Area of a trapezoid from given the two bases and diagonals Find the area of trapezoid with bases $7$ cm and $20$ cm and diagonals $13$ cm and $5\sqrt{10} $ cm.
My approach:
Assuming that the bases of the trapezoid are the parallel sides, the solution I can think of is a bit ugly,


*

*Find the  other two non-parallel sides of the trapezoid by using this formula.

*Find the height using this $$ h= \frac{\sqrt{(-a+b+c+d)(a-b+c+d)(a-b+c-d)(a-b-c+d)}}{2(b-a)}$$


Now, we can use $\frac12 \times$ sum of the parallel sides $\times$ height.
But, this is really messy and I am not sure if this is correct or feasible without electronic aid, so I was just wondering how else we could solve this problem?

Here is a very short solution: Consider a point B' on the line (A,B) to the right of point B such that |BB'|=b. Then consider the triangle ACB'. It is easy to see that the area of triangle ACB' is equal to the area of our trapezoid ABCD since the area of triangle ACD is equal to the area of trianlge BCB'. For the triangle ACB' it is easy to find the area from the Heron formula since its three sides are known: 27, 13, and 5$\sqrt{10}$. Plugging in these numbers we find that the area is 67.5.
