Mensuration Class 8 Notes for Chapter 11

What is Mensuration

Mensuration is one of the important chapters of class 8 maths. It consists the topics about measurement of geometric figures and their parameters like surface area, length, volume, shape, etc.

Perimeter & Area

Perimeter is the total distance along the boundary of any shape while the area of any shape is the region covered by it.

Formula for perimeter and area

Let us see the formula for perimeter and area of basic shapes.

• Perimeter of a square with side a: 4a
• Area of a square with side a: a2
• Perimeter of a rectangle with length l and breadth b: 2(l + b)
• Area of a rectangle with length l and breadth b: l × b
• Area of a triangle with base b and height h: 1/2 × b × h
• Area of a parallelogram with base b and height h: b × h
• Perimeter of a circle with radius r = 2 × π × r
• Area of a circle with radius r = π × r2

Perimeter and Area of Quadrilaterals

Area of a trapezium = 1/2 × (sum of parallel sides) × height
Area of a rhombus = 1/2 × product of the diagonals
Area of a quadrilateral = 1/2 × diagonal × sum of heights of triangles formed by the diagonal

Let us see an example.

Find the area of the trapezium whose parallel sides are 50 m and 45 m respectively and the distance between the parallel sides is 30 m.

Sum of parallel sides = 50 + 45 = 95 m
Distance between parallel sides is the height of the trapezium, that is, h = 30 m.

NCERT Solutions for Class 8 Maths Chapter 11
NCERT Solutions Class 8 Maths Chapter 11
NCERT Solutions Class 8 Maths Exercise 11.1
NCERT Solutions Class 8 Maths Exercise 11.2
NCERT Solutions Class 8 Maths Exercise 11.3
NCERT Solutions Class 8 Maths Exercise 11.4

Surface Area

Surface area is the sum of the areas of all faces of a 3D shape.

Cuboid

To understand the surface area of cuboids, we will take an example of a room. In a room, there are four walls and two faces, one face at the bottom and another face at the top. The top and the bottom faces are identical which contain length and breadth.

Similarly, opposite walls are identical, which contain the length and the height, and another pair of opposite walls contains the breadth and the height.

Total Surface Area = Area of two rectangles with length and breadth + area of two rectangles with breadth and height + area of two rectangles with length and height.

Let the length be l, breadth be b and height be h.
Total Surface Area = 2lb + 2bh + 2hl = 2 (lb + bh + hl)
The lateral surface area of a cuboid is the sum of the areas of the four walls or side faces (without the area of the roof and the floor).

Hence, Lateral Surface Area = Area of four walls = 2lh + 2bh = 2(l + b)h

Cube

If all the 3 dimensions of the cuboid are equal, then it will become a cube. In a cube, length = breadth = height.

Let the length, breadth, and height of a cube be a.
Total surface area of a cuboid is 2(lb + bh + hl)
Substitute, l = b = h = a,
Total Surface Area = 2 (a × a + a × a + a × a) = 6a2
Lateral Surface Area = 4a2

Cylinder

Consider a can that is in the shape of a right circular cylinder. If we wrap a rectangular sheet around its curved surface, the length of the rectangular sheet will be the circumference of the circle and the breadth of the sheet will be the height of the cylinder.

Let the height of the cylinder be h, and the radius be r.
Curved Surface Area = Area of Rectangle
Length of Rectangle (l) = Circumference of Circle = 2πr
Breadth of Rectangle (b) = Height of Cylinder = h
Area of Rectangle (lb) = 2πr × h = 2πrh
Curved Surface Area = 2πrh

For the total surface area of the cylinder, we need to add the areas of two circles that are present on the top and the bottom of the cylinder.

Total surface area = 2πrh + πr2 + πr2 = 2πrh + 2πr2
Total Surface Area = 2πr(r + h)


✍ Also learn: Find Surface Area of Cylinder

Let us see an example.

Find the curved surface area and the total surface area of the cylinder with radius 21 cm and height 25 cm. (take π = 22/7)

Here, r = 21 cm and h = 25 cm

Total surface area = 2πr(r + h)

= [2 ×22/7 × 21 (21 + 25)] cm2
= 132 (46) cm2
= 6072 cm2

Curved surface area = 2πrh

= [2 × 22/7 × 21 × 25] cm2
= 3300 cm2

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Volume

The space occupied by a three-dimensional object or the space enclosed by a boundary of a 3D object is called its volume.

Formulae

• For a cuboid with length l, breadth b, and height h, volume = lbh
• For a cube with side a, volume = a3
• For a cylinder with radius r, and height h, volume = πr2h

Examples:

How many bricks are required to construct a wall of dimension 155 cm × 55 cm × 30 cm? The dimension of each brick is 5 cm × 5 cm × 5 cm.

Volume of the wall, V = L × B × W

V = 155 cm × 55 cm × 30 cm
V = 255750 cm3
Volume of a brick, v = l × b × w
v = 5 cm × 5 cm × 5 cm
v = 125 cm3

Number of bricks required



What is the height of cylindrical vessel whose volume and radius are 88 cm3 and 2 cm respectively?

Height of the cylinder, h = ?
Radius of the cylinder, r = 2 cm
Volume of the cylinder, V = 88 cm3
Volume of the cylinder, V = πr2h
88 cm3 = 22/7 × (2 cm)2 × h
88 cm3 = 22/7 × 4 cm2 × h

Hence, the height of the cylindrical vessel is 7 cm.

Links to Other Resources for Class 8 Maths

Other Class 8 Maths Related Links
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