NCERT Class X Chapter 11: Area Related To Circles Exercise 11.1 Question 5

NCERT Class X Chapter 11: Area Related To Circles

Question:

Unless stated otherwise, use \( \pi = \frac{22}{7} \). In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find:
(i) the length of the arc
(ii) area of the sector formed by the arc
(iii) area of the segment formed by the corresponding chord

Given:

Radius \( r = 21 \) cm
Central angle \( \theta = 60^\circ \)

To Find:

(i) Length of the arc
(ii) Area of the sector
(iii) Area of the segment formed by the corresponding chord

Formula:

(i) Length of arc:   \( L = \frac{\theta}{360^\circ} \times 2\pi r \)
(ii) Area of sector:   \( A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 \)
(iii) Area of segment:   \( A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} \)

Solution:

Step 1: Find the length of the arc using the formula.

$$ L = \frac{\theta}{360^\circ} \times 2\pi r = \frac{60^\circ}{360^\circ} \times 2 \times \frac{22}{7} \times 21 $$ $$ = \frac{1}{6} \times 2 \times \frac{22}{7} \times 21 $$ $$ = \frac{1}{6} \times 2 \times 22 \times 3 $$ $$ = \frac{1}{6} \times 132 $$ $$ = 22 \text{ cm} $$

Step 2: Find the area of the sector using the formula.

$$ A_{\text{sector}} = \frac{\theta}{360^\circ} \times \pi r^2 = \frac{60^\circ}{360^\circ} \times \frac{22}{7} \times (21)^2 $$ $$ = \frac{1}{6} \times \frac{22}{7} \times 441 $$ $$ = \frac{1}{6} \times 22 \times 63 $$ $$ = \frac{1}{6} \times 1386 $$ $$ = 231 \text{ cm}^2 $$

Step 3: Find the area of the triangle formed by the two radii and the chord (using the formula for area of an isosceles triangle with angle \( \theta \) at the center).

$$ \text{Area of } \triangle = \frac{1}{2} r^2 \sin\theta $$ $$ = \frac{1}{2} \times 21^2 \times \sin 60^\circ $$ $$ = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} $$ $$ = \frac{441}{4} \sqrt{3} $$ $$ \approx 190.95 \text{ cm}^2 $$

Step 4: Find the area of the segment by subtracting the area of the triangle from the area of the sector.

$$ A_{\text{segment}} = A_{\text{sector}} - A_{\triangle} $$ $$ = 231 - 190.95 $$ $$ \approx 40.05 \text{ cm}^2 $$

Result:

(i) Length of arc = \( 22 \) cm
(ii) Area of sector = \( 231 \) cm\(^2\)
(iii) Area of segment ≈ \( 40.05 \) cm\(^2\)

Next question solution:

NCERT Class X Chapter 11: Area Related To Circles Exercise 11.1 Question 6

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