# Right Triangle

• Mar 16th 2008, 01:57 PM
Vitamin
Right Triangle
One leg of a right triangle is 24.0 cm longer than the other leg. If the hyptoenuse is to be greater than 65.0 cm, what values for the length of the other leg are permissible?

My work is:

$\displaystyle (25 + x)^2 + x^2 = 65^2$

$\displaystyle 625 + 50x + x^2 + x^2 = 65^2$

$\displaystyle 2x^2 + 50x - 3600 = 0$

$\displaystyle x = \frac{-50 \pm \sqrt{50^2 - 4(2)(-3600)}}{2(2)}$

$\displaystyle x = \frac{-50 \pm 176.92}{4}$

$\displaystyle x = 31.73\; or\; x = -56.73$

The $\displaystyle x = -56.73$ gets crossed out because its negative which leaves the $\displaystyle x = 31.73$ as the answer right? I don't think I right it just as $\displaystyle x = 31.73$ as the answer right?
• Mar 16th 2008, 05:20 PM
Soroban
Hello, Vitamin!

Quote:

One leg of a right triangle is 24.0 cm longer than the other leg.
If the hyptoenuse is to be greater than 65.0 cm,
what values for the length of the other leg are permissible?

Let $\displaystyle x + 24$ = length of "one leg"
Let $\displaystyle x$ = length of "other leg".

Then we have: .$\displaystyle x^2 + (x+24)^2 \: > \;65^2$

. . which simplifies to: .$\displaystyle 2x^2 + 48x - 2449 \;>\;0$

This is an up-opening parabola which is negative between its x-intercepts
. . and positive "outside" its x-intercepts.

For its x-intercepts: .$\displaystyle 2x^2 + 48x - 2449 \;=\;0$

. . Quadratic Formula: .$\displaystyle x \;=\;\frac{-24 \pm\sqrt{5474}}{2}$

The roots are: .$\displaystyle x\;=\;\begin{array}{ccccc}\frac{-24 + \sqrt{5474}}{2} & = & 24.99324263 \\ \frac{-24-\sqrt{5474}}{2} &=&-48.99324263 \end{array}$

Therefore: .$\displaystyle x \;\geq \;25.0$

• Mar 16th 2008, 05:30 PM
Vitamin
Oh my bad i wrote it wrong. its not 24 cm its 25 cm. Is my answer right then? So the answer would be $\displaystyle x > 31.73$ or $\displaystyle x \ge 31.73$ ?