NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 11: Area Related To Circles - Exercise 11.3

Question 9

Step 1: Find area of the outer rectangle

Length = 26 m, Width = 12 m

Area = length × width = \(26 \times 12 = 312\,\text{m}^2\)

Step 2: Understand the inner shape

The inner figure is like a small "stadium".

• Its total length = 20 m
• Its width = 4 m

This shape is made up of:

• a rectangle in the middle
• and two semicircles (together making one full circle) at the ends

Step 3: Radius of the semicircles

Width = 4 m, so diameter = 4 m

Radius = diameter ÷ 2 = \(4 \div 2 = 2\,\text{m}\)

Step 4: Length of the rectangle part inside

Total length = 20 m

But at both ends, circles take up \(2r = 4\,\text{m}\).

So rectangle length = total length − 2r = \(20 − 4 = 16\,\text{m}\)

Step 5: Area of the inner stadium shape

• Rectangle area = length × width = \(16 \times 4 = 64\,\text{m}^2\)
• Circle area = \(\pi r^2 = \pi \times 2^2 = 4\pi\,\text{m}^2\)

Total inner area = rectangle + circle = \(64 + 4\pi\,\text{m}^2\)

Step 6: Area of shaded region

Shaded = Outer rectangle − Inner shape

= \(312 − (64 + 4\pi) = 248 − 4\pi\,\text{m}^2\)

Step 7: Approximate value

Take \(\pi = 3.1416\)

\(248 − 4\pi = 248 − 12.5664 ≈ 235.43\,\text{m}^2\)