Question

Three circles each of radius r units are drawn inside an equilateral triangle of side a units, such that each circle touches the other two and two sides of the triangle as shown in the figure, (P, Q and R are the centres of the three circles). Then relation between r and a is

• A. $a=2(\sqrt{3}+1)r$
• B. $a=(\sqrt{3}+1)r$
• C. $a=(\sqrt{3}+2)r$
• D. $a=2(\sqrt{3}+2)r$

Answer 1

The correct option is A $a=2(\sqrt{3}+1)r$

$\angle A=\angle B=\angle C={60}^o$
$MN=QR=2r$ in quadrilateral BTQM
$\angle B+\angle BMQ+\angle MQT+\angle QTB=360+60+90+90+\angle QTB={360}^o$
$\Rightarrow \angle QTB=120$
$\Delta BTQ\cong \Delta BQM$
$\Rightarrow \angle TQB=\angle MQB={60}^o$
in $\Delta BQMtan{60}^o=\frac{BM}{QM}=\frac{x}{r}$
$\sqrt{3}=\frac{x}{r}\Rightarrow x=r\sqrt{3}$
Similarly $CN=r\sqrt{3}$
$BC=M+MN+CN=a$
$2x+2r=a$
$\Rightarrow 2(x+r)=a$
$\Rightarrow 2(r\sqrt{3}+r)=a\Rightarrow a=2r(\sqrt{3}+1)$