How do you solve the area of a trapezoid using diagonals The height of a trapezoid is $10$ cm. The lengths of the two diagonals of the trapezoid are $30$ cm and $50$ cm. Calculate the area of the trapezoid. 

On the homework I solved this using $${D_1D_2\over 2}$$ and my teacher marked me wrong. So I don't know what I did wrong. Please help. I know I can only use the formula if the diagonals are $90$ degrees. But how do I check that ?
The area will be $\frac12\cdot 10\cdot (y+x+y+z)=5(x+2y+z)$

Now, $(y+z)^2+10^2=50^2$ and $(x+y)^2+10^2=30^2$

$(y+z)=\sqrt{50^2-10^2} CM=20\sqrt6 CM$

$(x+y)=\sqrt{30^2-10^2} CM=20\sqrt2 CM$

SO, the area will be $5(20\sqrt2(\sqrt3+1)) CM^2$
the formula $\frac{d_1\cdot d_2}2$ is the area of the special quadrilateral called rhombus but not the trapezium..

If you want to find the area of the trapezium the like the above mentioned using the Pythogoras theorem. Find the sides and form an equation and with that solve the linear equation then you can solve the area of the trapezium..
finally you will get 

area$=\frac{h(a+b)}2$