Question : Three circles of radius 6 cm each touch each other externally. Then the distance of the centre of one circle from the line joining the centres of the other two circles is equal to:
Option 1: $6\sqrt{5}\;\mathrm{cm}$
Option 2: $6\sqrt{3}\;\mathrm{cm}$
Option 3: $6\sqrt{2}\;\mathrm{cm}$
Option 4: $6\sqrt{7}\;\mathrm{cm}$
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Correct Answer: $6\sqrt{3}\;\mathrm{cm}$
Solution :
Let the three circles form an equilateral triangle $\triangle \mathrm{ABC}$ with each side equal to the diameter of a circle, which is $12\;\mathrm{cm}$. $\mathrm{BC = BD + DC = 6 + 6 }=12\;\mathrm{cm}$ The altitude $(h)$ of an equilateral $\triangle \mathrm{ABC}$, $h = \frac{\sqrt{3} a}{2}$ where $a$ is the length of a side of the triangle. $h = \frac{12\sqrt{3}}{2} = 6\sqrt{3}$ Hence, the correct answer is $ 6\sqrt{3}\;\mathrm{cm}$.
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