Question

In 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 (in $c{m}^2$).

• A. 100
• B. 98
• C. 96
• D. 95

Answer 1

The correct option is B

98

Area of quadrant = $\frac{90}{360}\times \pi \times 14\times 14$

= $\frac{90}{360}\times \frac{22}{7}\times 14\times 14$

= 154 sq. cm

Area of $△ABC$ = $\frac{1}{2}\times {14}^2$ = 98 $c{m}^2$

In $△ABC$ ; $A{B}^2+A{C}^2=B{C}^2$ ( using Pythagoras Theorem )

Given the BC is the diameter of the semi-circle. And AB = AC = 14 cm

$\therefore$ diameter (BC) = 14 × $\sqrt{2}$ = 19.8 cm.

Hence radius of the semicircle = 9.9 cm

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

= $\frac{22}{7}\times \frac{1}{2}\times {9.9}^2$

= 154.01 ~ 154

= 154 $c{m}^2$

The area of segment =154 - 98 = 56$c{m}^2$

The required area =154 - 56 = 98 $c{m}^2$.