There is a shaded square inside a circle represented by the equation x^2 + y^2 = 9. Find the area of the square in the circle.
I seek the first-two steps.
I am assuming the square is inscribed within the circle, and so the diagonal of the square is equal to the diameter of the circle.
Let $\displaystyle D=2r$ by the diameter of the circle. The area $\displaystyle A$ of the square is:
$\displaystyle A=s^2$
And we know:
$\displaystyle 2r=\sqrt{2}s\implies s=\sqrt{2}r\implies A=2r^2$
Can you proceed?
If the circle is inscribed within the square, then the side length of the square will be equal to the diameter of the circle, thus:
$\displaystyle A=(2r)^2=4r^2$
Now, given the result for the original problem you posted, we now know:
$\displaystyle 2r^2<\pi r^2<4r^2$
Or:
$\displaystyle 2<\pi<4$
How could we improve these bounds on $\displaystyle \pi$?