Question

Area of minor segment if a chord of a circle of radius 10 cm subtends a right angle at the centre is (use \( \pi = 3.14 \)):

• A. 28 cm\(^2\)
• B. 28.5 cm\(^2\)
• C. 27 cm\(^2\)
• D. 27.5 cm\(^2\)

Hint:

For a right-angled triangle in a circle with the center at the vertex, the area of the minor segment is given by the difference between the area of the sector and the area of the triangle.

Answer 1

The Correct Option is B

Step 1: Understand the problem setup
We are given a circle with a radius of 10 cm.
A chord of the circle subtends a right angle at the center of the circle.
Step 2: Find the area of the sector
The formula for the area of a sector of a circle is given by:
\[ \text{Area of Sector} = \frac{\theta}{360^\circ} \times \pi r^2 \] where \( \theta = 90^\circ \) (since the chord subtends a right angle), and \( r = 10 \, \text{cm} \). Substitute the values: \[ \text{Area of Sector} = \frac{90^\circ}{360^\circ} \times 3.14 \times 10^2 \] \[ \text{Area of Sector} = \frac{1}{4} \times 3.14 \times 100 = 78.5 \, \text{cm}^2 \] Step 3: Find the area of the triangle
The area of the triangle formed by the two radii and the chord can be found using the formula for the area of a right triangle:
\[ \text{Area of Triangle} = \frac{1}{2} \times \text{base} \times \text{height} \] Here, the base and height are both the radius \( r = 10 \, \text{cm} \), since the triangle is isosceles and the angle at the center is \( 90^\circ \). \[ \text{Area of Triangle} = \frac{1}{2} \times 10 \times 10 = 50 \, \text{cm}^2 \] Step 4: Find the area of the minor segment
The area of the minor segment is the area of the sector minus the area of the triangle: \[ \text{Area of Minor Segment} = \text{Area of Sector} - \text{Area of Triangle} \] \[ \text{Area of Minor Segment} = 78.5 \, \text{cm}^2 - 50 \, \text{cm}^2 = 28.5 \, \text{cm}^2 \] Thus, the area of the minor segment is \( 28.5 \, \text{cm}^2 \).