Question

a sign in the shape of a right triangle has one leg that is 7 inch(es) longer than the other leg. what is the length of the shorter leg if the perimeter is 30 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 $x + 7$ inches. By the Pythagorean theorem, the hypotenuse $h=\sqrt{x^{2}+(x + 7)^{2}}$.

Step2: Set up the perimeter equation.

The perimeter $P=x+(x + 7)+\sqrt{x^{2}+(x + 7)^{2}}$. Given $P = 30$, we have $x+(x + 7)+\sqrt{x^{2}+(x + 7)^{2}}=30$.
Simplify the left - hand side: $2x+7+\sqrt{x^{2}+(x + 7)^{2}}=30$.
Then $\sqrt{x^{2}+(x + 7)^{2}}=30-(2x + 7)=23 - 2x$.

Step3: Square both sides.

$x^{2}+(x + 7)^{2}=(23 - 2x)^{2}$.
Expand: $x^{2}+x^{2}+14x + 49=529-92x + 4x^{2}$.
Combine like terms: $2x^{2}+14x + 49=529-92x + 4x^{2}$.
Rearrange to get a quadratic equation: $2x^{2}-106x + 480 = 0$.
Divide by 2: $x^{2}-53x + 240=0$.

Step4: Solve the quadratic equation.

Use the quadratic formula $x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}$ for the quadratic equation $ax^{2}+bx + c = 0$. Here $a = 1$, $b=-53$, $c = 240$.
$\Delta=b^{2}-4ac=(-53)^{2}-4\times1\times240=2809 - 960=1849$.
$\sqrt{\Delta}=43$.
$x=\frac{53\pm43}{2}$.
We get two solutions: $x_1=\frac{53 + 43}{2}=\frac{96}{2}=48$ and $x_2=\frac{53 - 43}{2}=5$.
But if $x = 48$, then $2x+7=2\times48+7=103>30$, which is not possible for the perimeter.
So we take $x = 5$.

Answer:

5