Question

Let the centroid of an equilateral triangle$ABC$ be at the origin. Let one of the sides of the equilateral triangle be along the straight line $x+y=3$. If $R\&r$ be the radius of circumcircle and incircle respectively of $\Delta ABC$ , then $(R+r)$ is equal to:

• A. $2\sqrt{2}$
• B. $3\sqrt{2}$
• C. $7\sqrt{2}$
• D. $\frac{9}{\sqrt{2}}$

Answer 1

The correct option is D

$\frac{9}{\sqrt{2}}$

Explanation for the correct option:

Finding the value of $(R+r)$:

Illustrating a figure according to given data

Given that one of the sides of the equilateral triangle be along the straight line $x+y=3$

And we know in an equilateral triangle each side have an angle $60°$

So, considering the given line bisects the angle.

Therefore,

So, from the figure

$\Rightarrow r=OM=$perpendicular distance between $O\left(0,0\right)\&x+y=3$

$\Rightarrow r=\left|\frac{(0+0–3)}{\sqrt{2}}\right|=\frac{3}{\sqrt{2}}...\left(i\right)$

$$\begin{gathered} \because \sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}} \\ \therefore \sin 30°=\frac{1}{2}=\frac{r}{R} \\ \Rightarrow R=\frac{6}{\sqrt{2}}....\left(ii\right) \end{gathered}$$

Adding the equation $\left(i\right)\&\left(ii\right)$

$$\begin{gathered} \Rightarrow r+R=\frac{3}{\sqrt{2}}+\frac{3}{\sqrt{2}} \\ =\frac{9}{\sqrt{2}} \end{gathered}$$

Hence, option (D) is the correct answer.