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Question

An equilateral triangle ABC is inscribed in a circle of radius 10 cm, which is centered at O, as shown below. Then the length of the side of this triangle is equal to


A

253 cm

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B

153 cm

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C

103 cm

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D

53 cm

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Solution

The correct option is C

103 cm


Given, ABC is equilateral and is inscribed in a circle.

Let OE be the radius of this circle which is perpendicular to BC and cuts BC at point D.

Let the length of side BC be x cm.

We know that the perpendicular from the centre of the circle to chord bisects the chord.

So, BD = x2 cm

Also, we know that a side of an equilateral triangle drawn with vertices on a circle bisects the radius perpendicular to it.

So, OD = 5 cm ( OE = 10 cm and OD = OE/2)

Applying Pythagoras theorem to BOD, we get,

BO2=OD2+BD2

102=52+(x2)2

(x2)2=75

x2=75

x=225×3

x=103 cm


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