Question

Find the area of the shaded region where ABC is a quadrant of radius 5 cm and a semicircle is drawn with BC as diameter.

• A. 19.64 $c{m}^2$
• B. 12.5 $c{m}^2$
• C. 7.14 $c{m}^2$
• D. 8.8 $c{m}^2$

Answer 1

The correct option is B

12.5 $c{m}^2$

Area of the shaded region = Area of semicircle-Area of segment BXC
Here, AB = AC = 5 cm
Then, by Pythagoras theorem,
$B{C}^2=A{B}^2+A{C}^{2}B{C}^2=5^2+5^{2}B{C}^2=50BC=5\sqrt{2}cm$

Area of the semicircle with BC as diameter
$=\frac{1}{2}\times \frac{22}{7}\times \frac{5\sqrt{2}}{2}\times \frac{5\sqrt{2}}{2}=\frac{1}{2}\times \frac{22}{7}\times \frac{5}{\sqrt{2}}\times \frac{5}{\sqrt{2}}$
$=19.64c{m}^2...\left(i\right)$

Area of segment BXC = Area of quadrant - Area of ΔABC
$=\frac{90°}{360°}\times \frac{22}{7}\times 5^2-\frac{1}{2}\times 5\times 5$
$=19.64-12.5$
$=7.14c{m}^2...\left(ii\right)$

Area of the shaded region = Area of semicircle with BC as diameter - Area of Segment BXC
$=19.64-7.14$
$=12.5c{m}^2$