Question

In the given figure, A is the centre of the circle. $\angle$ABC = 45^° and AC = 7$\sqrt{2}$cm. Find the area of segment BXC.

Answer 1

In ∆ABC,

AB = AC (Radii of the circle)

∴ ∠ACB = ∠ABC = 45º (Equal sides have equal angles opposite to them)

Using angle sum property, we have

∠ACB + ∠ABC + ∠BAC = 180º

∴ 45º + 45º + ∠BAC = 180º

⇒ ∠BAC = 180º − 90º = 90º

Here,

Radius of the circle, r = $7\sqrt{2}$ cm

Measure of arc BXC, θ = 90º

∴ Area of segment BXC

$$\begin{gathered} =r^2\left(\frac{\mathrm{\pi }\theta }{360°}-\frac{\sin \theta }{2}\right) \\ ={\left(7\sqrt{2}\right)}^2\left(\frac{22}{7}\times \frac{90°}{360°}-\frac{\sin 90°}{2}\right) \\ =98\times \frac{22}{7}\times \frac{1}{4}-98\times \frac{1}{2} \\ =77-49 \\ =28{\mathrm{cm}}^2 \end{gathered}$$
Thus, the area of the segment BXC is 28 cm².