# Solution to the [[Five Squares]] Puzzle +-- {.image} [[FiveSquares.jpeg:pic]] > Five squares. What’s the total shaded area? =-- ## Solution by [[Similar Triangles]] +-- {.image} [[FiveSquaresLabelled.jpeg:pic]] =-- With the points labelled as above, angles $A \hat{C} B$ and $D \hat{C} E$ are the same since they are [[vertically opposite]]. Then as triangles $A B C$ and $C D E$ are [[right-angled triangles|right-angled]] and have an angle in common, they are [[similar]]. Let $a$, $b$, and $c$ be the lengths of line segments $C E$, $C D$, and $E D$ respectively. Then the lengths of line segments $A C$, $C B$, and $A B$, are, respectively, $c - b$, $\frac{10 - a}{2}$, and $5$. So by similarity: $$ \begin{aligned} \frac{c}{5} &= \frac{a}{c - b} \\ \frac{10 - a}{10} &= \frac{b}{c} \end{aligned} $$ The second of these rearranges as follows: $$ \begin{aligned} c(10 - a) &= 10 b \\ 10 c - a c &= 10 b \\ a c &= 10 c - 10 b = 10 (c - b) \\ \frac{a c}{c - b} &= 10 \end{aligned} $$ Then using the first: $$ 10 = \frac{a c}{c - b} = \frac{c}{5} \times c $$ So $c^2 = 5 \times 10 = 50$. Since $c$ is the side length of the tilted squares, the total shaded area is $100$. These rearrange to $c(c - b) = 5 a$ and $c(10 - a) = 10 b$.