**Mensuration Formula PDF Download**

**Mensuration Formula PDF Download**

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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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**Mensuration Formula PDF Download**

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### Mensuration Formulas for 2-D Figures

Mensuration Formulas for 2-Dimensional Figures | ||

Shape | Area | Perimeter |

Circle | πr² | 2 π r |

Square | (side)² | 4 × side |

Rectangle | length × breadth | 2 (length + breadth) |

Scalene Triangle | √[s(s−a)(s−b)(s−c), Where, s = (a+b+c)/2 | a+b+c (sum of sides) |

Isosceles Triangle | ½ × base × height | 2a + b (sum of sides) |

Equilateral Triangle | (√3/4) × (side)² | 3 × side |

Right Angled Triangle | ½ × base × hypotenuse | A + B + hypotenuse, where the hypotenuse is √A²+B² |

Parallelogram | base × height | 2(l+b) |

Rhombus | ½ × diagonal1 × diagonal2 | 4 × side |

Trapezium | ½ h(sum of parallel sides) | a+b+c+d (sum of all sides) |

### Mensuration Formulas for 3-D Figures

Mensuration Formulas for 3-Dimensional Figures | |||
---|---|---|---|

Shape | Area | Curved Surface Area (CSA)/ Lateral Surface Area (LSA) | Total Surface Area (TSA) |

Cone | (1/3) π r² h | π r l | πr (r + l) |

Cube | (side)³ | 4 (side)² | 6 (side)² |

Cuboid | length × breadth × height | 2 height (length + breadth) | 2 (lb +bh +hl) |

Cylinder | π r² h | 2π r h | 2πrh + 2πr² |

Hemisphere | (2/3) π r³ | 2 π r² | 3 π r² |

Sphere | 4/3πr³ | 4πr² | 4πr² |

### Differences between 2-Dimensional and

### 3-Dimensional Figures

2-Dimensional Figures | 3-Dimensional Figures |

As the name suggests, a 2D shape means that it will have only 2 dimensions which are length and breadth. | Here, a 3D shape means that this figure will have 3 dimensions, which are length, breadth, and height. |

For 2D shapes, we can calculate 2 things i.e. area and perimeter. | For 3D shapes, we can calculate their volume, total surface area, and curved/lateral surface area. |

2D shapes are flat as they do not have depth and also, these cannot be held physically because of the lack of depth. | 3D shapes contain a depth so they can be held physically and are not flat like 2D shapes. |

Example: Square, Rectangle, Triangle, Circle, etc. | Example: Cone, Cylinder, Sphere, Cube, Prism, Pyramid, etc. |

### 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 Formula PDF Download**

**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

=> 5.85 cm (Mensuration Formula PDF Download)

**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

Area of circle with radius, 21cm = 1386 cm² (Mensuration Formula PDF Download)

**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.

Area of rhombus formula is ½ (d1xd2) (Mensuration Formula PDF Download)

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)

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**Mensuration Questions Answers:-**

**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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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}

h = 10800 cm = 108 m (Mensuration Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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 (Mensuration Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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 (Mensuration Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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 Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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 (Mensuration Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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 (Mensuration Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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 (Mensuration Formula PDF Download)

=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

(Mensuration Formula PDF Download)

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}

=2\sqrt5 (Mensuration Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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 (Mensuration Formula PDF Download)

**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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

(Mensuration Formula PDF Download)

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

**Mensuration Formula PDF Download**

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