Question

In the given figure, $△ABC$ is inscribed in a semicircle. If $AB=AC$, and $BC=14cm$, what will be the area ($inc{m}^2$)of the green region? Consider $\pi =\frac{22}{7}$

• A. 28
• B. 56
• C. 105
• D. 52.5

Answer 1

The correct option is A 28


Let O be the centre of the semicircle.
Given, BC = 14 cm
i.e. Diameter of the semicircle = 14 cm
Radius of the semicircle = $\frac{14}{2}$ = 7 cm

In $△ABC$, AO is the median onto the side BC. (Since O is the midpoint of the side BC)
Since AB = AC, the median and altitude between AB and AC are the same i.e. AO.
AO = radius of the semicircle = 7 cm

Area of the semicircle = $\frac{\pi r^2}{2}$
= $\frac{\frac{22}{7}\times 7^2}{2}$
= 77 $c{m}^2$

Area of $△ABC$ = $\frac{1}{2}\times BC\times AO$
=$\frac{1}{2}\times 14\times 7$
= 49 $c{m}^2$

Area of the green region = Area of the semicircle - Area of $△ABC$
= 77 - 49
= 28 $c{m}^2$