Question

The area of the triangle whose vertices are $(a,\theta )$, $(2a,\theta -\frac{\pi }{3})$and $(3a,\theta +\frac{2\pi }{3})$ is

• A. $\frac{\sqrt{3}}{4}{a}^2$
• B. $3\frac{\sqrt{3}}{4}{a}^2$
• C. $5\frac{\sqrt{3}}{4}{a}^2$
• D. $7\frac{\sqrt{3}}{4}{a}^2$

Answer 1

Answer: (C)

$5\frac{\sqrt{3}}{4}{a}^2$

Let$(r_1,{\theta }_1)(r_2,{\theta }_2)(r_3,{\theta }_3)$ be the polar co-ordinates of vertices.

Comparing them with $(a,\theta ),(2a,\theta -\frac{\pi }{3})(3a,\theta +\frac{2\pi }{3})$

We get,

$r_1=a,r_2=2a,r_3=3aand{\theta }_1=\theta ,{\theta }_2=\theta -\frac{\pi }{3},{\theta }_3=\theta +\frac{2\pi }{3}$

Now,

Area of triangle

$=\frac{1}{2}|r_2{r}_3\sin ({\theta }_3-{\theta }_2)+r_3{r}_1\sin ({\theta }_1-{\theta }_3)-r_1{r}_2\sin ({\theta }_1-{\theta }_2)|$

$=\frac{1}{2}|2a.3a\sin (\theta +\frac{2\pi }{3}-\theta +\frac{\pi }{3})+3a.a\sin (\theta -\theta -\frac{2\pi }{3})-a.2a\sin (\theta -\theta +\frac{\pi }{3})|$

$=\frac{1}{2}|6a^2\sin \left(\pi \right)+3a^2\sin (-\frac{2\pi }{3})-2a^2\sin \left(\frac{\pi }{3}\right)|$

$=\frac{1}{2}|-\frac{3\sqrt{3}{a}^2}{2}-\frac{2\sqrt{3}{a}^2}{2}|$

$=\frac{5\sqrt{3}}{4}{a}^2$