I tried using x+2 as the longer side of the large unshaded rectangle, and subtracted the right triangles to get 192. My friend tells me this is incorrect, and I was wondering how to get the correct answer.
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4Use Pythagorean theorem wisely – Atmos Mar 31 '21 at 18:34
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5Hint: the triangle with sides $2$ and $6$ is similar to the triangle with sides $9$ and $l$, where $l$ is the horizontal distance over to where the rectangle hits the bottom edge. – Théophile Mar 31 '21 at 18:34
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By the Pythagorean theorem, we can clearly see that $EF = 2\sqrt{10}$. Also, note that $\triangle EAF \sim \triangle FBG$, so $\frac{AF}{EF} = \frac{BG}{FG}$, meaning that $\frac{2}{2\sqrt{10}} = \frac{9}{FG}$, which means $FG = 9\sqrt{10}$. We know that the sides of the rectangle are $EF = 2\sqrt{10}, $$ \space FG = 9\sqrt{10}$, thus the area is $9\cdot2\cdot10 =180$.
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$\triangle CAB \sim \triangle EDC \implies DE = 27 $
The other lengths shown are easy to compute.
$area \square ADFG = 29 \cdot 15 =435$
The sum of the areas of the two smaller triangles is $6 \cdot 2 = 12$
The sum of the areas of the two larger triangles is $9 \cdot 27 = 243$
The area of the inscribed rectangle is $435 - 243 - 12 = 180$
Steven Alexis Gregory
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