Question

An equilateral triangle has area $A\sqrt{3}$. Three circles are drawn with their centres at the vertices of the triangle. Diameter of each circle is equal to the length of each side of the triangle. The area of the triangle NOT included in any of the three circles is

• A. $A(\sqrt{3}-\frac{\pi }{6})$
• B. $A(\pi -\sqrt{3})$
• C. $A(3\pi -\sqrt{3})$
• D. $A(\sqrt{3}-\frac{\pi }{2})$

Answer 1

Answer: (D)

$A(\sqrt{3}-\frac{\pi }{2})$

Area of equilateral triangle $=A\sqrt{3}c{m}^2$
side$=\sqrt{A\times 4}=\sqrt{4A}cm$
$\therefore$Radius of each circle$=\frac{\sqrt{4A}}{2}cm$
Area of 3 sector$=3\times \frac{60}{360}\pi r^2$
$=3\times \frac{1}{6}\times \pi \times {\left(\frac{\sqrt{4}A}{2}\right)}^2$
$=A\frac{\pi }{2}c{m}^2$
Area of the triangle not included in circle$=A\sqrt{3}-A\frac{\pi }{2}$
$=A(\sqrt{3}-\frac{\pi }{2})c{m}^2$