# A Square, Triangle, and Circle in a Square +-- {.image} [[ASquareTriangleandCircleinaSquare.png:pic]] > The square, circle and triangle are stacked inside a larger square. What's the area of the circle? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[ASquareTriangleandCircleinaSquareLabelled.png:pic]] =-- Let $a$, $b$, $c$ be the lengths of the sides of triangle $A B C$, so that $A B$ has length $a$, $B C$ of length $b$, and $A C$ of length $c$. Since the central square has area $100$, $c = 10$. The area of the triangle is given by $\frac{1}{2} a b$ so $a b = 48$. Applying [[Pythagoras' theorem]] to this triangle shows that $a^2 + b^2 = 100$. Putting these together, $(a + b)^2 = 100 + 2 \times 48 = 196$, so $a + b = 14$. Triangle $C E G$ is [[congruent]] to triangle $A B C$. Inside that triangle, $C D$ and $C H$ have the same length, as to $G H$ and $G F$. So the combined lengths of $F E$ and $E D$ give the difference between the combination of $C E$ and $E G$ and the length of $C G$. So $F E$ has length $\frac{14 - 10}{2} = 2$ and so the area of the circle is $4 \pi$.