# Triangles with a Parallelogram +-- {.image} [[TriangleswithaParallelogram.png:pic]] > What's the area of the parallelogram? =-- ## Solution by [[Similar Triangles]] and [[Area Scale Factor]] +-- {.image} [[TriangleswithaParallelogramLabelled.png:pic]] =-- With the points labelled as above, the area of the [[parallelogram]] can be found by multiplying the lengths of $F G$ and of $I H$. The line segments $F C$ and $A E$ are [[parallel]], and angles $F \hat{B} C$ and $E \hat{B} A$ are [[vertically opposite]] and so are equal. Therefore, triangles $A B E$ and $F B C$ are [[similar]]. The [[area scale factor]] is $9$, so the length [[scale factor]] is $3$. This means that the length of $I H$ is $\frac{4}{3}$ times that of $B I$. A similar argument applied to triangles $F C B$ and $G C D$ shows that $G F$ is $\frac{5}{3}$ of $F C$. Since the area of triangle $F B C$ is $45$, multiplying the lengths of $F C$ and $B I$ gives $90$. Therefore the area of the parallelogram is $\frac{4}{3} \times \frac{5}{3} \times 90 = 200$.