Question

A triangle is inscribed in a circle. The vertices of the triangle divide the circle into three arcs of length 3, 4 and 5 units. Then area of the triangle is equal to

• A. $\frac{9\sqrt{3}(1+\sqrt{3})}{{\pi }^2}$
• B. $\frac{9\sqrt{3}(\sqrt{3}-1)}{{\pi }^2}$
• C. $\frac{9\sqrt{3}(1+\sqrt{3})}{2{\pi }^2}$
• D. $\frac{9\sqrt{3}\left(\sqrt{3-1}\right)}{2{\pi }^2}$

Answer 1

The correct option is A $\frac{9\sqrt{3}(1+\sqrt{3})}{{\pi }^2}$

Circumference $=3+4+5=2\pi r$

$⟹r=\frac{6}{\pi }$

The arcs subtend angles that are proportionate to their lengths. i.e. $3\theta ,4\theta ,5\theta$

$3\theta +4\theta +5\theta ={360}^{\circ }$

$\theta ={30}^{\circ }$

$⟹$ The $3$ triangles subtend $90,120,150$ angles.

Area is given by $\frac{1}{2}ab\sin C$

Total Area $=\frac{1}{2}{r}^2\sin {\theta }_1+\frac{1}{2}{r}^2\sin {\theta }_2+\frac{1}{2}{r}^2\sin {\theta }_3$

$=\frac{1}{2}\times \frac{36}{{\pi }^2}(1+\frac{\sqrt{3}}{2}+\frac{1}{2})$

$=\frac{9}{{\pi }^2}(3+\sqrt{3})$