Right Triangle

**Vitamin** · March 16th 2008, 01:57 PM

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:

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

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

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

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

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

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

The $x = -56.73$ gets crossed out because its negative which leaves the $x = 31.73$ as the answer right? I don't think I right it just as $x = 31.73$ as the answer right?

**Soroban** · March 16th 2008, 05:20 PM

Hello, Vitamin!

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

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

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

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

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

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

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

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

**Vitamin** · March 16th 2008, 05:30 PM

Oh my bad i wrote it wrong. its not 24 cm its 25 cm. Is my answer right then? So the answer would be $x > 31.73$ or $x \ge 31.73$ ?

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