Question

a sign in the shape of a right triangle has one leg that is 51 inch(es) longer than the other leg. what is the length of the shorter leg if the perimeter is 168 inches? the length of the shorter leg is inches.

Explanation:

Step1: Let the shorter leg be $x$.

Let the shorter leg length be $x$ inches. Then the longer leg is $x + 51$ inches. Let the hypotenuse be $h$ inches. The perimeter $P=x+(x + 51)+h$. Given $P = 168$, so $x+(x + 51)+h=168$, which simplifies to $2x+h=117$, and $h = 117 - 2x$.

Step2: Apply Pythagorean theorem.

By the Pythagorean theorem $x^{2}+(x + 51)^{2}=h^{2}$. Substitute $h = 117 - 2x$ into it: $x^{2}+x^{2}+102x + 2601=(117 - 2x)^{2}$.
Expand the right - hand side: $x^{2}+x^{2}+102x + 2601=13689-468x + 4x^{2}$.
Combine like terms: $2x^{2}+102x + 2601=13689-468x + 4x^{2}$.
Rearrange to get a quadratic equation: $2x^{2}-570x + 11088 = 0$. Divide through by 2: $x^{2}-285x+5544 = 0$.

Step3: Solve the quadratic equation.

For a quadratic equation $ax^{2}+bx + c = 0$ ($a = 1$, $b=-285$, $c = 5544$), use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$.
First, calculate the discriminant $\Delta=b^{2}-4ac=(-285)^{2}-4\times1\times5544=81225 - 22176=59049$.
Then $x=\frac{285\pm\sqrt{59049}}{2}=\frac{285\pm243}{2}$.
We have two solutions: $x_1=\frac{285 + 243}{2}=\frac{528}{2}=264$ and $x_2=\frac{285 - 243}{2}=\frac{42}{2}=21$.
Since $x_1 = 264$ would make $h=117-2\times264=117 - 528=-411$ (not possible for a length), we take $x = 21$.

Answer:

21