An equilateral \(\triangle\) inscribed in a circle, the altitudes of from each vertex meets at a common point

O. The distance from the tip of the vertex to the point O is the radius of the circle

Here,

For easy calculation, let take the radius = 2 ( it can be any value)
MO = radius of the circle = 2

since, MNP is equilateral \(\triangle\)

so, \(\angle\) OMP = 30

\(\angle\) OBM = 90
\(
\angle\) MOB = 60

Hence, the \(\triangle\) MOB is a 30:60:90 triangle and the sides are divided in the ratio as 1 : \(\sqrt{3}\) : 2

We know, OM = 2

therefore, MB = \(\sqrt{3}\)

and OB = 2

Now, area of the circle = \(\pi * 2^2\) = \(4\pi\)

Area of the equilateral \(\triangle \)= \(\frac{{\sqrt{3}}}{4}\) * \({side}^2\)

= \(\frac{{\sqrt{3}}}{4}\) * \({2\sqrt{3}}^2\)

= \(3\sqrt{3}\)


Area of the shaded region = \(4\pi\) - \(3\sqrt{3}\)

QTY A = \(3\sqrt{3}\)

QTY B = \(4\pi\) - \(3\sqrt{3}\)

adding \(3\sqrt{3} \)

QTY A = \(6\sqrt{3}\)

QTY B = \(4\pi\)


Hence QTY A < QTY B