NCERT Class X Chapter 12: Surface Areas And Volumes Example 2

NCERT Class X Chapter 12: Surface Areas and Volumes

Question:

The decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Use \( \pi = \frac{22}{7} \))

Given:

Edge of cube, \( a = 5 \) cm
Diameter of hemisphere, \( d = 4.2 \) cm
Radius of hemisphere, \( r = \frac{4.2}{2} = 2.1 \) cm

To Find:

Total surface area of the block

Formula:

Surface area of cube: \( 6a^2 \)
Curved surface area of hemisphere: \( 2\pi r^2 \)
Area of base of hemisphere: \( \pi r^2 \)
Total surface area of block:
\( \text{Surface area of cube} + \text{Curved surface area of hemisphere} - \text{Area of base of hemisphere} \)

Solution:

Step 1: Find the surface area of the cube.

$$ \text{Surface area of cube} = 6a^2 = 6 \times (5)^2 = 6 \times 25 = 150 \ \text{cm}^2 $$

Step 2: Find the curved surface area of the hemisphere.

$$ \text{Curved surface area of hemisphere} = 2\pi r^2 = 2 \times \frac{22}{7} \times (2.1)^2 $$ $$ = 2 \times \frac{22}{7} \times 4.41 $$ $$ = 2 \times \frac{22 \times 4.41}{7} $$ $$ = 2 \times \frac{97.02}{7} $$ $$ = 2 \times 13.86 = 27.72 \ \text{cm}^2 $$

Step 3: Find the area of the base of the hemisphere (which is not exposed).

$$ \text{Area of base of hemisphere} = \pi r^2 = \frac{22}{7} \times (2.1)^2 $$ $$ = \frac{22}{7} \times 4.41 $$ $$ = \frac{22 \times 4.41}{7} $$ $$ = \frac{97.02}{7} = 13.86 \ \text{cm}^2 $$

Step 4: Calculate the total surface area of the block.

$$ \text{Total surface area} = \text{Surface area of cube} + \text{Curved surface area of hemisphere} - \text{Area of base of hemisphere} $$ $$ = 150 + 27.72 - 13.86 $$ $$ = 163.86 \ \text{cm}^2 $$

Result:

The total surface area of the block is approximately \( 163.86 \ \text{cm}^2 \).

Next question solution:

Example 3
© Kaliyuga Ekalavya. All rights reserved.

Comments

Post a Comment