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`
