    Question

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

A

23 cm

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B

22 cm

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C

33 cm

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D

3 cm

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Solution

## The correct option is A 2√3 cm Construct an altitude from point A to point D in ΔABC. Given: ABC is an equilateral triangle of side 6 cm, We know that, perpendicular drawn from the cneter of the circle bisects the chord. Here, BC is a chord of the circle, as OD is perpendiular from the centre to BC. Hence it bisects BC i.e., BD = DC = 3cm In ΔABD (AB)2=(BD)2+(AD)2 ⟹(6)2=(3)2+(r+x)2 ⟹36=9+(r+x)2 ⟹(r+x)=3√3)−−−−−−−−(1) InΔBOD ⟹(BO)2=(OD)2+(BD)2 ⟹(r)2=(x)2+(3)2 ⟹(r)2−(x)2=9 ⟹(r+x)(r−x)=9 We know that (r+x)=3√3 ∴(r−x)=9r+x=93√3 ⟹r−x=√3−−−−−−−−(2) Adding (1) and (2), we get r=2√3 cm which is the radius of the circle.  Suggest Corrections  0      Similar questions
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