To solve this problem, let's use the properties of a right triangle, specifically the Pythagorean Theorem. The Pythagorean Theorem states that in a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the length of the hypotenuse.
The problem gives us various relationships between the lengths of the sides:
1. The longer leg is 4 inches more than twice the shorter leg.
2. The hypotenuse is 6 inches more than the shorter leg.
Let's denote the length of the shorter leg as [tex]\(x\)[/tex].
Using these relationships, we can write the lengths of the sides as:
- Shorter leg = [tex]\(x\)[/tex]
- Longer leg = [tex]\(2x + 4\)[/tex]
- Hypotenuse = [tex]\(x + 6\)[/tex]
According to the Pythagorean Theorem:
[tex]\[
\text{(Shorter leg)}^2 + \text{(Longer leg)}^2 = \text{(Hypotenuse)}^2
\][/tex]
Substitute the expressions we have for the sides:
[tex]\[
x^2 + (2x + 4)^2 = (x + 6)^2
\][/tex]
Expand and simplify the equation:
[tex]\[
x^2 + (2x + 4)^2 = (x + 6)^2
\][/tex]
[tex]\[
x^2 + (4x^2 + 16x + 16) = x^2 + 12x + 36
\][/tex]
Combine like terms:
[tex]\[
x^2 + 4x^2 + 16x + 16 = x^2 + 12x + 36
\][/tex]
[tex]\[
5x^2 + 16x + 16 = x^2 + 12x + 36
\][/tex]
Subtract [tex]\(x^2 + 12x + 36\)[/tex] from both sides:
[tex]\[
5x^2 + 16x + 16 - x^2 - 12x - 36 = 0
\][/tex]
Combine like terms again:
[tex]\[
4x^2 + 4x - 20 = 0
\][/tex]
Simplify by dividing everything by 4:
[tex]\[
x^2 + x - 5 = 0
\][/tex]
Solve this quadratic equation using the quadratic formula [tex]\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)[/tex], where [tex]\(a = 1\)[/tex], [tex]\(b = 1\)[/tex], and [tex]\(c = -5\)[/tex]:
[tex]\[
x = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(-5)}}{2(1)}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{1 + 20}}{2}
\][/tex]
[tex]\[
x = \frac{-1 \pm \sqrt{21}}{2}
\][/tex]
This gives us two solutions, but we only need the positive one since a length cannot be negative:
[tex]\[
x = \frac{-1 + \sqrt{21}}{2}
\approx 1.79128784747792
\][/tex]
So, the length of the shorter leg is approximately [tex]\(1.791\)[/tex] inches.
Now, let's find the lengths of the longer leg and the hypotenuse using this value for [tex]\(x\)[/tex]:
[tex]\[
\text{Longer leg} = 2x + 4 = 2(1.79128784747792) + 4
\approx 7.58257569495584 \text{ inches}
\][/tex]
[tex]\[
\text{Hypotenuse} = x + 6 = 1.79128784747792 + 6
\approx 7.79128784747792 \text{ inches}
\][/tex]
So, the side lengths of the triangle are approximately:
- Shorter leg: [tex]\(1.791\)[/tex] inches
- Longer leg: [tex]\(7.583\)[/tex] inches
- Hypotenuse: [tex]\(7.791\)[/tex] inches
