Question

An equilateral triangle of side $9cm$ is inscribed in a circle. Height of an equilateral triangle is

• A. $\frac{3\sqrt{3}}{2}cm$
• B. $9\sqrt{3}cm$
• C. $\frac{9\sqrt{3}}{2}cm$
• D. $\frac{5\sqrt{3}}{2}cm$

Answer 1

The correct option is A $\frac{3\sqrt{3}}{2}cm$

Given: $△ABC$ is an equilateral triangle.

$AB=BC=CA=9cm$

$O$ is the circumcentre of $△ABC$.

$\therefore OD$ is the perpendicular bisector of the side $BC$. ($O$ is the point of intersection of the perpendicular bisectors of the sides of the triangle)

In $△OBD$ and $△OCD$,

$OB=OC$ (Radius of the circle)

$BD=DC$ ($D$ is the mid point of $BC$)

$OD=OD$ (Common)

$\therefore △OBD\cong △OCD$ ($SSS$ congruence criterion)

$\Rightarrow \angle BOD=\angle COD$ (CPCT)

and, $\angle BOC=2\angle BAC=2\times {60}^{\circ }={120}^{\circ }$ ( 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)

$\therefore \angle BOD=\angle COD=\frac{\angle BOC}{2}=\frac{{120}^{\circ }}{2}={60}^{\circ }$

and $BD=DC=\frac{BC}{2}=\frac{9}{2}cm$

In $△BOD$, we have

$sin\angle BOD=sin{60}^{\circ }=\frac{BD}{OB}$

$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\frac{9}{2}cm}{OB}$

$\Rightarrow OB=3\sqrt{3}cm$

Thus, the radius of the circle $=3\sqrt{3}cm$.

Now, In $△BOD$ by Pythagoras theorem, we have

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

$\Rightarrow O{D}^2=(3\sqrt{3}{)}^2-(\frac{9}{2}{)}^2$

$\Rightarrow O{D}^2=27-\frac{81}{4}$

$\Rightarrow O{D}^2=\frac{108-81}{4}$

$\Rightarrow O{D}^2=\frac{27}{4}$

$\Rightarrow OD=\frac{3\sqrt{3}}{2}$

Thus, Height of triangle $=\frac{3\sqrt{3}}{2}cm$