Question

An equilateral triangle of side $9cm$ is inscribed in a circle. Find the radius of the circle.

Answer 1

Given $BCD$ is a equilateral triangle inscribed in a circle with sides length $CB=BD=DC=9cm$.
Now, take a look on the triangles $\Delta AOC$ and $\Delta AOB$,

OC=OB(D is the mid-point of BC)

AB=AC(Radius of the circle)

AO=AO(Common side of the triangle)

Therefore, using the SSS rule of the congruency, we can say that the triangles AOC and AOB are congruent.

Thus, $\Delta AOC\cong$ $\Delta AOB$,
Then we have $\angle CAO=\angle BAO$.
We know that the angle subtended by an arc at the centre is double the angle subtended by it at any point on the remaining part of the circle.
$CAB=2\angle CDB$

Since, $\Delta BCD$ is an equilateral triangle, then each of its angles has a measure of ${60}^{\circ }$.
SO, $\angle CDB={60}^{\circ }$

Therefore, $\angle CAB=2\times {60}^{\circ }={120}^{\circ }$

Now, we can easily, break the angle ∠CAB in the sum of two angles as:
∠CAB=∠CAO+∠BAO

Since, we have ∠CAO=∠BAO, then

∠CAB=∠CAO+∠CAO

∠CAB=2∠CAO

$\angle CAO=\frac{1}{2}\left(\angle CAB\right)$

Substitute the value $\angle CAB={120}^{\circ }$ into the equation:

$\Rightarrow \angle CAO=12({120}^{\circ }$

$\therefore \angle CAO={60}^{\circ }$.
Since, the altitude from the vertex D bisects the side BC,
So, $CO=BO=\frac{1}{2}\times BC$
$\Rightarrow CO=\frac{9}{2}$
From triangle $\Delta AOC$, we have $\sin {90}^{\circ }=\frac{CO}{AC}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{CO}{AC}$
$\Rightarrow AC=\frac{\sqrt{3}}{2}\times CO$
$\Rightarrow AC=\frac{2}{\sqrt{3}}\times \frac{9}{2}$
$\therefore AC=3\sqrt{3}$ is the required radius.