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Q5(iii):
In a circle of radius 21 cm, an arc subtends an angle of 60° at the centre. Find: (iii) area of the segment formed by the corresponding chord (Unless stated otherwise, use $\pi = \frac{22}{7}$)

Solution :

Given:

  • Radius of the circle ($r$) = $21\text{ cm}$
  • Angle subtended by the arc at the centre ($\theta$) = $60^\circ$
  • Value of $\pi = \frac{22}{7}$

To Find:

The area of the segment formed by the corresponding chord.

O A B 60° 21 cm

Step 1: Understanding the Formula

The area of a segment of a circle is calculated by subtracting the area of the triangle formed by the two radii and the chord from the area of the corresponding sector.

Formula: $\text{Area of Segment} = \text{Area of Sector} - \text{Area of } \triangle OAB$

Step 2: Calculating the Area of the Sector

The formula for the area of a sector is $\frac{\theta}{360^\circ} \times \pi r^2$.

Substituting the given values:

$\text{Area of Sector} = \frac{60}{360} \times \frac{22}{7} \times 21 \times 21$

$\text{Area of Sector} = \frac{1}{6} \times 22 \times 3 \times 21$ [Since $\frac{21}{7} = 3$]

$\text{Area of Sector} = \frac{1}{2} \times 22 \times 3 \times 7$ [Simplifying $\frac{3}{6} = \frac{1}{2}$]

$\text{Area of Sector} = 11 \times 3 \times 7 = 231\text{ cm}^2$

Step 3: Calculating the Area of $\triangle OAB$

Since the central angle $\theta = 60^\circ$ and the two sides $OA = OB = r = 21\text{ cm}$, the triangle is an equilateral triangle.

The formula for the area of an equilateral triangle is $\frac{\sqrt{3}}{4} \times (\text{side})^2$.

$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times (21)^2$

$\text{Area of } \triangle OAB = \frac{\sqrt{3}}{4} \times 441$

$\text{Area of } \triangle OAB = 110.25\sqrt{3}\text{ cm}^2$

Step 4: Calculating the Area of the Segment

$\text{Area of Segment} = \text{Area of Sector} - \text{Area of } \triangle OAB$

$\text{Area of Segment} = 231 - 110.25\sqrt{3}$

Using $\sqrt{3} \approx 1.732$:

$\text{Area of Segment} = 231 - 110.25(1.732)$

$\text{Area of Segment} = 231 - 190.953 = 40.047\text{ cm}^2$

Final Answer: The area of the segment is $(231 - 110.25\sqrt{3})\text{ cm}^2$ or approximately $40.05\text{ cm}^2$.


More Questions from Class 10 Mathematics Areas Related to Circles EXERCISE 11.1


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