# Parallelogram in a Rectangle +-- {.image} [[ParallelograminaRectangle.png:pic]] > What fraction is shaded? =-- ## Solution by [[Similar Triangles]] and [[Area of a Triangle]] +-- {.image} [[ParallelograminaRectangleLabelled.png:pic]] =-- With the points labelled as above, triangles $A D C$ and $G D E$ are [[similar]]: angles $D \hat{A} C$ and $D \hat{G} E$ are equal as they are [[alternate angles]], the same applies to angles $A \hat{C} D$ and $G \hat{E} D$. Since $E G$ has half the length of $A C$, the length of $D F$ is half of that of $B D$. Therefore the length of $B D$ is $\frac{2}{3}$rds of the length of $B F$. The area of triangle $A C D$ is then $\frac{1}{3}$ of the area of the lower half of the square. Hence the area of the parallelogram is $\frac{1}{3}$rd of the area of the square. ## Solution by [[Tessellation]] or [[Dissection]] +-- {.image} [[ParallelograminaRectangleTessellation.png:pic]] =-- The above diagram shows either by [[tessellation]] or [[dissection]] that the square has the same area as three parallelograms.