In the below figure, $ A B C $ is a right-angled triangle, $ \angle B=90^{\circ}, A B=28 \mathrm{~cm} $ and $ B C=21 \mathrm{~cm} $. With $ A C $ as diameter a semicircle is drawn and with $ B C $ as radius a quarter circle is drawn. Find the area of the shaded region correct to two decimal places."

Given:

\( A B C \) is a right-angled triangle, \( \angle B=90^{\circ}, A B=28 \mathrm{~cm} \) and \( B C=21 \mathrm{~cm} \).

With \( A C \) as diameter a semicircle is drawn and with \( B C \) as radius a quarter circle is drawn.

To do: 

We have to find the area of the shaded region correct to two decimal places.

Solution:

In right angled triangle $ABC$, by Pythagoras theorem,

$\mathrm{AC}^{2}=\mathrm{AB}^{2}+\mathrm{BC}^{2}$

$=(28)^{2}+(21)^{2}$

$=784+441$

$=1225$

$=(35)^{2}$

$\Rightarrow \mathrm{AC}=35 \mathrm{~cm}$

Therefore,

Radius of the quadrant $\mathrm{BCD}=21 \mathrm{~cm}$

Radius of the semicircle on $AC$ as diameter $=\frac{35}{2} \mathrm{~cm}$

Area of $\triangle \mathrm{ABC}=\frac{1}{2} \mathrm{AB} \times \mathrm{BC}$

$=\frac{1}{2} \times 28 \times 21$

$=294 \mathrm{~cm}^{2}$

Area of the quadrant $\mathrm{BCD}=\frac{1}{4} \pi r^{2}$

$=\frac{1}{4} \times \frac{22}{7} \times (21)^2$

$=\frac{693}{2}$

$=346.5 \mathrm{~cm}^{2}$

Area of semicircle on $\mathrm{AC}$ as diameter $=\frac{1}{2} \pi \mathrm{R}^{2}$

$=\frac{1}{2} \times \frac{22}{7} \times (\frac{35}{2})^2$

$=\frac{1925}{4}$

$=481.25 \mathrm{~cm}^{2}$

Therefore,

Area of the shaded region $=$ Area of $\Delta \mathrm{ABC}+$ Area of semicircle $-$ Area of the quadrant

$=294+481.25-346.50$

$=428.75 \mathrm{~cm}^{2}$

The area of the shaded region correct to two decimal places is $428.75 \mathrm{~cm}^{2}$.

Tutorialspoint

Updated on: 10-Oct-2022

75 Views

Kickstart Your Career

Get certified by completing the course

Get Started