Notes
squares meeting in a circle solution

Squares Meeting in a Circle

Solution by Pythagoras' Theorem and Area of a Triangle

Let $x$ be the length of the sides of the squares. As $A B$ is the height of a triangle with base $x$ and area $5$, it has length $\frac{10}{x}$. Similarly, $C D$ has length $\frac{24}{x}$. Triangles $A O B$ and $O D C$ are congruent and are right-angled. Let $r$ be the radius of the circle, then by Pythagoras' theorem, $r^2 = \frac{100}{x^2} + \frac{576}{x^2} = \frac{676}{x^2}$. So $x r = 26$. The pink triangle therefore has area $\frac{1}{2} x r = 13$.