Question

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

Answer 1

given: AC=AB=BC= 9 cm

to find: the radius.

construction: draw AD being the perpendicular bisector of side BC.

proof: we know that that ∠ADC would be equal to 90° by construction.

And BD = DC

Now,

∠B = 60° { given }
∠OBD = 30° { half of 60° as BO bisects ∠B }

in triangle OBD

$cos30°=\frac{base}{hypotenuse}$

$\Rightarrow \frac{\surd 3}{2}=\frac{4.5}{hypotenuse}$

$Hypotenuseorradius=3\surd 3cm.$