Question

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

• A. $\frac{r}{4}\sqrt{3}$
• B. $2r\sqrt{3}$
• C. $\frac{r}{2}\sqrt{3}$
• D. $r\sqrt{3}$

Answer 1

The correct option is D

$r\sqrt{3}$

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 $x$ units.

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

So, BD = $\frac{x}{2}$ units

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

So, OD = $\frac{r}{2}$ units ($\because$ OE = r units and OD = OE/2)

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

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

$⟹r^2=(\frac{r}{2}{)}^2+(\frac{x}{2}{)}^2$

$⟹(\frac{x}{2}{)}^2=r^2-(\frac{r}{2}{)}^2$

$⟹(\frac{x}{2}{)}^2=r^2-\frac{r^2}{4}$

$⟹(\frac{x}{2}{)}^2=\frac{3r^2}{4}$

$⟹\frac{x}{2}=\sqrt{\frac{3r^2}{4}}$

$⟹x=2\sqrt{\frac{3r^2}{4}}$

$⟹x=r\sqrt{3}$