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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.

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HINT

  • 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 $H$)
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    $\begingroup$ would it be $27 \sqrt{7}$? $\endgroup$ – jjhh Mar 1 '18 at 14:44
  • $\begingroup$ What did you obtain for H? $\endgroup$ – gimusi Mar 1 '18 at 14:56
  • $\begingroup$ Umm $3\sqrt{7}$ $\endgroup$ – jjhh Mar 1 '18 at 15:01
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    $\begingroup$ I think you forgot the $1/2$ part? $\endgroup$ – jjhh Mar 1 '18 at 15:14
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    $\begingroup$ It should be x=10, then $27 \sqrt 7$ it's ok! $\endgroup$ – gimusi Mar 1 '18 at 15:35
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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: $16x=80$.

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