Question

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

• A. $\frac{2x}{3\sqrt{3}}$ units
• B. $\frac{x}{2\sqrt{3}}$ units
• C. $\frac{2x}{\sqrt{3}}$ units
• D. $\frac{x}{\sqrt{3}}$ units

Answer 1

The correct option is D

$\frac{x}{\sqrt{3}}$ units

Given, $△$ABC is an equilateral triangle each side of which is $x$ units and is inscribed in a circle which is centered at O.

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

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

So, we must have, BD = $\frac{x}{2}$ units ($\because$ Given, BC = $x$ 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

By applying Pythagoras theorem to $△$BOD, we get,

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

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

$⟹r^2-\frac{r^2}{4}=\frac{x^2}{4}$

$⟹\frac{3r^2}{4}=\frac{x^2}{4}$

$⟹r^2=\frac{x^2}{3}$

$⟹r=\frac{x}{\sqrt{3}}$ units