# Solution to the [[Two Triangles in a Rectangle]] Puzzle +-- {.image} [[TwoTrianglesinaRectangle.png:pic]] > Show that the two shaded areas cover the same proportion of the rectangle. =-- ## Solution by [[Similar Triangles]] and [[Area of a Triangle]] +-- {.image} [[TwoTrianglesinaRectangleAnnotated.png:pic]] =-- With the diagram labelled as above, let $a$ and $b$ be the lengths of the sides of the outer rectangle, with $a$ the length of $A C$ and $b$ the length of $A G$. Then $D I$ has length $a - b$, so the area of the orange triangle, $C I E$, is $\frac{1}{2} b ( a - b)$. Let $c$ be the length of $B I$, so $F I$ has length $b - c$. The area of the yellow triangle is then $\frac{1}{2} a (b - c)$. Triangle $A B I$ is [[similar]] to triangle $A C E$ since both share angle $I \hat{A} B$ and both are [[right-angled triangles]]. Therefore, $\frac{c}{b} = \frac{b}{a}$ which rearranges to $a c = b^2$. This means that $a (b - c) = a b - a c = a b - b^2 = b(a - b)$. Hence the area of the yellow triangle is $\frac{1}{2} a (b - c) = \frac{1}{2} b (a - b)$ which is the area of the orange triangle. ## Solution by [[Area of a Triangle]] With the same diagram as above, consider the triangles $G J I$ and $I J C$. To work out their areas, we need the [[perpendicular height]] of $I$ above $G C$ and the lengths of $G J$ and $J C$. Since $J$ is the [[midpoint]] of $G C$, these lengths will be the same and so the areas of $G J I$ and $I J C$ are the same. Then triangles $G J E$ and $E J C$ also have the same area since each is a quarter of the full rectangle. Therefore, the orange and yellow areas are the same. (A similar result follows by considering triangles $A G I$ and $A C I$ and noting that the [[perpendicular height]] of $C$ above $A I$ is the same as that of $G$.)