Triangle in a Right Trapezium

Most important, draw the diagram and refer to it as you read. It can be easily proved that triangles ABE and DEC are similar. Thus the ratio of their heights(from AB and DC) will be equal to the ratio of their bases AB and DC. We know the total height is $AD=75$ . Thus we get height of triangle $EDC$ from $DC$ to be 60. Triangles $DBC$ and $ADC$ are between the same set of parallel lines $AB$ and $CD$ and have the same base $DC$ . Thus they have the same area (by area theorems). Thus area of triangle BEC=Areas of triangles(ADC-DEC).Find it: You Know all the bases and heights. Area of BEC will be $((AD*DC)-(DC*60))/2 = 210$ .

Let $x=[ADE]=[BEC]$ , $y=[AEB]$ , $z=[DCE]$ . We have $x:z=AE:EC=y:x$ , hence

$x^2=yz\tag{1}$

Furthermore

$x+y=\frac{7 \cdot 75}{2} \tag{2}$

$x+z=\frac{28 \cdot 75}{2} \tag{3}$

From (1), (2) and (3) it follows

$x^2=\left(\frac{525-2x}{2}\right)\left(\frac{2100-2x}{2}\right) \qquad \Leftrightarrow \qquad x=\boxed{210}$

$[ABCD]=\dfrac{AB+DC}{2}\cdot AD=\dfrac{2625}{2}$

$\Delta AEB\sim DEC$ since $\text{alternate}\angle ABE= \text{alternate}\angle EDC$ $\angle AEB=\angle DEC$

Therefore,

$\dfrac{AB}{BE}=\dfrac{DC}{ED}$

$\dfrac{7}{BE}=\dfrac{28}{ED}$

$\dfrac{ED}{BE}=4$

From this,we can deduce that $\Delta DEC$ has 4 times the side lengths of $\Delta AEB$ . From this,we can infer that the height of $\Delta DEC$ is $4$ times the height of $\Delta AEB$ . Let the height of $\Delta AEB$ be $h$ .

$\therefore h+4h=75$ $h=15$

Therefore, $[ABE]=\dfrac{1}{2}\cdot7\cdot h$ and $[ADC]=\dfrac{28\cdot 75}{2}=1050$

$\therefore [ABE]+[ADC]=\dfrac{2205}{2}$

Now, $[BEC]=[ABCD]-([ABE]+[ADC])=\boxed{210}$