Question

The area of the equilateral triangle, in which three coins of radius 1 cm are placed, as shown in the figure, is

• A. $(6+\sqrt3)$ sq cm
• B. $(4\sqrt3-6)$ sq cm
• C. $(7+4\sqrt3)$ sq cm
• D. $4\sqrt3$ sq cm

Answer 1

Answer: (A)

$(6+\sqrt3)$ sq cm

Answer 2

Since, tangents draw n from external points to the circle
subtends equal angle at the centre.
$\therefore$ $\angle O_1 BD=30^0$
In $? O_1 BD, tan 30^0=\frac{O_1D}{BD}$ $\Rightarrow$ = $\sqrt3$ cm
Also, $DE=O_1O_2=2$ cm and EC = $\sqrt3$ cm
Now, BC = BD + DE + E C =$2+2\sqrt3$
$\Rightarrow$ Area of $?$ ABC = $\frac{\sqrt3}{4}$ $(BC)^2$ = $\frac{\sqrt3}{4}.4(1+ \sqrt3)^2$
= $(6+4\sqrt3)$ sq cm