In the following figure, *ABC* is a right-angled triangle, ∠*B* = 90°, *AB* = 28 cm and *BC* = 21 cm. With *AC* as diameter a semicircle is drawn and with *BC* as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places.

#### Solution

We have given two semi-circles and one circle.

Area of the shaded region = area of semicircle with diameter AC + area of right angled triangle ABC − area of sector

First we will find the hypotenuse of right angled triangle ABC.

`AC^2=AB^2+BC^2`

`∴ AC^2=28^2+21^2`

`∴ AC^2=784+441`

`∴ AC^2=1225`

`∴ AC=35`

`∴ "Area of the shaded region"=(pixx17.5xx17.5)/2+1/2xx28xx21-θ/360xxpixx21xx21`

`∴ "Area of the shaded region"=(pixx17.5xx17.5)/2+14xx21-1/4xxpixx21xx21`

Substituting `pi=22/7`we get,

`∴ "Area of shabed region"=(22/7xx17.5xx17.5)/2+14xx21-1/4xx22/7xx21xx21`

`∴ "Area of shabed region"=962.5/2+14xx21-1/2xx11xx3xx21`

`∴ "Area of shabed region"=481.25+294-346.5`

`∴ "Area of shabed region"=428-75`

Therefore, area of shaded region is `428.75 cm^2`