# Systems of equations finding right triangles

I need help setting up the equation for the question, "Find all right triangles for which the perimeter is $24$ units and the area is $24$ square units."

I know that the area is $A = \frac12 b h$ and perimeter is $P = a + b + h$.

Using this, would the system look something like?

$a + b + c = 24$
$\frac12b c = 24$

(I'm using $c$ for consistency)

Then continue on using substitution?

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The $h$ in the perimeter should be the hypotenuse length ($\sqrt{b^2+h^2}$), not the height of the triangle (assuming the shorter sides are $b$ and $h$). –  David Mitra Dec 20 '11 at 18:11

You might be more comfortable using $a$ and $b$ for the two "legs" of the triangle, and $c$ for the hypotenuse.

Then, exactly as you did, we obtain $$a+b+c=24 \qquad \text{and}\qquad \frac{1}{2}ab=24.\qquad\qquad(1)$$

By the Pythagorean Theorem, we have $$a^2+b^2=c^2.\qquad\qquad(2)$$ Now what? One approach (but not the only one) is to start by writing $c=24-(a+b)$, square both sides, and use Equation $(2)$ to eliminate $c$ and obtain a simple equation that involves only $a$ and $b$. We have $$c^2=(24-(a+b))^2=24^2-48(a+b)+a^2+2ab+b^2=a^2+b^2.$$ There is a fair bit of cancellation. Note that $2ab=96$. So we get $$48(a+b)=24^2+96$$ and therefore $a+b=14$.

Now we could write $b=14-a$ and substitute into $ab=48$ to get a quadratic equation in $a$. But the following is I think prettier. From $(a+b)^2=196$ and $4ab=192$ we conclude by subtraction that $(a-b)^2=4$, so $a-b=\pm 2$. We can decide now that without loss of generality $a>b$. So $a-b=2$. From $a+b=14$, by adding and dividing by $2$, we find that $a=8$ and $b=6$. And of course $c=24-(a+b)=10$.

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Do you mean c^2 = (24 - (a + b))^2 giving me 48a-a^2+48b-2ab-b^2-576? –  erimar77 Dec 20 '11 at 22:33
Finished calculation in main post. Note that your expansion had a sign glitch. –  André Nicolas Dec 20 '11 at 23:04
Yes, as I worked back through it, I see my error. Thank you for taking the time do elaborate. When you set that equation equal to a^2 + b^2, you can really see where the author of the book was headed. –  erimar77 Dec 20 '11 at 23:24

Let A=24. I am assuming c as hypotenuse and a, b the other two sides of the triangle $\implies c^2=a^2+b^2$.

We now have $1/2*a*b= A$ and $a+b+c=A$

$(a+b)^2=(A-c)^2$ $\implies a^2+b^2+2ab= A^2-2Ac+c^2$ $\implies c=\dfrac{A^2-2ab}{2A}=10$

There we have ab=48 and c=10. So the possible right triangle is 6,8,10.

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I'll fix up some standard notatation. Let $\Delta$ABC be right angled at $A$. Then, by standard notation, the hypotenuse is $a$ units long, the side $AC$ considered the height is of length $b$ units. The base, $AB$ is of length $c$ units.
$$\frac{1}{2} bc = 24$$ $$a+b+c=24$$ $$a^2=b^2+c^2$$
Now, you see that, by adding, $2 \times 48$ to both sides of last equation in two different ways, you have, by setting $a =\alpha$ and $b+c = \beta$, you'll have (do some algera here!) $\alpha =10$ and $\beta = 14$. Now, use the first equation and $\beta$ to get, $b=6,8$ and $c=8,6$. You are through.