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The question:

Suppose there is a right triangle with sides $a$ and $b$ and hypotenuse $c$. Its perimeter is the same as its area, and $b = 6$. What are its side lengths?

I just cannot figure out how to do this! The second sentence isn't particularly helpful:

$$a + c + 6 = \frac{6a}{2}$$ $$a + c + 6 = 3a$$ $$???$$

And I can't get anywhere with the Pythagorean Theorem either:

$$a^2 + 36 = c^2$$ $$c^2 - 36 = a^2$$ $$(c + 6)(c - 6) = a^2$$ $$???$$

How do I solve this puzzle?

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Do you have an error with your formula? if a =6, why should you mention a and 6? Do you imply that b is also = 6. IF a = 6, then you can use this fact to have only 2 unknowns. That should simplify the formulas somewhat. –  cuabanana Sep 11 '13 at 23:14
You seem to have set $b = 6$ on the LHS of the first equation. Do so also on the right. –  Daniel Fischer Sep 11 '13 at 23:14
@cuabanana Whoops, typo. I meant $b = 6$. –  Doorknob 冰 Sep 11 '13 at 23:15
@DanielFischer Okay, that still didn't help :/ –  Doorknob 冰 Sep 11 '13 at 23:16
You get $a = \frac{c+6}{2}$. So $c^2 = 36 + \frac{(c+6)^2}{4}$. $4(c+6)(c-6) = (c+6)^2$, $4(c-6) = (c+6), 3c = 30$. –  Daniel Fischer Sep 11 '13 at 23:22

1 Answer 1

up vote 2 down vote accepted

If $a=6$, then area is equal to $(6b)/2=3b$. If that is equal to the perimeter, then $a+b+c=3b$ so $a+c=2b$ so $6+c=2b$. You also know that $a^2+b^2=c^2$ so $36+b^2=c^2$.

From the $6+c=2b$ you get $c=2b-6$, and so $36+b^2=(2b-6)^2$. Expand and you get a quadratic on $b$.

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Sorry, I made a typo. I meant $b = 6$, not $a = 6$. –  Doorknob 冰 Sep 11 '13 at 23:17
Its the same. Just change $a$ for $b$ and $b$ for $a$ everywhere above lol –  Daniel Montealegre Sep 11 '13 at 23:23
Oh, alright, well thanks then! +1 –  Doorknob 冰 Sep 11 '13 at 23:26

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