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: $6\sqrt{7}\;\mathrm{cm}$
Latest: SSC CGL preparation tips to crack the exam
Don't Miss: SSC CGL complete guide
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}$.
Related Questions
Know More about
Staff Selection Commission Combined Grad ...
Answer Key | Eligibility | Application | Selection Process | Preparation Tips | Result | Admit Card
Get Updates Brochure
Your Staff Selection Commission Combined Graduate Level Exam brochure has been successfully mailed to your registered email id “”.
