Question

an equilateral triangle of side 9 cm is inscribed in a circle find the radius

Answer 1

Answer:

$Radius \: of \: the \: circle (r)= 3\sqrt{3}\:cm$

Step-by-step explanation:

∆ABC is an equilateral triangle.

AB = BC = CA = 9 cm

<BOC = 2<BAC

= 2× 60°

= 120°

OD bisects BC perpendicularly.

In ∆BOC ,

<OBD = 30° ,

BD = BC/2 = 9/2 cm

$cos 30\degree = \frac{BD}{OB}\\=\frac{\frac{9}{2}}{r}$

$\implies \frac{\sqrt{3}}{2} = \frac{\frac{9}{2}}{r}$

$\implies r = \frac{9}{2} \times \frac{2}{\sqrt{3}}\\=\frac{9}{\sqrt{3}}\\=\frac{9\sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\=\frac{9\sqrt{3}}{3}=3\sqrt{3}$

Therefore,

$Radius \: of \: the \: circle (r)= 3\sqrt{3}\'cm$

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Answer 2

Answer:

∆ABC is an equilateral triangle.

AB = BC = CA = 9 cm

<BOC = 2<BAC

= 2× 60°

= 120°

OD bisects BC perpendicularly.

In ∆BOC ,

<OBD = 30° ,

BD = BC/2 = 9/2 cm

Step-by-step explanation:

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