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
 A: Let $ABCD$ the trapezoid. The area can be partitioned in a triangle and a rectangle:

Then the area of the triangle can be calculated by Heron's Formula, because the sides are $14, 16$ and $40-30=10$.
$$[A_1D_1B_2]=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{20\cdot 6 \cdot 4 \cdot 10}=40\sqrt{3}$$
Finally the height of the triangle is $$h=\frac{2[A_1D_1B_2]}{A_1B_2}=8\sqrt{3}$$
From here you are almost done.
A: Area of a trapezium without knowing the height:
(a+c) / 4(a-c) * √(a+b-c+d)(a-b-c+d)(a+b-c-d)(-a+b+c+d)
Where a>c and 'a' is parallel to 'c'. 'b' and 'd' are the 'diagonals'.
Therefore,
(40+30) / 4(40-30) * √(40+14-30+16)(40-14-30+16)(40+14-30-16)(-40+14+30+16)
70 / 40 * √(40)(12)(8)(20)
7/4 * 277.128129...
484.974
A: 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.
