Question

In a circular table cover of radius $32$cm, a design is formed leaving an equilateral triangle ABC in the middle as shown in figure. Find the area of the design (shaded region)

• A. $(\frac{16148}{7}-218\sqrt{3})cm$
• B. $(\frac{19221}{7}-572\sqrt{3})cm$
• C. $(\frac{6288}{7}-672\sqrt{3})cm$
• D. None of these

Answer 1

Answer: (D)

None of these

Refer fig.

$\cos \theta =\frac{AD}{OA}$

$\cos {30}^{\circ }=\frac{\sqrt{3}}{2}$

$AD=\frac{\sqrt{3}}{2}\times 32$

$AD=16\sqrt{3}cm$

$AB=32\sqrt{3}cm$

shaded region $=$ Area of circle - Area of triangle

$=\pi (32{)}^2-3\times \frac{1}{2}\times 32\sqrt{3}\times 16$

$=(\frac{22578}{7}-768\sqrt{3})c{m}^2$