Question

Three circles of unit radii are inscribed in an equilateral $\Delta$ touching the sides of the $\Delta$. Then the area of the triangle is?

• A. $6+4\sqrt{3}$
• B. $12+8\sqrt{3}$
• C. $7+4\sqrt{3}$
• D. $4+\frac{7}{2}\sqrt{3}$

Answer 1

The correct option is A $6+4\sqrt{3}$

Let $ABC$ be the equliatrial triangle mentioned in the question Let $C_1,C_2,C_3$ be the centers of the there circles inscribed in the triangle , each with a unit radius.

Since all the angles of an equilateral triangle are ${60}^0$;

$\angle C_{2}BP={30}^0,C_{2}P=1$ (radius of circle)

$\therefore \tan {30}^0$ Let $BP=x$

$\therefore \tan {30}^0=\frac{1}{x}\Rightarrow \frac{1}{\sqrt{3}}=\frac{1}{x}\Rightarrow x=\sqrt{3}$

$BC=BP+PQ+QC$

$=x+(1+1)+x$

$=2x+2=2\sqrt{3}+2$

$\therefore$ Area of $\Delta ABC=\frac{\sqrt{3}}{4}{a}^2=\frac{\sqrt{3}}{4}(2\sqrt{3}+2{)}^2$

$=\frac{\sqrt{3}}{4}\times 4(\sqrt{3}+1{)}^2$

$=\sqrt{3}(3+1+2\sqrt{3})$

$=3\sqrt{3}+\sqrt{3}+6$

$=6+4\sqrt{3}$