# Squares Meeting in a Circle +-- {.image} [[SquaresMeetinginaCircle.png:pic]] > The two squares are identical, and the pink line is a diameter of the circle. What's the area of the pink triangle? =-- ## Solution by [[Pythagoras' Theorem]] and [[Area of a Triangle]] +-- {.image} [[SquaresMeetinginaCircleLabelled.png:pic]] =-- 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 triangle|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 $\frac{1}{2} x r = 13$.