Question

$\Delta$ABC 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 B

Option (b)

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

Area of Δ BDO =$\frac{3}{4}$ * 49

Area of sector OBD = $\frac{49}{6}$

Hence area of the shaded region

= 2[$\frac{49}{6}$ – $\frac{3}{4}$*49]

= 49[$\frac{1}{3}$ – $\frac{3}{2}$]

Shortcut

By graphical division, there are 3 equilateral triangles of areas of $\frac{\sqrt{3}}{4}$ * 49

Area of interest = (area of semi circle [$\frac{r^2}{2}$] - area of three triangles) = (49*$\frac{1}{2}$ - 3*3*$\frac{49}{4}$)*($\frac{2}{3}$) = 49*($\frac{1}{3}$ – $\frac{3}{2}$)