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Q4(ii):
A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding : (ii) major sector. (Use $\pi = 3.14$)

Solution :

Given:

Radius of the circle ($r$) = $10\text{ cm}$

Angle subtended by the chord at the centre ($\theta$) = $90^\circ$

Value of $\pi$ = $3.14$

To Find:

Area of the major sector.

10 cm 10 cm 90° O

Step 1: Calculate the area of the entire circle.

The formula for the area of a circle is $A = \pi r^2$.

$A = 3.14 \times (10)^2$

$A = 3.14 \times 100$

$A = 314\text{ cm}^2$

Step 2: Calculate the area of the minor sector.

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

$\text{Area}_{\text{minor}} = \frac{90^\circ}{360^\circ} \times 314$

$\text{Area}_{\text{minor}} = \frac{1}{4} \times 314$

$\text{Area}_{\text{minor}} = 78.5\text{ cm}^2$

Step 3: Calculate the area of the major sector.

The area of the major sector is the difference between the area of the circle and the area of the minor sector.

$\text{Area}_{\text{major}} = \text{Area}_{\text{circle}} - \text{Area}_{\text{minor}}$

$\text{Area}_{\text{major}} = 314 - 78.5$

$\text{Area}_{\text{major}} = 235.5\text{ cm}^2$

Alternative Method (Using the reflex angle):

The angle of the major sector is $360^\circ - 90^\circ = 270^\circ$.

$\text{Area}_{\text{major}} = \frac{270^\circ}{360^\circ} \times \pi r^2$

$\text{Area}_{\text{major}} = \frac{3}{4} \times 314$

$\text{Area}_{\text{major}} = 3 \times 78.5 = 235.5\text{ cm}^2$

Final Answer: The area of the major sector is 235.5 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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