Area Ques 11
The area of the equilateral triangle, in which three coins of radius $1 \mathrm{cm}$ are inscribed, is
$(2005,1 \mathrm{M})$
(a) $(6+4 \sqrt{3})$ $ \mathrm{sq}$ $ \mathrm{cm}$
(b) $(4 \sqrt{3}-6) $ $\mathrm{sq} $ $\mathrm{cm}$
(c) $(7+4 \sqrt{3}) $ $\mathrm{sq} $ $\mathrm{cm}$
(d) $4 \sqrt{3} $ $\mathrm{sq} $ $\mathrm{cm}$
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Answer:
Correct Answer: 11.(a)
Solution: (a) Since tangents drawn from external points to the circle subtend equal angles at the centre.
$\therefore \quad \angle O_1 B D=30^{\circ}$
In $ \quad \triangle O_1 B D, \tan 30^{\circ}=\frac{O_1 D}{B D} \Rightarrow B D=\frac{O_1 D}{\tan 30^{\circ}}$
Also, $ \quad D E=O_1 O_2=2 \mathrm{cm} \text { and } E C=\sqrt{3} \mathrm{cm} $
Now, $ \quad B C=B D+D E+E C=2+2 \sqrt{3} $
$\Rightarrow \quad$ Area of $\triangle A B C=\frac{\sqrt{3}}{4}(B C)^2=\frac{\sqrt{3}}{4} \cdot 4(1+\sqrt{3})^2$
$ = \quad(6+4 \sqrt{3}) \mathrm{sq} \mathrm{cm} $
BETA
