# Five Rectangles in a Square +-- {.image} [[FiveRectanglesinaSquare.png:pic]] > What fraction of the square is covered by these $5$ identical rectangles? =-- ## Solution by [[Lengths in a Square]] +-- {.image} [[FiveRectanglesinaSquareLabelled.png:pic]] =-- Let the sides of the rectangle be $a$ and $b$, with $b$ the short side and $a$ the long. The side of the outer square is then $a + b$ so its area is $(a + b)^2$. Each rectangle has area $a b$, so the fraction of the square covered by the rectangles is: $$ \frac{5 a b}{(a + b)^2} $$ Triangle $F D E$ is an [[isosceles]] [[right-angled triangle]] so the length of $F E$ is $\sqrt{2}$ times the length of $D E$. Similarly, the length of $E H$ is $\sqrt{2}$ times the length of $E G$. Since $F E$ has length $a$ and $E H$ has length $b$, the sum of these lengths is $a + b$ and this is $\sqrt{2}$ times the length of $D G$, which is $a - b$. That is to say, $a + b = \sqrt{2}(a - b)$. Squaring both sides shows that: $$ \begin{aligned} a^2 + 2 a b + b^2 &= 2 a^2 - 4 a b + 2 b^2 \\ 8 a b &= a^2 + 2 a b + b^2 \\ &= (a + b)^2 \end{aligned} $$ So the fraction of the square covered by the rectangles is $\frac{5}{8}$.