Question

ΔABC is an equilateral triangle of side 14 cm. A semi circle on BC as diameter is drawn to meet AB at D, and AC at E. Find the area of the shaded region.

• A.
• B.
• C. 49 cm²
• D. 49sqrt(2) cm²

Answer 1

The correct option is

C: $\frac{49}{2}(\frac{\pi }{3}-\frac{\sqrt{3}}{2})c{m}^2$

(b)

O is the centre of the circle and the mid-point of BC. DO is parallel to AC. So, <DOB = 60°

Area of Δ BDO =$\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}\cdot 49$

Area of sector OBD = $\frac{\pi .r^2}{6}=\frac{\pi \cdot 49}{6}$

Hence area of the shaded region

= $\pi \cdot \frac{49}{6}-\frac{\sqrt{3}}{4}\cdot 49$

= $\frac{49}{2}(\frac{\pi }{3}-\frac{\sqrt{3}}{2})c{m}^2$