Question

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

[Use $\pi$ = 3.14]

Answer 1

In right triangle ABC

$B{C}^2=A{B}^2+A{C}^2=7^2+{24}^2=49+576=625\therefore B{C}^2=625\Rightarrow BC=25$
So,
Area of a circle $=\pi \times r^2=\frac{22}{7}\times {12.5}^2$
$=491.07c{m}^2$
Also,
Area of triangle $ABC=\frac{1}{2}\times (AC\times AB)$
$=\frac{1}{2}\times (24\times 7)=84c{m}^2$
Since, $\angle BOD=90°$ and $\angle COD=\angle BOD-90°$ (Linear pair)
So, $\angle COD=90°$
$OC=\frac{BC}{2}=\frac{25}{2}=12.5cm$
So, sector $OCDO$ is a quadrant.
Area of sector $OCDO=\frac{1}{4}\times (\pi \times r^2)$
$=\frac{1}{4}\times \frac{22}{7}\times 12.5\times 12.5$ $=122.7c{m}^2$.
Area of the shaded region = area of the circle - area of triangle $ABC$ - area of sector $OCDO$.
$=491.07-84-122.7=284.37c{m}^2$
So, the area of the shaded region is approx $284.37c{m}^2$