Question

In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and $△$ BOD = 90${}^{\circ }$. Find the area of shaded region.

[Use $\pi$ = 3.14]

Answer 1

In right triangle ABC

$B{C}^2=A{B}^2+A{C}^2=7^2+{24}^2=49+576=625\therefore ,B{C}^2=625\Rightarrow BC=25$

Now,$\angle COD+\angle BOD={180}^o$ ( Linear pair angles )

$\Rightarrow \angle COD={180}^o–{90}^o={90}^o$

Now, the area of the shaded region = Area of the sector having the central angle ${360}^o–{90}^o$ – Area of triangle ABC

$=\frac{{270}^o}{{360}^o}\times \pi (\frac{BC}{2}{)}^2–\frac{1}{2}\times AB\times AC=\frac{3}{4}\times 3.14(\frac{25}{2}{)}^2–\frac{1}{2}\times 7\times 24=367.97–84=283.97c{m}^2$

Hence, the area of the shaded region is $283.97c{m}^2$