Area Of A Segment Of A Circle

Let's find the area of the minor segment when a chord of a circle with radius 10 cm subtends a right angle at the centre. First, calculate the area of the sector using the angle $90^\circ$. The sector area is $90/360 \times$$\times 10^2 = 78.5\text{ cm}^2$. Next, find the area of the central triangle $\Delta OAB$. Because it's a right triangle, we can directly use the formula $\frac{1}{2} \times \text{base} \times \text{height}$. Here, both the base and height are the radius, so the area is $\frac{1}{2} \times 10 \times$$= 50\text{ cm}^2$. Finally, we calculate the area of the minor segment by the triangle area from the sector area. The final area is $78.5 - 50 = 28.5\text{ cm}^2$.