# Solution to the [[Three Congruent Rectangles III]] Puzzle +-- {.image} [[ThreeCongruentRectanglesIII.png:pic]] > The three green rectangles are congruent. What fraction of the square do they cover? =-- ## Solution by [[Pythagoras' Theorem]] and [[Congruency]] +-- {.image} [[ThreeCongruentRectanglesIIILabelled.png:pic]] =-- The first step is to establish that the tilted rectangle would fit in the middle of the square. That is, with the points labelled as above then rectangles $B J F L$ and $B C F G$ are [[congruent]]. This is a consequence of [[Pythagoras' theorem]]. The green rectangles are all congruent, so $F C$ and $B J$ have the same length. Therefore, triangles $B F C$ and $B F J$ are [[right-angled triangles]] that have two sides of the same length, and therefore have all three sides the same. Triangles $B C I$ and $F J I$ are therefore [[congruent]] since both are [[right-angled triangle|right-angled]], the lengths $B C$ and $F J$ are the same, and the angles $B \hat{I} C$ and $F \hat{I} J$ are equal since they are [[vertically opposite]]. Therefore, the lengths of $F I$ and $B I$ are the same. Each of the three rectangles consists of one third of the outer square, so the ratio of the side lengths of the rectangles is $1 : 3$. Let $B C$ have length $1$. Then $C F$ has length $3$, so the sum of the lengths of $C I$ and $I B$ is $3$. Let $C I$ have length $x$, then by [[Pythagoras' Theorem]]: $$ 1^2 + x^2 = (3 - x)^2 = 9 - 6 x + x^2 $$ Hence $x = \frac{4}{3}$. The area of triangle $B C I$ is therefore $\frac{4}{6}$ and so the white area is $\frac{4}{3}$. The total area of the square is $9$, so the area that is shaded is, as a fraction of the whole: $$ \frac{9 - \frac{4}{3}}{9} = \frac{25}{27} $$