# Find the exact area of a trapezium in which three sides have length of $8$ and a diagonal has a length $12$.

Find the exact area of a trapezium in which three sides have length of $8$ and a diagonal has a length $12$.

I'm trying to use separate the trapezoid into two triangles — one isosceles with sides $8,8,12$ and the remaining one — and use $(1/2)ab\sin C$ and the law of cosines to get the other unknown side of the trapezoid or the angles, but it is not working very well for me.

• find the area $A$ of the triangle with sides $8,8,12$ and then the height of the trapezium $H=\frac{A}4$
• consider the other triangle with side $x,8,12$ and find the main base $x$ (we know $H$)
• would it be $27 \sqrt{7}$? – jjhh Mar 1 '18 at 14:44
• Umm $3\sqrt{7}$ – jjhh Mar 1 '18 at 15:01
• I think you forgot the $1/2$ part? – jjhh Mar 1 '18 at 15:14
• It should be x=10, then $27 \sqrt 7$ it's ok! – gimusi Mar 1 '18 at 15:35
Let $h$ be the height of the trapezium and $2x$ the unknown base. By Pythagoras' theorem we then have the two equations: $$(4+x)^2=12^2-h^2\\ (4-x)^2=8^2-h^2\\$$ Subtracting the second equation from the first one, we get: $16x=80$.