Question

The radius of the circumcircle of the triangle PRS with the three sides as $2\sqrt{3}$, $6\sqrt{2}$ and $10$ is

• A. $5$
• B. $3\sqrt{3}$
• C. $3\sqrt{2}$
• D. $2\sqrt{3}$

Answer 1

The correct option is B $3\sqrt{3}$

Given $a=10,b=2\sqrt{3},c=6\sqrt{2}$

Let $R$ be the radius of the circumcircle

We know that circumradius $R=\frac{abc}{4\Delta }$

where $\Delta$ is the area of the triangle.

Now we have $\Delta =\frac{1}{2}bc\sin A$

So $R=\frac{abc}{4\left(\frac{1}{2}bc\sin A\right)}$ $\Rightarrow R=\frac{a}{2\left(\sin A\right)}$

We have $\cos A=\frac{b^2+c^2-a^2}{2bc}$ $\Rightarrow \cos A=\frac{72+12-100}{2\times 12\sqrt{6}}=-\frac{2}{3\sqrt{6}}$

$\Rightarrow \sin A=\sqrt{1-{(-\frac{2}{3\sqrt{6}})}^2}=\frac{5}{3\sqrt{3}}$

Now $R=\frac{10}{2\times \left(\frac{5}{3\sqrt{3}}\right)}=3\sqrt{3}$