Area of a trapezoid without the height How would I find the area of a non-iscoceles trapezoid and without the height? The trapezoid's bases are $30$ and $40$, and the legs $14$ and $16$. Thanks

Let $ABCD$ be the trapezoid with $ AB || CD, |AB|=30, |BC|=16, |CD|=40, |DA|=14$.$|DA|=14, |AY|=16, |YD|=10$ (the latter is the difference between the bases because $ABCY$ is a parallelogram).  Since $|AY|^2<|DA|^2+|YD|^2$, the angle at $D$ is acute, therefore $X$ lies between $D$ and $Y$.
We then define $|DX|=a$, and with $X$ between $D$ and $Y$ we then have $|XY|=10-a$.  Apply the Pythagorean Theorem to each of the right triangles $AYX, ADX$:
$h^2+(10-a)^2=256$
$h^2+a^2=196$
The difference between these equations gives a linear equation for $a$ alone, to wit, $100-20a=60$, thus $a=2$.  Substitute this into either Pythagorean relation (or both, to check!) and thus obtain $h^2=192, h=8\sqrt{3}$.  With $h$ known the area is then easy to get from the usual formula.
