1
$\begingroup$

I've been trying to find the solution for this problem for a long time... but I can't seem to do it. I'm not asking for the full solution, maybe just a hint or where to start off from.

Question: Let ABCD be a trapezium with AB||CD. M and N are the midpoints of AB and CD respectively. If AC=6 BD=8 and MN=4 Find the area of ABCD.

  • 0
    A bit ambiguous. Are the vertices named cyclically? In other words are AB and CD the slanted sides or the diagonals? – 2017-01-08
  • 0
    AB and CD are slanted sides – 2017-01-08
  • 0
    You wrote $AB\parallel CD$, so I presume $AB$ and $CD$ are the bases. while $AC$ and $BD$ are the diagonals, right? – 2017-01-08
  • 0
    Yes you are right – 2017-01-08

1 Answers 1

0

HINT.

Extend base $AB$ by $BB'=CD$, so that $CB'=BD=8$. Triangle $AB'C$ has the same area as the trapezium and its median $CM'$ is congruent to $MN$.

So the problem amounts at finding the area of a triangle, with two sides of length $6$ and $8$, and the median relative to the third side of length $4$.

SECOND HINT.

To find the area of such a triangle, let $AM'=B'M'=x$, $\angle AM'C=\alpha$ and $\angle B'M'C=\pi-\alpha$ (see diagram below). By the cosine rule you then get $$ \begin{align} x^2+16-8x\cos\alpha=36,\\ x^2+16+8x\cos\alpha=64.\\ \end{align} $$ Summing these you immediately obtain $x=\sqrt{34}$.

enter image description here

  • 0
    How do we know Triangle AB'C has the same area as the trapezium – 2017-01-09
  • 0
    Because triangles $ACD$ and $CBB'$ have the same area (equal bases and same altitude as the trapezium). – 2017-01-09