Question

What is the area of circle whose radius is diagonal of a square whose area is 9?

Answer 1

Area of the square is 9,therefore side length of square is 3 units.
length of diagonal of suare is =sqrt{3^2+3^2} =3sqrt{2}
Area of circle=pi*[3sqrt{2}]^2
                   =18pi
                   =18*22/7
                   =56.57(approx.)

Answer 2

We know the formula for the area of a circle is $\pi r^2$ and for a square it is $s^2$.

The "r" is the diagonal. Now let's find the side of the square. It is $(\sqrt{9}=3)$. Use trigonometry to find the diagonal.

We know two angles, namely the right angle and the 45.We know on side and considering the right angle is $\alpha$, the hypotenuse is the diagonal and the other side of the square is the opposite. Find the sine.

$sin( \alpha )=sin(45)= 0.707(degrees)$
$h=sin( \alpha )*6$
$h=\sqrt{18}$

Therefore the diagonal is $\sqrt{18}$ . Now put in the area

$A=\pi r^2 \\ A=\pi\sqrt{18}^2 \\ A=18\pi \\ A=56.54$

The Area is 56.54 or $18\pi$.