    Question

# An equilateral triangle ABC, whose side is 6 cm, is inscribed in a circle. Find the radius of the circle.

A

2√3 cm

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B

2√2 cm

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C

3√3 cm

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D

3 cm

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Solution

## The correct option is B 2√3 cm Given, that ABC is an equilateral triangle of side 6 cm Let 'r' be the radius of the circle. Construct an altitude from point 'A' to point 'D' in ΔABC In an equilateral triangle the altitute will also be the perpendicular bisector. Perpendicular bisectors intersect at circumcentre. So, in the below fig. 'O' is the centre of the circle. In ΔABD AB2=BD2+AD2 (by Pythagoras theorem) ⇒62=32+(r+x)2 ⇒36=9+(r+x)2 ⇒r+x=3√3 ... (i) OD = x = (r+x)−r = 3√3−r In △BOD, BO2=OD2+BD2 ⇒r2=(3√3−r)2+32 ⇒r2=27−6√3r+r2+9 ⇒6√3r=36 ⇒r=366√3 ⇒r=2√3 cm which is the radius of the circle. Alternate: In △BOD, ∠OBD=30∘ cos30∘=BDOB ∵cos30∘=√32 ⇒√32=3r ⇒ r=6√3 ⇒ r=2√3 cm  Suggest Corrections  0      Similar questions  Related Videos   Circles and Their Chords - Theorem 3
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