I drew a square of side 5 together with its circumcircle. At the middle of the arc between a side and the circle, I drew a smaller square with two vertices on the side, and the other two on the circle.

What was the area of the smaller square?

Let the equation of the circle be **x^2+y^2=25/2 (eq1).**

Then the coordinates of the point where the small square touches the circle in the 1st quadrant are: (x, y)

2x and y-2.5 are the sides of the small square.

**2x= y-2.5** (eq2) squared and subtracted from 4* **(eq1)** leads to

**y^2-y-8.75=0** , **(eq3)**

(y-3,5)*(y+2,5)=0

The positive answer is y=3.5, so 2x=1 and y-2,5=1

and the area of the smaller square is **1*1=1.**

*Edited on ***December 7, 2014, 10:27 pm**