Question

In the following figure, ABC is a right angled triangle in which ∠A = 90°, AB = 21 cm and AC = 28 cm. Semi-circles are described on AB, BC and AC as diameters. Find the area of the shaded region.

Answer 1

We have given three semi-circles and one right angled triangle.

`"∴ Area ofshaded region=Area of semi-circle with AB as a diameter"`

   `"+ Area of semi-circle with AC as a diameter"`

    `"+ Area of right angled ABC"`

   ` "-Area of semi-circle with BC as a diameter"`

Let us calculate the area of the semi-circle with AB as a diameter. 

`∴"Area of semi-circle with AB as a diameter"=(pir^2)/2`

`∴ "Area of semi-circle with AB as a diameter"=pi/2(21/2)^2`

` "Area of semi-circle with AB as a diameter"=pi/2(21/2)^2`

Now we will find the area of the semi-circle with AC as a diameter.

`"Area of semi-circle with AC as a diameter"=pir^2/2`

`"Area of semi-circle with AC as a diameter"pi(28/2)^2/2`

`"Area of semi-circle with AC as a diameter"pi/2(28/2)^2`

Now we will find the length of BC.

In right angled triangle ABC, we will use Pythagoras theorem, 

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

`∴ BC^2=21^2+28^2`

`∴BC^2=441+784`

`∴BC^2=1225`

`∴ BC=35`

Now we will calculate the area of the right angled triangle ABC.

`A(ΔABC)=1/2xxABxxAC`

`∴ A(ΔABC)=1/2xx21xx28`

`∴A(ΔABC)=21xx14`

`∴ A(ΔABC)=294` 

Now we will find the area of the semi-circle with BC as a diameter. 

`"Area of semi-circle with BC as a diameter"=`(pi r^2)/2`

`∴" Area of semi-circle with AB as a diameter"=pi(35/2)^2/2`

`∴ " Area of semi-circle with AB as a diameter"=pi/2(35/2)^2`

Now we will substitute all these values in equation (1).

`∴ "Area of the shaded region"=pi/2(21/2)^2+pi/2(28/2)^2+294-pi/2(35/2)^2`

`∴ "Area of the shaded region"=pi/8(21^2+28^2-35^2)+294` 

`∴ "Area of the shaded region"=pi/8(441+784-1225)+294`

`∴"Area of the shaded region"pi/8(1225-1225)+294`

`∴"Area of the shaded region"=294`

Therefore, area of shaded region is `294 cm^2`