Question

An equilateral triangle of side $9cm$ inscribed in a circle The radius of the circle is:

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

Answer 1

Answer: (C)

$△$ABC is an equilateral triangle

$AB=BC=CA=9cm$

O is the circumcentre of $△$ABC

$\therefore$OD id the perpendicular of the side BC

In $△$OBD and $△$ODC

$OB=OC$ (Radius of the circle)

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

$OD=OD$ (common)

$\therefore △OBD=△ODC$

$\Rightarrow \angle BOD=\angle COD$

$\angle BOC=2\angle BAC=2\times {60}^{\circ }={120}^{\circ }$ (The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle)

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

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

In $△BOD$

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

$\therefore \frac{\sqrt{3}}{2}=\frac{\frac{9}{2}}{OB}$

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