The bases of a trapezoid have lengths 50 and 75. Its diagonals have lengths 35 and 120. Find the area of the trapezoid.
+605 Let $x$ be the distance from the lower-left vertex to the intersection of the diagonals. Let $y$ be the same distance just for the lower right. Similar triangles gives
$$\frac{x}{35-x}=\frac{y}{120-y}=\frac{75}{50}\implies x=21, y=72$$
Heron's formula tells us that the area of the lower similar triangle (larger one) is $756$. Hence the height of the triangle is $\frac{504}{25}$. Then the height of the other one is $\frac{336}{25}$. So the whole height is $\frac{168}{5}$. The average of the bases is $\frac{125}{2}$. Thus the area of the trapezoid is $2100$.
