Question

In the adjoining figure, three congruent circles are touching each other. Triangle ABC circumscribes all the three circles. Triangle PQR is formed by joining the centres of the circle. There is a third triangle DEF. Points A,D, P and B, E, Q and C,F,R lie the same straight line respectively.


What is the ratio of perimeters of $\Delta ABC:\Delta DEF:\Delta PQR$

• A. $3\sqrt{2}:2\sqrt{2}:1$
• B. $2(4+\sqrt{3}):(2+\sqrt{3}):3$
• C. $2(1+\sqrt{3}):(2+\sqrt{3}):2$
• D. $2(1+\sqrt{3}):2\sqrt{3}:\sqrt{3}$

Answer 1

The correct option is C

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

Let the radius of each circle be r unit then
$PQ=QR=PR=2r\angle PDM=\angle QEN={30}^{\circ }$




$\therefore \frac{DM}{DP}=cos{30}^{\circ }DM=DP\times \frac{\sqrt{3}}{2}[DP=QE=(r\left)\right]DM=\frac{r\sqrt{3}}{2}\therefore DE=DM+MN+NE=\frac{r\sqrt{3}}{2}+2r+\frac{r\sqrt{3}}{2}=(2+\sqrt{3})rDE=DF=EF=(2+\sqrt{3})r$
Again $\angle PAM=\angle QBN={30}^{\circ }\therefore \frac{PM}{AM}=tan{30}^{\circ }=\frac{1}{\sqrt{3}}\frac{r}{AM}=\frac{1}{\sqrt{3}}$



$\Rightarrow AM=r\sqrt{3}=BN\therefore AB=AM+MN+NB=r\sqrt{3}+2r+r\sqrt{3}=2r(1+\sqrt{3})AB=BC=AC=2r(1+\sqrt{3})$
Ratio of perimeter of equilateral triangle = ratio of their sides
$\therefore$ Ratio of perimeter of $\Delta ABC:\Delta DEF:\Delta PQR=2(1+\sqrt{3}):(2+\sqrt{3}):2$