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In isosceles trapezoid $ABCD$ the median $MN$ cuts diagonals $AC$ and $BD$ at $P$ and $Q$, respectively. If $PQ$ is $4$ find the perimeter of the trapezoid. I found that $AC=BD=8$ and I know that $PQ$ is half of $AB-CD$ but I can't find $CD$. Any help would be appreciated. Angle A is 60 degrees. $MN$ is 8.

Edit: I wasn't given that $MN$ is 8. Now when I know that I can find the perimeter. $P=a+b+2(a-b)=32$

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Two diagonals from the same vertex $A$? – enzotib Sep 26 '12 at 20:22
Sorry, my fault. The diagonals are AC and BD. OP edited. – lam3r4370 Sep 26 '12 at 20:32
How exactly did you find that the length of the diagonals are $8$? – EuYu Sep 26 '12 at 23:14
up vote 2 down vote accepted

The problem is ill-posed; the perimeter isn't determined by the given information. If we move $D$ and $C$ outward, as they come to lie vertically above $A$ and $B$, respectively, $P$ and $Q$ converge at the centre. Thus the trapezoid has to be scaled up without bounds for $PQ$ to retain the value $4$, and thus it can have arbitrarily large perimeters.

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