Question

Two circles are placed in an equilateral triangle as shown in Fig. What is the ratio of the area of smaller circle to that of equilateral triangle?

• A.
  
• B.
  
• C.
  
• D.
  
• E.
  

Answer 1

The correct option is A ($\pi :27\sqrt{3}$)

Since, the length of the tangents from an external point to a circle are equal.
From the big circle:
Let $B$ the external point then the length of the tangents the from the external point $B$ are equal, so $BD=BU$,
similarly from the point $C$, we have $VC=DC$ and
From the point point $A$, we have $AU=AV$.
As the circles inscribed in an equilateral triangle, so $AU=AV=BD=BU=VC=DC$
Similarly for the small circle, we have $A$ be an external point and the length of the tangents of from point $A$ are $SA=AT$
Similarly for the point $P$, we have $SP=PR$
for the point $Q$, we have $RQ=TQ$
Since, the small circle inscribed in equilateral triangle
So, $SA=AT=SP=PR=RQ=TQ$
And also $AV=AT+TQ+QV$
$\Rightarrow AV=3RQ$ [Since, $AT+TQ+QV=RQ$]
Thus, $AC=2AV$ as $AV=VC$ $\Rightarrow V$ is mid point of $AC$.
So, $AC=2\left(RQ\right)$
$\therefore AC=6RQ$......(i)
Consider from right-angled triangle $OQR$, we have
$\tan {30}^{\circ }=\frac{OR}{RQ}$, where $OQ$ is bisector of an angle $\angle Q={60}^{\circ }$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{r}{RQ}$
$\therefore RQ=r\sqrt{3}$
Thus, the length of the each side of triangle $AC=6r\sqrt{3}$ [From equation(i)]
Now, area of the smaller circle$=\pi r^2$
Area of the triangle $ABC=\frac{\sqrt{3}}{4}A{C}^2$
$=\frac{\sqrt{3}}{4}(6r\sqrt{3}{)}^2$
$=\frac{\sqrt{3}}{4}\times 36r^2\times 3$
$=27\sqrt{3}{r}^2$
Therefore, the ratio of the area of smaller circle to that of equilateral triangle$=\pi r^2:27\sqrt{3}{r}^2$
$=\pi :27\sqrt{3}$