Question

In the given figure, A is the centre of the circle. $\angle ABC={45}^o$ and $AC=7\sqrt{2}cm.$ Find the area of segment BXC. (Use $\pi =\frac{22}{7}$ )

• A. $26c{m}^2$
• B. $27c{m}^2$
• C. $28c{m}^2$
• D. $29c{m}^2$

Answer 1

The correct option is C $28c{m}^2$

In $△ABC,$
$AB=AC=7\sqrt{2}cm$
[Radii of the circle]

$\therefore \angle ACB=\angle ABC={45}^o$
[Equal sides have equal angles opposite to them]

Using angle sum property, we have

$\angle ACB+\angle ABC+\angle BAC={180}^o$

$\therefore {45}^o+{45}^o+\angle BAC={180}^o$

$\Rightarrow \angle BAC={180}^o-{90}^o={90}^o$

Area of $△ABC$
$=\frac{1}{2}\times base\times height$
$=\frac{1}{2}\times (2\times ABcos45)\times ABsin45)$
$=\frac{1}{2}\times 2\times 7\times 7$
$=49c{m}^2$

Here,
Radius of the circle, $r=7\sqrt{2}cm$
Measure of arc $BXC,\theta ={90}^o$

$\therefore$ Area of segment BXC
$=\pi r^2\left(\frac{\theta }{360}\right)-Areaof△ABC$
$=\frac{22}{7}\times \left(7\sqrt{2}{)}^2\right(\frac{90}{360})-49$
$=98\times \frac{22}{7}\times \frac{1}{4}-49$
$=77-49$
$=28$

Thus, the area of the segment BXC is $28c{m}^2.$