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The length of one leg of a right triangle is $(x - 6)$ centimeters, and the area is $(\frac12 x^2 - 7x + 24)$ square centimeters. What is the length of the other leg?

I think the equation that I need to solve is $(\frac12 x^2 - 7x + 24) = [\frac12 (x - 6) (h)]$ but I'm not sure if that's right or where to begin on solving it.

If I have the equation right could you please explain how to solve it?

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  • $\begingroup$ Presumably by 1/2x^2 you mean $\frac 12x^2$ and not $\frac 1{2x^2}$ Please use parentheses when writing slash fractions. $\endgroup$ – Ross Millikan Feb 4 '15 at 2:01
  • $\begingroup$ possible duplicate of Help solving for h? $\endgroup$ – Daniel W. Farlow Feb 4 '15 at 3:13
  • $\begingroup$ @induktio This one has answers; close in the other direction. $\endgroup$ – user147263 Feb 4 '15 at 4:27
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@Kara, Like you recognized, the equation needs to be solved for $h$ in terms of $x$. Here is my solution to your question: $$\frac{x^2}{2}-7x+24=\frac{x-6}{2}h$$ Move all $x$ terms to one side, to obtain $h$ explicitly. $$h=\frac{2}{x-6}*\frac{\frac{x^2}{2}-7x+24}{1}$$ Simplify by multiplying the $2$ through the equation. $$h=\frac{x^2-14x+48}{x-6}$$ Factor the quadratic, in order to obtain the linear factors. Let's use the quadratic formula. If we find the $x$ values for which this equation equals zero, we have the factors. $$x^2-14x+48=0$$ $$x=\frac{-(-14)\pm\sqrt{(-14)^2-4(1)(48)}}{2(1)}$$ $$x=\frac{14\pm\sqrt{196-192}}{2}$$ $$x=\frac{14\pm\sqrt{4}}{2}$$ $$x=\frac{14\pm2}{2}$$ $$x=6,8$$ Therefore, the factors are: $(x-6),(x-8)$. Let's put that back into the original equation and see what we can cancel. $$h=\frac{(x-6)(x-8)}{x-6}$$ Notice the $(x-6)$ in the numerator and in the denominator. Cancel it. $$h=(x-8)$$

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Yes, that is the equation to solve. Did you try polynomial division?

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The first step would be to multiply the left side by 2 to get rid of the 1/2 on the right side. Then you could factor the left side and the (x-6)'s should be able to cancel if you factored correctly leaving you with h as an isolated term.

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Saying

The length of one leg of a right triangle is $x−6$ centimeters, and the area is $\frac12x^2−7x+24$ square centimeters. What is the length of the other leg?

is the same as saying

One dimension of a rectangle is $x-6$ centimeters, and the area is $x^2-14x+48$ square centimeters. What is the other dimension? And ...

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Never forget that $x$ is just a number you haven't given a name to--

Your instincts were correct but you seem to have been cowed by the surfeit of letters.

I suggest trying "base cases" if you're ever put off by the variables.

Let's take off $x$'s mask and suppose $x=10$. Then your question reads:

The length of one leg of a right triangle is 4 centimeters, and the area is 4 square centimeters. What is the length of the other leg?

This is a much simpler problem--we need only solve $4=\frac{1}{2}*4*h$, from which it is immediate that

$h=2$ (which equals $x-8$, of course).

One base case should give us confidence to do polynomial division--the last snag of the problem was noticing a shared factor.

Here again bases cases might help--if you try a few other cases and go through the same procedure, you may notice the pattern:

x 11 20 30

h 9 18 28

Once you see the pattern, you should be alerted that your complicated-looking fraction may not be fully simplified.

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