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