area of trapezium A trapezium has perpendicular diagonals and altitude 4. If one of the diagonals has length 5, find the area of the trapezium A 12 B 50/3 C 25/2 D 40/5 I guess the answer is C. I divided the trapezium into two triangles and then it is 5(x+y)/2, in which x+y=length of the diagonal, which is 5, so it's 5(5)/2. But what if it is a irregular trapezium? The diagonals will not be the same. I didn't use the altitude so I think my answer is wrong. What should be the answer?

!Trapezium

$AH=4$, $AC=5$ $\rightarrow$ $HC=3$.

$\displaystyle \frac{AB}{AO}=\frac{DC}{OC}=\frac{AC}{HC}=\frac{5}{3}$.

$\displaystyle AB+DC=(AO+OC)\cdot \frac{5}{3} = \frac{25}{3}$.

$\displaystyle S_{ABCD}=AH\cdot \frac{AB+DC}{2}=4\cdot \frac{25}{6} = \frac{50}{3}$.