Inscribed Squares

A square is inscribed in a circle or a polygon if its four vertices lie on the circumference of the circle or on the sides of the polygon. Figure A shows a square inscribed in a circle. Figure B shows a square inscribed in a triangle. Figure C shows a square inscribed in a quadrilateral.

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.

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SOLUTION

Let $r$ be the radius of the circle, and $x$ the side length of the square, then the area of the square is $x^2$. By the Pythagorean theorem, we have $(2r)^2=x^2+x^2.$

Simplifying further, we get $x^2=2r^2.\qquad (1)$

The area of the shaded region is equal to the area of the circle minus the area of the square, so we have

$\begin{aligned} 25\pi -50 &=\pi r^2 - 2r^2\\ &=r^2(\pi-2)\\ r^2&=\dfrac{25\pi -50}{\pi -2}\\ &=25.\qquad (2) \end{aligned}$

Now substituting (2) into (1) gives $x^2=2\times 25=50.\ _\square$

A square with side length $a$ is inscribed in a circle. Express the radius of the circle in terms of $a$.

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SOLUTION

The diagonal of the square is the diameter of the circle. Let $d$ and $r$ be the diameter and radius of the circle, respectively. Then by the Pythagorean theorem, we have

$\begin{aligned} d^2&=a^2+a^2\\ &=2a^2\\ d&=\sqrt{2a^2}\\ &=a\sqrt{2}. \end{aligned}$

We know that the diameter is twice the radius, so

$r=\dfrac{d}{2}=\dfrac{a\sqrt{2}}{2}.\ _\square$