Question

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

• A. $(6+\sqrt{3})$ sq cm
• B. $(4\sqrt{3}-6)$ sq cm
• C. $(7+4\sqrt{3})$ sq cm
• D. $4\sqrt{3}$ sq cm

Answer 1

Answer: (A)

Since, tangents draw n from external points to the circle
subtends equal angle at the centre.
$\therefore$ $\angle O_{1}BD=30^0$
In $\Delta O_{1}BD,tan30^0=\frac{O_{1}D}{BD}$ $\Rightarrow$ = $\sqrt{3}$ cm
Also, $DE=O_1{O}_2=2$ cm and EC = $\sqrt{3}$ cm
Now, BC = BD + DE + E C =$2+2\sqrt{3}$
$\Rightarrow$ Area of $\Delta$ ABC = $\frac{\sqrt{3}}{4}$ $(BC{)}^2$ = $\frac{\sqrt{3}}{4}.4(1+\sqrt{3}{)}^2$
= $(6+4\sqrt{3})$ sq cm