CBSE · Class 10 · Mathematics · Areas Related to CirclesA chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding minor segment. (Use $\pi = 3.14$)
Given:\nRadius ($r$) = 10 cm\nCentral angle ($\theta$) = 90° \nStep 1: Find the area of the corresponding sector.\nArea of sector = $\frac{\theta}{360^{\circ}} \times \pi r^2$\nArea of sector = $\frac{90^{\circ}}{360^{\circ}} \times 3.14 \times (10)^2$\nArea of sector = $\frac{1}{4} \times 3.14 \times 100$\nArea of sector = $0.25 \times 314 = 78.5 \text{ cm}^2$ \nStep 2: Find the area of the triangle formed by the two radii and the chord.\nSince the angle is 90°, the triangle is a right-angled isosceles triangle with base and height equal to the radius ($r$).\nArea of triangle = $\frac{1}{2} \times \text{base} \times \text{height}$\nArea of triangle = $\frac{1}{2} \times 10 \times 10 = 50 \text{ cm}^2$ \nStep 3: Calculate the area of the minor segment.\nArea of minor segment = Area of sector - Area of triangle\nArea of minor segment = $78.5 - 50 = 28.5 \text{ cm}^2$ \nTherefore, the area of the corresponding minor segment is $28.5 \text{ cm}^2$.