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explanation problem I've gotten to the point where I can solve for the third side, of course, and drawing in diagonals to form right triangles to find the area of the quadrilateral, but the only possible way of solving this that I can think of is going through all possible numbers 1-30 for the two sections of the hypotenuse and seeing whether or not the triangles are even possible, but this is a question in a test of 20, which you have only 50 minutes to complete, so this isn't a option. Can anyone help with finding a quick/ moderately quick way to do this?

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    Hint: $HTG$ and $HIR$ are similar triangles and $IGTR$ is a cyclic quadrilateral, whose area just depends on the ratio $\frac{HT}{HI}$. – 2017-02-22
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    You didn't actually tell us your result for $|RH|$. The area of the quadrilateral is just the difference of the two triangle areas $|HRI|-|HGT| $ – 2017-02-22
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    Son of a jesus I can't believe I missed that hahaaaa thank you so much. – 2017-02-22

2 Answers 2

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There are 4 distinct areas which are very easy to calculate: enter image description here

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    Very nice solution! – 2017-02-24
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You seem to have found that $|HR|=30$ from your remarks about integers to test, which is correct, with an area of $18\cdot 24 / 2 = 216$ for the large triangle $HRI$.

So you have an $18:24:30$ right triangle, which is of course in $3:4:5$ ratio, and since the smaller rectangle also has a right angle and the angle at $H$ it is similar: also a $3:4:5$ triangle. Given that $HG:HT$ is in the ratio $5:4$ and all lengths are integer, $HG$ must be a multiple of $5$: $\{ 5,10,15,20\} $ being the only values that will fit in.

For $HG=5$, the area of $\triangle HTG$ is $3\cdot4/2=6$ so the quadrilateral has area $216-6=210$.
And similarly for the other cases.