Question

In the figure, ABC is a quadrant of a circle of radius 14 cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. [2 MARKS]

Answer 1

Concept: 1 Mark
Application: 1 Mark

Area of quadrant
$=\frac{90}{360}\times \pi \times 14\times 14=153.94c{m}^2$

Area of $\Delta ABC=0.5\times {14}^2=98c{m}^2$

The diameter $=14\times \sqrt{2}=19.8cm$.

Area of semi-circle
$=0.5\pi \times {9.9}^2=153.95c{m}^2$

The area of segment
$=153.94-98=55.94c{m}^2$

The required area
$=153.95-55.94=98.01c{m}^2$.