It looks like there was a misunderstanding. Let's solve the problem step-by-step to find the side lengths of the triangle.
Given:
1. The longer leg of a right triangle is 7 meters longer than the shorter leg.
2. The hypotenuse is 9 meters longer than the shorter leg.
We need to find the lengths of the sides of this right triangle.
Let [tex]\( x \)[/tex] be the length of the shorter leg.
Then:
- The length of the longer leg is [tex]\( x + 7 \)[/tex].
- The length of the hypotenuse is [tex]\( x + 9 \)[/tex].
Using the Pythagorean theorem which states that in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides:
[tex]\[ (x + 9)^2 = x^2 + (x + 7)^2 \][/tex]
Expanding both sides:
[tex]\[ (x + 9)^2 = x^2 + (x + 7)^2 \][/tex]
[tex]\[ x^2 + 18x + 81 = x^2 + x^2 + 14x + 49 \][/tex]
Combining like terms:
[tex]\[ x^2 + 18x + 81 = 2x^2 + 14x + 49 \][/tex]
Rearranging to form a standard quadratic equation:
[tex]\[ 2x^2 + 14x + 49 - x^2 - 18x - 81 = 0 \][/tex]
[tex]\[ x^2 - 4x - 32 = 0 \][/tex]
Solving this quadratic equation using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex], where [tex]\( a = 1 \)[/tex], [tex]\( b = -4 \)[/tex], and [tex]\( c = -32 \)[/tex]:
[tex]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 1 \cdot (-32)}}{2 \cdot 1} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 + 128}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{144}}{2} \][/tex]
[tex]\[ x = \frac{4 \pm 12}{2} \][/tex]
This gives us two potential solutions:
[tex]\[ x = \frac{4 + 12}{2} = 8 \][/tex]
[tex]\[ x = \frac{4 - 12}{2} = -4 \][/tex]
Since a length cannot be negative, we take the positive solution:
[tex]\[ x = 8 \][/tex]
Therefore, the lengths of the sides are:
- The shorter leg: [tex]\( 8 \)[/tex] meters
- The longer leg: [tex]\( 8 + 7 = 15 \)[/tex] meters
- The hypotenuse: [tex]\( 8 + 9 = 17 \)[/tex] meters
Thus, the side lengths of the triangle are:
- Shorter leg: [tex]\( 8 \)[/tex] meters
- Longer leg: [tex]\( 15 \)[/tex] meters
- Hypotenuse: [tex]\( 17 \)[/tex] meters
