# Inscribed Squares

## Sample Problems

## Square \(ABCD\) is inscribed in a circle with center at \(O\) as shown in the figure. If the area of the shaded region is \(25\pi -50\), find the area of the square. Take \(\pi=3.1416\).

We let \(r\) be the radius of the circle and \(x\) be the side length of the square, so the area of the square is \(x^2\). By pythagorean theorem , we have\((2r)^2=x^2+x^2\)

Simplifying further, we get

\(x^2=2r^2\) (Equation 1)

The area of the shaded region is equal to the area of the circle minus the area of square. We have

\(25\pi -50=\pi r^2 - 2r^2\)

\(25\pi -50=r^2(\pi-2)\)

\(r^2=\dfrac{25\pi -50}{\pi -2}\) (Equation 2)

Now substitute (Equation 2) in (Equation 1) and at the same time substitute \(\pi=3.1416\), we have

\(x^2=2\left[\dfrac{25(3.1416)-50}{3.1416-2}\right]=50\) \( _\square \)

## A square with side length \(a\) is inscribed in a circle. Expressed the radius of the circle in terms of \(a\).

The diagonal of the square is the diameter of the circle. We let \(d\) and \(r\) be the diameter and radius of the circle, respectively. By pythagorean theorem , we have\(d^2=a^2+a^2\)

\(d^2=2a^2\)

\(d=\sqrt{2a^2}=a\sqrt{2}\)

We know that the diameter is twice the radius. We have

\(r=\dfrac{d}{2}=\dfrac{a}{2}\sqrt{2}\) \( _\square \)

**Cite as:**Inscribed Squares.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/inscribed-squares/