Question

In the given figure, O is the centre of the circle with AC = 24 cm, AB = 7 cm and $\angle BOD={90}^0$. Find the area of the shaded region.

Answer 1

Area of the shaded region $=$ area of the circle $-$area of the triangle ABC$-$area of the quadrant COD

Here, in triangle ABC $\angle CAB={90}^o$ (angle in a semicircle)

Hence triangle ABC is right-angled at A

Then applying Pythagoras theorem,

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

$B{C}^2={24}^2+7^2$

$B{C}^2=625$

$BC=25cm$

Therefore diameter of circle $=25cm$

i.e., Radius$=\frac{25}{2}=12.5cm$

Then area of circle $=\pi r^2=490.625c{m}^2$

Area of triangleABC$=\frac{1}{2}\times base\times height$

$=84c{m}^2$

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

$=122.65625c{m}^2$

Therefore area of shaded region $=490.625-84-122.65625=283.96875c{m}^2$