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From a rectangular piece of paper, a triangular corner is cut off resulting in a pentagon.If the sides of the pentagon have lengths 10,17,18,24 and 39 in some order.Find the sides of the rectangle and the sides of the triangle cut off.

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  • $\begingroup$ how exactly would this fall in the combinatorics category? $\endgroup$ – Varun Iyer Nov 16 '14 at 11:46
  • $\begingroup$ I have posted a solution below $\endgroup$ – Varun Iyer Nov 16 '14 at 12:04
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Let us consider a rectangle with side lengths of $x$ and $y$, with $x <y$.

Then consider a right triangle with legs of $a$ and $b$, with $a > b$.

If this triangle is cut off from the rectangle, then it would result in a pentagon with side lengths of $y-a$, $x-b$, $y$, $x$, and $\sqrt{a^2+b^2}$


Now, we are given the lengths to be: $10$, $17$, $18$, $24$, and $39$.

First, note that $\sqrt{a^2+b^2}$ must be one of the integers, so immediately Pythagorean triples comes to mind.

One that immediately comes to mind is $(8, 15, 17)$

Therefore, we can see that $\sqrt{a^2+b^2}$ could be $17$, and $a$ and $b$ can be $15$ and $8$, respectively.

Also, note that the values of $x$ and $y$ must be greater than $a$ and $b$, otherwise the triangle cut off would not form a pentagon.

Therefore, the values of $x$ and $y$ are $24$ and $39$ respectively.

Thus, $x -b = 24 - 8 = 16$.

But wait! notice that $16$ isn't a number.

Therefore, if we simply set $y$ equal to $39$, then $y - a = 39 - 15 = 24$

Therefore, $x$ must equal $18$

And $x -b = 18 - 8 = 10$


Our solution is complete:

The sides of the rectangle is $18$ x $39$

And the sides of the triangle is $8$ x $15$ x $17$.

This was a lengthy derivation, so if you have questions, comment below.

Hope this helped.

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