NCERT Class X Chapter 11: Area Related To Circles Example 2
NCERT Class X Chapter 11: Area Related To Circles Example 2
Question:
Find the area of the segment AYB shown in Fig. 11.6, if radius of the circle is 21 cm and \( \angle AOB = 120^\circ \). (Use \( \pi = \frac{22}{7} \))
Given:
Radius of the circle, \( r = 21 \) cm
Central angle, \( \angle AOB = 120^\circ \)
\( \pi = \frac{22}{7} \)
To Find:
Area of the segment AYB
Formula:
Area of segment = Area of sector − Area of triangle
Area of sector = \( \frac{\theta}{360^\circ} \times \pi r^2 \)
Area of triangle (when two sides and included angle are known):
\( = \frac{1}{2} r^2 \sin\theta \)
Solution:
Step 1: Find the area of sector AOB.
$$ \text{Area of sector} = \frac{120^\circ}{360^\circ} \times \pi \times (21)^2 $$Step 2: Simplify the fraction and substitute the value of \( \pi \).
$$ = \frac{1}{3} \times \frac{22}{7} \times 441 $$Step 3: Calculate the area of the sector.
$$ = \frac{1}{3} \times \frac{22}{7} \times 441 \\ = \frac{22}{7} \times 147 \\ = 22 \times 21 \\ = 462~\text{cm}^2 $$Step 4: Find the area of triangle AOB using the sine formula.
$$ \text{Area of } \triangle AOB = \frac{1}{2} \times 21 \times 21 \times \sin 120^\circ $$Step 5: Substitute \( \sin 120^\circ = \frac{\sqrt{3}}{2} \) and calculate.
$$ = \frac{1}{2} \times 441 \times \frac{\sqrt{3}}{2} \\ = \frac{441 \sqrt{3}}{4} $$Step 6: Find the approximate value of the area of triangle AOB.
$$ \sqrt{3} \approx 1.732 \\ \text{Area} = \frac{441 \times 1.732}{4} \\ = \frac{763.752}{4} \\ = 190.938~\text{cm}^2 \\ \text{(rounded to two decimal places: } 190.94~\text{cm}^2) $$Step 7: Find the area of the segment AYB.
$$ \text{Area of segment AYB} = \text{Area of sector} - \text{Area of triangle} \\ = 462 - 190.94 \\ = 271.06~\text{cm}^2 $$Result:
The area of the segment AYB is approximately 271.06 cm2.
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