Area of a square whose one part is in circle

Question

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 ?

Answer 1

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 $\color{blue}{64 \ \text{cm}^2}$.

Answer 2

Hint: Draw all the relevant radii, let $x$ be the side length of the square, and write down what you get from Pythagoras theorem.

Answer 3

Let $a$ be a side and $r$ the radius. Check that $$(a-r)^2+{a^2\over 4}=r^2$$ $\therefore$$${5a^2\over 4}-10a=0\implies a=8$$