# Solution to the Four Squares VI Puzzle +-- {.image} [[FourSquaresVI.png:pic]] > Four squares. What’s the total shaded area? =-- ## Solution by [[Pythagoras' Theorem]] +-- {.image} [[FourSquaresVILabelled.png:pic]] =-- As in the above diagram, let $a$, $b$, and $c$ be the side lengths of the squares, and $d$ is the difference between the top of the large yellow square and the lower white square. Applying [[Pythagoras' Theorem]] to the triangle shows that: $$ c^2 + d^2 = 2^2 = 4 $$ From the diagram, $2c = a + b$ and $d = a - c$, so $2d = a - b$. Therefore: $$ \begin{aligned} (2 c)^2 + (2 d)^2 &= (a + b)^2 + (a - b)^2 \\ &= a^2 + 2 a b + b^2 + a^2 - 2 a b + b^2 \\ &= 2 a^2 + 2 b^2 \end{aligned} $$ Therefore $$ a^2 + b^2 = 2 c^2 + 2 d^2 = 8 $$