source: #alcumus tags: #quadrilateral #trapezoid #area #proportion #ratio course: [[Geometry]] status: #solved ### Problem In a trapezoid $A B C D$ with $A B$ parallel to $C D$, the diagonals $A C$ and $B D$ intersect at $E$. If the area of triangle $A B E$ is 50 square units, and the area of triangle $A D E$ is 20 square units, what is the area of trapezoid $A B C D$? #### Solution ![[trapezoid-diagonals.svg]] Note that $\triangle ABE$ and $\triangle ADE$ share the same height $\overline{AF}$ where $F$ lies on $\overleftrightarrow{BD}$ This means that the ratio of the areas of $\triangle ABE$ and $\triangle ADE$ is the same as the ratio of their bases $\overline{BE}$ and $\overline{DE}$ $\frac{\|\triangle ABE\| }{\|\triangle ADE\|}=\frac{m\overline{BE}}{m\overline{DE}}=\frac{50}{20}=\frac{5}{2}$ This will be used as the scale factor for the similarity: $ \triangle ABE \sim \triangle CDE \text{ by AAA}\begin{cases}\angle ABE \cong\angle CDE : \text{Alternate Interior Angles}\\ \angle BAE \cong\angle DCE : \text{Alternate Interior Angles}\\ \angle BEA \cong\angle DEC : \text{Vertical Angles} \end{cases} $ by the [[Square-Cube law]], the ratio of the areas it the square of the ratio of the side lengths. $\frac{\|\triangle ABE\| }{\|\triangle CDE\|}=\left(\frac{m\overline{BE}}{m\overline{DE}}\right)^2=\left(\frac{5}{2}\right)^2=\frac{25}{4}$ The solving for $\|\triangle CDE\|$ $\frac{\|\triangle ABE\| }{\|\triangle CDE\|}=\frac{25}{4}$ $\frac{50}{\|\triangle CDE\|}=\frac{25}{4}$ $\|\triangle CDE\|=8$ By the same logic as the first step, $\|\triangle CDE\|$ and $\|\triangle CBE\|$ have the same height to $\overleftrightarrow{BD}$, meaning the ratio of their areas is the same as the ratio of sides $\frac{\|\triangle CBE\| }{\|\triangle CDE\|}=\frac{m\overline{BE}}{m\overline{DE}}$ $\frac{\|\triangle CBE\| }{8}=\frac{5}{2}$ $\|\triangle CBE \|= 20$ The total of the 4 triangles that make up the trapezoid is: $50+20+20+8=\boxed{98}$ ###### Note: Solving this problem should lead the student to realize something about the way diagonals cut up all quadrilaterals, but specifically the fact that trapezoids are cut in a way that has two triangles of equal area.