Problem 796

Let the height of trapezoid $A B C D$ be $h$.

Then its total area is $\frac{1}{2}(A B+C D) h=\frac{7}{2} h$.

Since $A B=2$ and $A X$ is parallel to $B C$, then $X C=2$.

Since $A B=2$ and $B Y$ is parallel to $A D$, then $D Y=2$.

Since $C D=5, X C=2$ and $D Y=2$, then $Y X=1$.

!

Now we want to determine the area of $\triangle A Z W$, so we will determine the areas of $\triangle A Z B$ and $\triangle A W B$ and subtract them.

First, we calculate the area of $\triangle A Z B$. Since $A B$ is parallel to $C D, \angle Z A B=\angle Z X Y$ and $\angle Z B A=\angle Z Y X$, so $\triangle A Z B$ is similar to $\triangle X Z Y$. Since the ratio of $A B$ to $X Y$ is 2 to 1 , then the ratio of the heights of these two triangles will also be 2 to 1 , since they are similar. But the sum of their heights must be the height of the trapezoid, $h$, so the height of $\triangle A Z B$ is $\frac{2}{3} h$. Therefore, the area of $\triangle A Z B$ is $\frac{1}{2}(2)\left(\frac{2}{3} h\right)=\frac{2}{3} h$.

Next, we calculate the area of $\triangle A W B$. Since $A B$ is parallel to $C D, \angle W A B=\angle W C Y$ and $\angle W B A=\angle W Y C$, so $\triangle A W B$ is similar to $\triangle C W Y$. Since the ratio of $A B$ to $C Y$ is 2 to 3 , then the ratio of the heights of these two triangles will also be 2 to 3 , since they are similar. But the sum of their heights must be the height of the trapezoid, $h$, so the height of $\triangle A W B$ is $\frac{2}{5} h$. Therefore, the area of $\triangle A W B$ is $\frac{1}{2}(2)\left(\frac{2}{5} h\right)=\frac{2}{5} h$.

Therefore, the area of $\triangle A Z W$ is the difference between the areas of $\triangle A Z B$ and $\triangle A W B$, or $\frac{2}{3} h-\frac{2}{5} h=\frac{4}{15} h$. Thus, the ratio of the area of $\triangle A Z W$ to the area of the whole trapezoid is $\frac{4}{15} h: \frac{7}{2} h=\frac{4}{15}: \frac{7}{2}=4(2): 7(15)=8: 105$