CBSE Notes For Class 8 Maths Chapter 9 Mensuration

1.0Introduction 

In your previous math classes, you would have learned to calculate the perimeter and area of simple plane closed figures. Here, we will focus on the same, but for other figures like quadrilaterals, along with the surface area and volume of solids like cubes, cylinders, and cuboids.

2.0CBSE Class 8 Math Chapter 9 Mensuration- Revision Notes

Area of a Polygon

• The area is the space enclosed within a simple closed shape
• A polygon can be divided into simpler shapes whose area can be calculated by the math formulas.
• For example, this polygon has been divided into 4 parts, that is \triangle A F B, Trapezium BFGC, \triangle C G D, and \triangle A E D. The sum of these individual areas would give the area of the pentagon.

3.0Solid Shapes

• The faces of 3D figures are formed by 2D shapes. For example, the faces of a 3D cube are formed by 2D squares.
• Congruent faces are the faces of a solid shape that are formed by the same two-dimensional shape. A cube is formed by 6 congruent squares.
• Cylinders have parallel congruent circular faces. 

Surface Area of Cuboid

• The area of this cuboid can be calculated as the sum of I, II, III, IV, V, and VI. These areas can be calculated by using the math formula for the area of a rectangle, $\ell \times b$

Surface Area of Cube

• For a cube, simply add the area of all the congruent squares forming it. Alternatively, multiply the area of one square by 6.

Surface Area of the Cylinder

• A cylinder can be divided into two circles at both ends and a rectangle which is curved to form the tube.
• Hence, the surface area is given by $2\left(\pi r^2\right)+2\pi rh.$

Volume of Cuboid

Volume of a Cuboid = Length x breadth x height 

Volume of Cylinder

• The volume of a cylinder = Area of base x height =$\pi r^{2}h$
• Volume is the amount of space occupied, whereas capacity is the amount it can hold.

4.0Solving Questions Related to Mensuration

Question 1:  A cuboid-shaped tank is being filled with water at a rate of 60 liters per minute. If the total volume of the water reservoir is 108 m3, find the duration needed to fill the tank.

Solution: The volume of the reservoir is 108 cubic meters. Converted to liters, it would be 108 × 1000L = 108000L.

Since water is poured at 60 liters per minute, water poured in one hour is,

 60 × 60 L = 3600 L per hour.

If we take n as the number of hours, it can be derived by dividing the total volume of the tank by the amount of water poured.

So n = 108000/3600 = 30 hours.

Question 2: A coffee company sells its product in cylindrical containers with a base of diameter 14 cm and a height of 20 cm. There is a label around the surface of the cylinder, placed 2 cm from top and bottom. Find the area of the label.

Solution: Since the label is at a distance of 2 cm from both ends, its height will be 20 - 2 - 2 = 16cm.

The diameter of the label is the same as the container i.e., 14 cm.

So, the radius will be half of the diameter, which is 7 cm.

The area of the label is a cylinder, whose radius is 7 cm and height 16 cm. 

$A=2\times \pi \times 7\times 16$

$A=2\times \frac{22}{7}\times 7\times 16$

$A=704{\mathrm{cm}}^2$

5.0Key Features of CBSE Maths Notes for Class 8 Chapter 9

• Notes are aligned with the latest CBSE math curriculum.
• Easy to understand, with simple examples.
• Crisp and short explanations for quick revision.
• Notes are checked well and are 100% accurate.