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I'm given the diagram and the given statement as pictured, and I've tried a lot of methods including systems of 3 equations and cyclic quadrilateral stuff, but I can't get anywhere with this problem. Any hints or solutions would be great, thanks! (Correct answer is 3744)

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  • $\begingroup$ @JohnHughes very clever method, thanks, I initially moved F down to coincide with D, but never thought to do the same on the other side, thanks! $\endgroup$ – Nick Brown Feb 27 '17 at 0:57
  • $\begingroup$ Symmetry is often your best friend. :) $\endgroup$ – John Hughes Feb 27 '17 at 0:58
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    $\begingroup$ Correct answer is 3744 FWIW the square with perimeter $182$ has an area of just $2070.25\,$, and that's the largest area of any quadrilateral with the given perimeter, so $3744$ can't be the right answer. $\endgroup$ – dxiv Feb 27 '17 at 2:04
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    $\begingroup$ @dxiv Yep I went round that calc in ignoring the given answer. $\endgroup$ – Joffan Feb 27 '17 at 2:35
  • $\begingroup$ @dxiv Good point! $\endgroup$ – Nick Brown Feb 27 '17 at 2:38
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Hint:

$$ S_{ABCD} = 24 \cdot AB = 60 \cdot BC\;, \;\;AB+BC=182/2 $$

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  • $\begingroup$ Nice! That smacking sound you just heard was me hitting my forehead... $\endgroup$ – John Hughes Feb 27 '17 at 1:25
  • $\begingroup$ Why does AB=60*BC? $\endgroup$ – Nick Brown Feb 27 '17 at 1:26
  • $\begingroup$ @NickBrown Either split it into two congruent triangles of area $\frac{1}{2}\,60\cdot BC$ each, or use the trapezoid formula $60 \cdot (BC+AD)/2\,$. $\endgroup$ – dxiv Feb 27 '17 at 1:28
  • $\begingroup$ @JohnHughes Thanks. This was my laziness kicking in when I started getting too many equations... $\endgroup$ – dxiv Feb 27 '17 at 1:31
  • $\begingroup$ I've gotta say, I'm still not following your logic here, can you elaborate further? Thanks $\endgroup$ – Nick Brown Feb 27 '17 at 1:34
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Define $x$ as the length of $AB$ and $y$ as the length of $AD$. Since these two line segments are half the perimeter of the parallelogram, $x+y=91$.

Draw a line from $B$ to $AD$ parallel to $FG$, meeting $AD$ at $H$. $BH$ will also have length $60$.

Now $\triangle ABH$ and $\triangle AED$ are similar right triangles, so $\frac{\large x}{\large 60} = \frac{\large y}{\large 24}$. Thus $2x=5y$. Set $t=\frac x5$ then $x=5t, y=2t$ and $7t=91$ giving $t=13$ and $x=65$.

Then the area of the parallelogram is $24x = 1560$ - which is not what you gave as the answer.

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Suggestion: move $F$ down to coincide with $D$ and $G$ down by the same amount!); then draw another line parallel to $FG$, but passing through $B$. That divides $BC$ into two parts, $BG$ and $GC$ of lengths $u$ and $v$ respectively, and $AD$ into two parts congruent to those. Let $x$ be the length of $AB$. Then $2(x+u+v)=182$. Let $y=ED$. Then $24^2+y^2=(u+v^2)$, and $24/y=60/v$. And $v^2+60^2=x^2$, That should be enough to work from...4 equations in 4 unknowns. True, two are quadratics...but it's a start.

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  • $\begingroup$ I tried it out and my TI-89 has been trying to calculate it for the past 15 minutes no joke lol $\endgroup$ – Nick Brown Feb 27 '17 at 1:15
  • $\begingroup$ I'm pretty sure there's another method out there considering this is from a timed test where scientific calculators that can solve equations are not expected $\endgroup$ – Nick Brown Feb 27 '17 at 1:16

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