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.

#### Solution

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

\[= r^2 \left( \frac{\pi\theta}{360° } - \frac{\sin\theta}{2} \right)\]

\[ = \left( 7\sqrt{2} \right)^2 \left( \frac{22}{7} \times \frac{90°}{360°} - \frac{\sin90°}{2} \right)\]

\[ = 98 \times \frac{22}{7} \times \frac{1}{4} - 98 \times \frac{1}{2}\]

\[ = 77 - 49\]

\[ = 28 {cm}^2\]

Thus, the area of the segment BXC is 28 cm^{2}.