Mensuration: Formulas, Tricks, Examples and Online Test

Mensuration is the skill of measuring the length of lines, areas of surfaces, and volumes of solids from simple data of lines and angles. Mensuration in its literal meaning is to measure. It is generally used where geometrical figures are concerned, where one has to determine various physical quantities such as perimeter, area, volume or length. Measuring these quantities is called Mensuration.

  • Perimeter: A perimeter is a path that surrounds an area. The term may be used either for the path or its length – it can be thought of as the length of the outline of a shape. The perimeter of a circular area is called its circumference. It is measured in cm, m, etc.
  • Area: Area is a quantity that expresses the extent of a two-dimensional surface or shape in the plane. Area can be understood as the amount of material with a given thickness that would be necessary to fashion a model of the shape, or the amount of paint necessary to cover the surface with a single coat. This is measured in square unit like cm², m², etc.
  • Volume: Volume is the amount of space enclosed by a shape or object, how much 3-dimensional space (length, width, and height) it occupies. This is measured in cubic unit like cm³, m³, etc.

Basic Conversions

Units related to area

  • 1 Hactare = 10000 metre square
  • 1 kilometre square = 1000000 metre square
  • 1 Decametre = 100 metre square
  • 1 Decimeter square = 1/100 metre square
  • 1 Centimeter square = 1/10000 metre square
  • 1 Milimeter square = 1/000000 metre square

Units related to volume

  • 1 litre = 1000 cm³
  • 1 Hectometer³ = 10000 meter³
  • 1 Decameter³ = 100 meter³
  • 1 Meter³ = 1000000 cm³
  • 1 Decimeter³ = 100 cm³
  • 1 Milimeter³ = 1/100 cm³

Two Dimensional Figures

Triangle

mensuration-f-19774.png

Perimeter (P) = a + b + c

Area (A) mensuration-f-19783.png

where mensuration-f-19789.png

and a, b and c are three sides of the triangle.

Also, mensuration-f-19795.png; where b is base and h is altitude

Equilateral triangle

mensuration-f-19801.png

Perimeter = 3a

Area mensuration-f-19812.png; where a is a side

Right angled triangle

mensuration-f-19818.png

Area mensuration-f-19824.pngand h² = p² + b² (Pythagoras triplet)

where p is perpendicular, b is base and h is hypotenuse


Example 1: The base of a triangular field is 880 m and its height 550 m. Find the area of the field. Also calculate the charges for supplying water to the field at the rate of Rs. 24.25 per sq. hectometer.

Solution: Area of the field mensuration-f-19830.png

mensuration-f-19836.png sq.m. = 24.20 sq.hm

Cost of supplying water to 1 sq. hm = Rs. 24.25

∴ Cost of supplying water to the whole field

= 24.20 × 24.25 = Rs. 586.85


Example 2: Find the area of a triangle whose sides are 50 m, 78 m, 112 m respectively and also find the perpendicular from the opposite angle on the side 112 m.

Solution: Here a = 50 m, b = 78 m, c = 112m

mensuration-f-19843.png

s – a = 120 – 50 = 70 m

s – b = 120 – 78 = 42 m

s – c = 120 – 112 = 8 m

∴ Area mensuration-f-19853.png= 1680 sq.m.

mensuration-f-19859.png

∴ Perpendicular mensuration-f-19865.png

Rectangle

mensuration-f-19871.png

Area = Length × breadth

Perimeter = 2(Length + breadth)

Diagonal = mensuration-f-19877.png


Example 3: The length and breadth of a rectangle are in the ratio 9 : 5. If its area is 720 m², find its perimeter.

Solution: Let the length and breadth of a rectangle are 9x m and 5x m respectively.

In a rectangle, area = length × breadth

∴ 720 = 9x × 5x

or x² = 16

⇒ x = 4

Thus, length = 9 × 4 = 36 m

and breadth = 5 × 4 = 20 m

Therefore, perimeter of rectangle = 2(36 + 20) = 112 m

Square

mensuration-f-19884.png

Area = (side) ²

Perimeter = 4 × side

Diagonal = mensuration-f-19892.png× side


Example 4: How many squares are there in a 5 inch by 5 inch square grid, if the grid is made up one inch by one inch squares?

Solution: Required no. of squares mensuration-f-19899.png

Parallelogram

mensuration-f-19905.png

Area = base × height

Perimeter = 2 × (Side A + Side B)

Rhombus

mensuration-f-19911.png

Perimeter = 4 a

Area mensuration-f-19917.pngwhere a is side and d¹ and d²are diagonals.


Example 5: The perimeter of a rhombus is 146 cm and one of its diagonals is 55 cm. Find the other diagonal and the area of the rhombus.

Solution: Let ABCD be the rhombus in which AC = 55 cm.

mensuration-f-19923.png

and AB mensuration-f-19933.png

Also, AO mensuration-f-19940.png

∴ BO mensuration-f-19946.png= 24 cm

Hence, the other diagonal BD = 48 cm

Now, Area of the rhombus

mensuration-f-19956.png mensuration-f-19962.png = 1320 sq. cm.

Irregular Quadrilateral

mensuration-f-19968.png

Perimeter = AD + DC + BC + AB

Area = ½ × (DP + BQ) × AC


Example 6: Find the area of a quadrilateral piece of ground, one of whose diagonals is 60 m long and the perpendicular from the other two vertices are 38 and 22 m respectively.

Solution: Area mensuration-f-19986.png mensuration-f-19998.png

Trapezium

mensuration-f-20004.png

Perimeter = a + b + m + n

Area mensuration-f-20010.png;where a and b are two parallel sides;

m and n are two non-parallel sides; h is perpendicular to b


Example 6: A 5100 sq. cm trapezium has the perpendicular distance between the two parallel sides 60 m. If one of the parallel sides be 40m then find the length of the other parallel side.

Solution: Since, mensuration-f-20016.png

⇒ 5100 mensuration-f-20025.png⇒ 170 = 40 + x

∴ other parallel side = 170 – 40 = 130 m

Circle

mensuration-f-20031.png

Area = π × (Radus) ²

Circumference = 2π × Radius

Radius = mensuration-f-20052.png

π mensuration-f-20061.png or 3.14


Example 7: If area of a circular jogging track is 3850 sq. metres. What is the circumference of the jogging track?

Solution: Let the radius of the circular jogging track be r metre.

mensuration-f-20067.png

or mensuration-f-20074.png

or r² = mensuration-f-20080.png

∴ r = mensuration-f-20089.png = 35 metre

∴ Circumference = mensuration-f-20097.png= 220 metre


Example 8: The radius of a wheel is 42 cm. How many revolutions will it make in going 26.4 km?

Solution: Distance travelled in one revolution = Circumference of the wheel

= 2πr = mensuration-f-20107.png= 264 cm

∴ No. of revolutions required to travel 26.4 km

= mensuration-f-20113.png = 10000

Semicircle

mensuration-f-20119.png

Perimeter = πr + 2r Area mensuration-f-20128.png

Sector of a Circle

mensuration-f-20134.png

Area of sector mensuration-f-20150.png

Length of an arc (ℓ) =mensuration-f-20156.png

Area of segment = Area of sector – Area of triangle OAB

mensuration-f-20162.png

Perimeter of segment

= length of the arc + length of segment AB mensuration-f-20168.png


Example 9: Find the area of sector of a circle whose radius is 6 cm when:

  1. the angle at the centre is 35°
  2. when the length of arc is 22 cm

Solution:

1. Area of sector

= mensuration-f-20174.png

2. Here length of arc ℓ = 22 cm

mensuration-f-20181.png

Area of sector

= mensuration-f-20196.png

= mensuration-f-20202.png

Ring

mensuration-f-20208.png

Area of ring mensuration-f-20215.png


Example 10: The circumference of a circular garden is 1012 m. Find the area of outsider road of 3.5 m width runs around it. Calculate the area of this road and find the cost of gravelling the road at Rs 32 per 100 sqm.

Solution: A = πr², C = 2πr = 1012

mensuration-f-18740.png

mensuration-f-20222.png

∴ Area of garden

mensuration-f-20228.pngsq. m

Area of the road = area of bigger circle – area of the garden

Now, radius of bigger circle

= 161 + 3.5 mensuration-f-20234.png m

∴ Area of bigger circle

mensuration-f-20240.png= mensuration-f-20247.png

Thus, area of the road

mensuration-f-20253.png mensuration-f-20259.png sqm.

Hence, cost

mensuration-f-20266.png = Rs. 1145.76

Three Dimensional Figures

Cuboid

A cuboid is a three dimensional box.

mensuration-f-20272.png

If L, B and H are length, breadth and height of the cuboid, then Volume = L × B × H

Surface area = 2 (L × B + B × H + H + L)

Diagonal = mensuration-f-20281.png

Cube

A cube is a cuboid which has all its edges equal.

mensuration-f-20287.png

If a is the each side of cube, then

Volume = a × a × a = a³

Whole surface area = 2(a × a + a × a + a × a) = 6a²

Diagonal of cube

= mensuration-f-20293.png

Right Prism

A prism is a solid which can have any polygon at both its ends.

mensuration-f-20299.png

Lateral or curved surface area = Perimeter of base × height

Total surface area = Lateral surface area + 2 (area of the end)

Volume = Area of base × height

Right Circular Cylinder

It is a solid which has both its ends in the form of a circle.

mensuration-f-20306.png

If radius of cylinder is r and height or length is h, then

Volume = πr²h

Lateral surface Area = 2πrh

Whole surface area = (2πrh + 2πr²)


Example 11: A road roller of diameter 1.75 m and length 1 m has to press a ground of area 1100 sqm. How many revolutions does it make?

Solution: Area covered in one revolution = curved surface area

∴ Number of revolutions

mensuration-f-20316.png

mensuration-f-20322.png

mensuration-f-20328.png

= 200

Pyramid

A pyramid is a solid which can have any polygon at its base and its edges converge to single apex.

mensuration-f-20334.png

Lateral or curved surface area= 1/2 (perimeter of base) × slant height

Total surface area = lateral surface area + area of the base

Volume = 1/3 (area of the base) × height


Example 12: A right pyramid, 12 cm high, has a square base each side of which is 10 cm. Find the volume of the pyramid.

Solution: Area of the base = 10 × 10 = 100 sq.cm.

Height = 12 cm

∴ Volume of the pyramid

mensuration-f-20353.png= 400 cu.cm.

Sphere

It is a solid in the form of a ball with radius r.

mensuration-f-20428.png

If r is the radius of sphere, then volume = mensuration-f-20434.png

Surface Area = 4πr²

Hemisphere

It is a solid half of the sphere.

mensuration-f-20441.png

Volume = mensuration-f-20450.png

Surface area = 2πr²

Total surface area = 2πr² + πr² = 3πr²

Right Circular Cone

It is a solid which has a circle as its base and a slanting lateral surface that converges at the apex.

mensuration-f-20359.png

If base-radius, height and vertical height of a cone is r, h and ℓ respectively, then

Volume = mensuration-f-20365.png

Lateral surface area = πrl

Total surface area = πrl + πr²

Vertical height = ℓ = mensuration-f-20376.png


Example 13: A frustum of a right circular cone has a diameter of base 10 cm, top of 6 cm, and a height of 5 cm; find the area of its whole surface and volume.

Solution: Here r¹ = 5 cm, r² = 3 cm and h = 5 cm.

mensuration-f-20382.png mensuration-f-20388.png mensuration-f-20394.png cm = 5.385 cm

∴ Whole surface of the frustum mensuration-f-20400.png

mensuration-f-20406.png = 242.25 sq.cm.

Volume =mensuration-f-20413.png mensuration-f-20420.png= 256.67 cu. cm.

Frustum of a Cone

When a cone cut the left over part is called the frustum of the cone.

mensuration-f-20456.png

Curved surface area = πℓ (r₁ + r₂)

Total surface area

mensuration-f-20462.png

where mensuration-f-20469.png

Volume mensuration-f-20475.png


Example 14: The height of a bucket is 45 cm. The radii of the two circular ends are 28 cm and 7 cm, respectively. The volume of the bucket is :

  1. 38610 cm³
  2. 48600 cm³
  3. 48510 cm³
  4. None of these

Solution : (3): Here r¹ = 7 cm, r² = 28 cm and h = 45 cm

mensuration-f-20481.png

Volume of the bucket mensuration-f-20489.png

Hence, the required volume

mensuration-f-20540.png

= 48510 cm³

Area of Pathways

Running across the middle of a rectangle

mensuration-f-20552.png

A = a (ℓ + b) – a²; where ℓ → length

b → breadth,a →width of the pathway.

Pathways outside

mensuration-f-20558.png

A = (l + 2a) (b + 2a) – lb;where l → length

b → breadtha → width of the pathway

Pathways inside

mensuration-f-20570.png

A = lb – (l – 2a) (b – 2a) ; where l → length

b → breadth