pi r^3 = \pir2h
\frac 43 \times \pi \times 9^3 = \pix 0.3 x 0.3 x h
h =\frac {4\times9\times9\times9}{3\times0.3\times0.3}

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

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)

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

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

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 (Mensuration Formula PDF Download)

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

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 (Mensuration Formula PDF Download)

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

Ans : (D) 56
b – h = 17
Area = \frac12x 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 (Mensuration Formula PDF Download)

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

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

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

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

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

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}

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

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}

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

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
=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\sqrt3cm
(B) 4\sqrt5cm
(C) 3\sqrt5cm
(D) 2\sqrt5cm

Ans : (D) 2\sqrt5cm
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}

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

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}

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

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 (Mensuration Formula PDF Download)

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

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

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

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

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