Find the area of the circle in which a square of area $64c{{m}^{2}}$is inscribed.
Find the area of the circle in which a square of area $64c{{m}^{2}}$is inscribed.

As per the question, it is given that

Area of square inscribed the circle $=64c{{m}^{2}}$

$sid{{e}^{2}}=64$

Side $=8cm$

Thus, $AB=BC=CD=DA=8cm$

Using Pythagoras theorem in triangle ABC,

$A{{C}^{2}}=A{{B}^{2}}+B{{C}^{2}}$

$A{{C}^{2}}={{8}^{2}}+{{8}^{2}}$

$A{{C}^{2}}=64+64=128$

$AC=\sqrt{128}=8\sqrt{2}cm$

Then, as $\angle B={{90}^{\circ }}$ and AC being the diameter of the circle

The radius is $AC/2=8\sqrt{2}/2=4\sqrt{2}cm$

therefore, the area of the circle $=\pi {{r}^{2}}=3.14{{\left( 4\sqrt{2} \right)}^{2}}$

$=100.48c{{m}^{2}}$