Question

If the length of side of an equilateral triangle inscribed in a circle of radius $4$ cm is $4\sqrt{m}$.Find $m$

Answer 1

Given $△ABC$ is a equilateral triangle inscribed in a circle.
Since, $△ABC$ is a equilateral triangle,
$\angle C=60$
Using the property, Angle subtended by same arc to centre of circle is twice the angle subtended by same arc to any point on the circle.
$\therefore \angle AOB=2\times \angle C=2\times 60=120$
Drawing perpendicular line from centre O to AB.
Using the property, A perpendicular line joining centre and chord bisects the chord and angle at centre.
$\therefore AD=BD=x\angle AOD=\angle BOD=60$
In $△AOD$,
$\angle OAD+\angle AOD+\angle ODA=180$ (Angle sum property of a triangle)
$or,90+60+\angle OAD=180or,\angle OAD=180-90-60=30$
Now, In $△AOD$,
$\cos 30=\frac{x}{4}or,\frac{\sqrt{3}}{2}=\frac{x}{4},or,x=\frac{4\sqrt{3}}{2}=2\sqrt{3}or,AD=BD=2\sqrt{3}or,AB=2\times 2\sqrt{3}=4\sqrt{3}$
Side of equilateral triangle =$4\sqrt{3}$

$\therefore m=3$