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Q7:
A chord of a circle of radius 12 cm subtends an angle of 120° at the centre. Find the area of the corresponding segment of the circle. (Use $\pi = 3.14$ and $\sqrt{3} = 1.73$)

Solution :

Given:

Radius of the circle ($r$) = $12\text{ cm}$

Angle subtended by the chord at the centre ($\theta$) = $120^\circ$

Constants: $\pi = 3.14$, $\sqrt{3} = 1.73$

To Find:

Area of the corresponding segment of the circle.

O r=12 r=12 120°

Step 1: Formula for the Area of a Segment

The area of a segment of a circle is given by the formula:

$\text{Area of Segment} = \text{Area of Sector} - \text{Area of Triangle}$

$\text{Area of Segment} = \left( \frac{\theta}{360^\circ} \times \pi r^2 \right) - \left( \frac{1}{2} r^2 \sin \theta \right)$

Step 2: Calculating the Area of the Sector

Substitute the given values into the sector area formula:

$\text{Area of Sector} = \frac{120}{360} \times 3.14 \times (12)^2$

$\text{Area of Sector} = \frac{1}{3} \times 3.14 \times 144$

$\text{Area of Sector} = 3.14 \times 48$

$\text{Area of Sector} = 150.72\text{ cm}^2$

Step 3: Calculating the Area of the Triangle

The area of the triangle formed by two radii and the chord is $\frac{1}{2} r^2 \sin \theta$.

$\text{Area of Triangle} = \frac{1}{2} \times (12)^2 \times \sin(120^\circ)$

Since $\sin(120^\circ) = \sin(180^\circ - 60^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$:

$\text{Area of Triangle} = \frac{1}{2} \times 144 \times \frac{\sqrt{3}}{2}$

$\text{Area of Triangle} = 72 \times \frac{1.73}{2}$

$\text{Area of Triangle} = 36 \times 1.73$

$\text{Area of Triangle} = 62.28\text{ cm}^2$

Step 4: Calculating the Area of the Segment

Subtract the area of the triangle from the area of the sector:

$\text{Area of Segment} = 150.72 - 62.28$

$\text{Area of Segment} = 88.44\text{ cm}^2$

Final Answer: The area of the corresponding segment of the circle is 88.44 cm².


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


CBSE Solutions for Class 10 Mathematics Areas Related to Circles


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