Question

In $\Delta ABC,A=\frac{2\pi }{3},b-c=3\sqrt{3}cm$ and $ar\left(\Delta ABC\right)=\frac{9\sqrt{3}}{2}c{m}^2.$ Then $a$ is

• A. $6\sqrt{3}cm$
• B. $9cm$
• C. $18cm$
• D. none of these

Answer 1

Answer: (B)

$9cm$

Given $b-c=3\sqrt{3}\Rightarrow {(b-c)}^2=27\Rightarrow b^2+c^2-2bc=27\Rightarrow b^2+c^2=27+2bc$

From cosine rule we have

$\cos A=\frac{b^2+c^2-a^2}{2bc}$

Here, $A=\frac{2\pi }{3}\Rightarrow \cos A=\cos \frac{2\pi }{3}=-\frac{1}{2}$

$\Rightarrow -\frac{1}{2}=\frac{27+2bc-a^2}{2bc}\Rightarrow a^2=27+3bc$ ....... $\left(1\right)$

Now Area of triangle $\Delta =\frac{bc\sin A}{2}$

From given area of triangle, we get $\frac{9\sqrt{3}}{2}=\frac{bc\sin \left(\frac{2\pi }{3}\right)}{2}=\frac{bc\left(\frac{\sqrt{3}}{2}\right)}{2}$

$\Rightarrow bc=18$

Putting value of $bc$ in equation $\left(1\right)$ we get

$a^2=27+54=81\Rightarrow a=9$