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. $25\sqrt{3}$ cm
• B. $15\sqrt{3}$ cm
• C. $10\sqrt{3}$ cm
• D. $5\sqrt{3}$ cm

Answer 1

The correct option is C

$10\sqrt{3}$ 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 = $\frac{x}{2}$ 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 ($\because$ OE = 10 cm and OD = OE/2)

Applying Pythagoras theorem to $△$BOD, we get,

$B{O}^2=O{D}^2+B{D}^2$

$⟹{10}^2=5^2+(\frac{x}{2}{)}^2$

$⟹(\frac{x}{2}{)}^2=75$

$⟹\frac{x}{2}=\sqrt{75}$

$⟹x=2\sqrt{25\times 3}$

$⟹x=10\sqrt{3}$ cm