# Solution to the Subdivided Parallelogram Puzzle +-- {.image} [[SubdividedParallelogram.png:pic]] > Unshaded triangles are isosceles. What fraction of the parallelogram is shaded? =-- ## Solution by [[Similar Triangles]] and [[Angles in Parallel Lines]] +-- {.image} [[SubdividedParallelogramLabelled.png:pic]] =-- _Note: there is a hidden assumption here that the bases of the unshaded triangles lie along the edges of the outer parallelogram._ Consider the diagram as labelled above. As the outer shape is a [[parallelogram]], line segments $E C$ and $A B$ are [[parallel]]. Therefore by [[angles in parallel lines]], angles $ D \hat{B} A$ and $B \hat{D} C$ are equal, as are angles $D \hat{A} B$ and $A \hat{D} E$. The base angles of the [[isosceles]] triangles $A D B$, $E F D$, and $D G C$ are therefore all the same and so these triangles are all [[similar]]. The sides $E D$, $D C$, and $A B$ correspond, so the [[scale factor|scale factors]] from the larger triangle to each of the smaller are, respectively, $\frac{2}{5}$ and $\frac{3}{5}$. The corresponding area scale factors are $\frac{4}{25}$ and $\frac{9}{25}$. The area of triangle $A D B$ is half that of the parallelogram so the unshaded area is $$ \frac{1}{2} \big( 1 + \frac{4}{25} + \frac{9}{25} \big) = \frac{19}{25} $$ of the parallelogram. The shaded area is therefore $\frac{6}{25}$ths of the parallelogram.