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?
@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. $$h=(x-8)$$