Question

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

Answer 1

BC is a chord

BC passes through O. We know that the chord passing through the centre of a circle is the diameter.

$\Rightarrow$ BC is the diameter of the circle.

We know that angle in semicircle is ${90}^o$.

BAC is a semi circle.

$\angle BAC={90}^o$

$\therefore$ In right triangle BAC,

By using Pythagoras theorem

$A{B}^2+A{C}^2=B{C}^2$

$\Rightarrow (7{)}^2+(24{)}^2=B{C}^2$

$\Rightarrow 44+576=B{C}^2$

$\Rightarrow B{C}^2=625$

$\Rightarrow BC=\sqrt{625}$

$\therefore BC=25cm$

$OB=OC=$Radius $=\frac{1}{2}$.diameter

$\Rightarrow r=\frac{1}{2}\cdot BC$

$\Rightarrow r=1/2.25$

$\therefore r=12.5$cm

Ara of triangle ABC$=\frac{1}{2}.AB.AC$

$=\frac{1}{2}.7.24$

$\therefore \angle COD={90}^o$

Sector COD is a quadrant

$\therefore$ Area of quadrant$=\frac{1}{4}.\pi r^2$

$=\frac{1}{4}.\frac{22}{7}.\frac{25}{2}.\frac{25}{2}$

$=\frac{11\times 625}{14\times 4}$

$=122.76$

$\therefore$ Area of shaded region

$\Rightarrow$ Area of circle $-$ Area of $\Delta$ABC$-$Area of quadrant

$=\pi r^2-84-122.76$

$\frac{22}{7}.\frac{25}{2}.\frac{25}{2}-84-122.76$

$=\frac{13.750}{28}.206.76$

$=491.07-206.76$

$=284.31$

$\therefore$ Area of the shaded region $=284.31c{m}^2$.