In the figure, if the triangle$▵ABC$ is an isosceles triangle of perimeter 20, calculate the approximate area of the circle with center O

To calculate the circle's area, you must know its radius. The lengths of the two sides are equal in an isosceles triangle.
A circle is made up of all the points that are at a defined distance from a single fixed point. The centre is the fixed point. The radius r is the constant distance from the centre.
The circle's diameter and radius are related as follows: diameter is equal to twice the radius.

Every right angle triangle can be solved using Pythagoras' theorem. A rectangle's neighbouring sides meet at a 90° angle.

The base and perpendicular of a right-angled triangle are at a 90° angle to one another, and the hypotenuse is the longest side.

Triangle $\mathrm{ABC}$ is isosceles
$\therefore \mathrm{BC}=\mathrm{AC}=\frac{(20-4)}{2}=8$

In triangle $\mathrm{ACD}$, angle $\mathrm{ACD}$ is 90 degrees, because angle made by diameter at any point on circle is $90°$.
Therefore we get
${\mathrm{AD}}^2={\mathrm{AC}}^2+{\mathrm{CD}}^2$
$\Rightarrow {\mathrm{AD}}^2=64+9=73$
$\Rightarrow \mathrm{AD}=2\mathrm{r}=\sqrt{73}$
$\Rightarrow \mathrm{r}=\frac{\sqrt{73}}{2}$
Area of circle is$\frac{22}{7}\times {\left(\frac{\sqrt{73}}{2}\right)}^2=\frac{22\times 73}{7\times 4}=57.33$

Hence, the correct answer is option D.