Notes
three squares vii solution

Solution to the Three Squares VII Puzzle

Solution by Pythagoras' Theorem, Similar Triangles, and the Area of a Triangle

Label the points as above.

The side length of the outer square is $6$, and triangle $A I H$ is right-angled so Pythagoras' Theorem applies. This means that $A H$ has length given by:

So line segment $A F$ has length $\sqrt{10}$ and the area of the smaller squares is $10$. Triangle $A B E$ therefore has area $5$.

Angle $B \hat{A} C$ is equal to angle $H \hat{A} I$ since both are $90^\circ - C \hat{A} F$. Therefore, triangles $A B C$ and $A I H$ are similar. This means that the lengths of line segments $A B$ and $B C$ are in the ratio $3 : 1$, so then as $B E$ and $A B$ have the same length, $C$ divides $B E$ in the ratio $1 : 2$. This means that triangle $A C E$ has area two thirds of $A B E$, so has area $\frac{10}{3}$.

The scale factor from triangle $A I H$ to $A B C$ is $\frac{\sqrt{10}}{6}$, so line segment $A C$ has length:

By the area of a triangle, taking $A C$ as the base of triangle $A C E$, point $E$ therefore has height $2$ above $A C$, and therefore above $A D$. So then triangle $A E D$ has area $\frac{1}{2} \times 6 \times 2 = 6$.