Area of a square whose one part is in circle A square has two of its vertices on a circle and the other two on a tangent to the circle. If the diameter of the circle is $10$ cm, then what is the area of the square is?
My solution:
I figured out this diagram 

What to do after this ?

Observe the diagram below:

Let $x =$ the length of one side of the square.
Since the diameter of the circle is $10 \ \text{cm}$, we know that $OA=OC=OD=5 \ \text{cm}$, since they are all radii of circle $O$. Thus, $OB = x-5$. We also know that $AB = \frac{x}{2}$. Using the Pythagorean theorem:
$$
\begin{align*}
(AB)^2+(OB)^2&=(OA)^2 \\
\left(\frac{x}{2}\right)^2 + (x-5)^2&=5^2 \\
\frac{1}{4}x^2 + x^2-10x+25&=25 \\
\frac{5}{4}x^2-10x&=0 \\
x\left( \frac{5}{4}x-10 \right) &= 0 \\
x=0 \text{ or } x&=8
\end{align*}
$$
Thus one side of the square is $8 \ \text{cm}$, and the area of the square is $\color{blue}{64 \ \text{cm}^2}$.
