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Q. The area of the equilateral triangle, in which three
coins of radius 1 cm are placed, as shown in the figure, isMathematics Question Image

IIT JEEIIT JEE 2005

Solution:

Since, tangents draw n from external points to the circle
subtends equal angle at the centre.
$\therefore $ $\angle O_1 BD=30^0$
In $\Delta O_1 BD, tan 30^0=\frac{O_1D}{BD}$ $\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 $\Delta$ ABC = $\frac{\sqrt3}{4}$ $(BC)^2$ = $\frac{\sqrt3}{4}.4(1+ \sqrt3)^2$
= $(6+4\sqrt3)$ sq cm

Solution Image