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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 D - Mathematics

Sum

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.

 

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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`

 

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APPEARS IN

RD Sharma Class 10 Maths
Chapter 13 Areas Related to Circles
Exercise 13.4 | Q 40 | Page 62
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