Question

Find the area of the shaded region.

Answer 1

Hint:There are one triangle and three semicircles on each side of the right-angled triangle in the given figure, the diameter of the semicircles on each side would be the length of the side itself.

We know that radius is half of the diameter of a circle, so we can find out the area of the semicircles easily when their radius is known, thus on finding the length of BC by Pythagoras theorem, we can find out the area of the shaded region.

Complete step by step answer:
$\vartriangle ABC$ is a right-angled triangle.
Area of a right-angled triangle is, ${A_\vartriangle } = \dfrac{1}{2} \times base \times height =
\dfrac{1}{2} \times 3 \times 4 = 6c{m^2}$

To find the area of the semicircle, we have to find out its radius first.
As $\vartriangle ABC$ is a right-angled triangle, so by applying Pythagoras theorem, we get –
$
B{C^2} = A{B^2} + A{C^2} \\
B{C^2} = {3^2} + {4^2} \\
B{C^2} = 25 \\
\Rightarrow BC = 5cm \\
$

BC is the diameter of the semicircle on the side BC, we know that radius is half of the diameter of the
circle, so $r = \dfrac{5}{2}cm$

The area of the semicircle on side BC is ${A_{BC}} = \dfrac{{\pi {r^2}}}{2} = \dfrac{1}{2} \times
\dfrac{{22}}{7} \times {(\dfrac{5}{2})^2} = \dfrac{{275}}{{28}}c{m^2}$

The area of the semicircle on side AB is ${A_{AB}} = \dfrac{1}{2} \times \dfrac{{22}}{7} \times
{(\dfrac{3}{2})^2} = \dfrac{{99}}{{28}}c{m^2}$

The area of the semicircle on side AC is ${A_{AC}} = \dfrac{1}{2} \times \dfrac{{22}}{7} \times
{(\dfrac{4}{2})^2} = \dfrac{{176}}{{28}}c{m^2}$

The area of the shaded region is –
$
A = {A_{AB}} + {A_{AC}} - ({A_{BC}} - {A_\vartriangle }) = \dfrac{{99}}{{28}} + \dfrac{{176}}{{28}} -
(\dfrac{{275}}{{28}} - 6) \\
A = \dfrac{{275}}{{28}} - \dfrac{{275}}{{28}} + 6 \\
\Rightarrow A = 6c{m^2} \\
$

Hence the area of the shaded region is $6c{m^2}$ .

Note:When two of the sides of a right-angled triangle are known, the third side can be found out using the Pythagoras theorem that tells us the relation between the three sides of a right-angled triangle that is the base, the height and the hypotenuse; it states that the sum of the square of the base and height is equal to the square of the hypotenuse. Using this theorem we can find out the radius of the semicircle on the side BC of the right-angled triangle ABC.