# Solution to the [[Three Squares VII]] Puzzle +-- {.image} [[ThreeSquaresVII.jpeg:pic]] > Three squares. What’s the shaded area? =-- ## Solution by [[Pythagoras' Theorem]], [[Similar Triangles]], and the [[Area of a Triangle]] +-- {.image} [[ThreeSquaresVIILabelled.jpeg:pic]] =-- Label the points as above. The side length of the outer square is $6$, and triangle $A I H$ is [[right-angled triangle|right-angled]] so [[Pythagoras' Theorem]] applies. This means that $A H$ has length given by: $$ \sqrt{2^2 + 6^2} = \sqrt{40} = 2\sqrt{10} $$ 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: $$ \frac{\sqrt{10}}{6} \times 2 \sqrt{10} = \frac{10}{3} $$ 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 $\frac{1}{2} \times 6 \times 2 = 6$.