The length of the longer leg of a right triangle is 14 ft longer than the length of the shorter leg [tex]\( x \)[/tex]. The hypotenuse is 6 ft longer than twice the length of the shorter leg. Find the dimensions of the triangle.

Select one:
A. Short leg [tex]\( = 11 \)[/tex], long leg [tex]\( = 25 \)[/tex], hypotenuse [tex]\( = 28 \)[/tex]
B. Short leg [tex]\( = 10 \)[/tex], long leg [tex]\( = 24 \)[/tex], hypotenuse [tex]\( = 26 \)[/tex]
C. Short leg [tex]\( = 9 \)[/tex], long leg [tex]\( = 23 \)[/tex], hypotenuse [tex]\( = 24 \)[/tex]
D. Short leg [tex]\( = 9 \)[/tex], long leg [tex]\( = 23 \)[/tex], hypotenuse [tex]\( = 28 \)[/tex]



Answer :

To solve for the dimensions of the right triangle, we need to use the given relationships and the Pythagorean theorem. Let's break it down step by step.

1. Identify the Relationships:
- Let the length of the shorter leg be [tex]\( x \)[/tex] ft.
- The longer leg is 14 ft longer than the shorter leg, so the length of the longer leg will be [tex]\( x + 14 \)[/tex] ft.
- The hypotenuse is 6 ft longer than twice the length of the shorter leg, so the length of the hypotenuse will be [tex]\( 2x + 6 \)[/tex] ft.

2. Set Up 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 hypotenuse. Mathematically, this is:
[tex]\[ x^2 + (x + 14)^2 = (2x + 6)^2 \][/tex]

3. Expand and Simplify the Equation:
- First, expand both sides of the equation:
[tex]\[ x^2 + (x + 14)^2 = x^2 + 28x + 196 \][/tex]
[tex]\[ (2x + 6)^2 = 4x^2 + 24x + 36 \][/tex]
- So the equation becomes:
[tex]\[ x^2 + x^2 + 28x + 196 = 4x^2 + 24x + 36 \][/tex]

4. Combine Like Terms and Simplify:
- Combine like terms on both sides:
[tex]\[ 2x^2 + 28x + 196 = 4x^2 + 24x + 36 \][/tex]
- Move all terms to one side to set the equation to zero:
[tex]\[ 2x^2 + 28x + 196 - 4x^2 - 24x - 36 = 0 \][/tex]
[tex]\[ -2x^2 + 4x + 160 = 0 \][/tex]
- Multiply through by -1 to make it positive:
[tex]\[ 2x^2 - 4x - 160 = 0 \][/tex]

5. Solve the Quadratic Equation:
Using the quadratic formula [tex]\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)[/tex]:
[tex]\[ a = 2, \quad b = -4, \quad c = -160 \][/tex]
[tex]\[ x = \frac{-(-4) \pm \sqrt{(-4)^2 - 4 \cdot 2 \cdot (-160)}}{2 \cdot 2} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{16 + 1280}}{4} \][/tex]
[tex]\[ x = \frac{4 \pm \sqrt{1296}}{4} \][/tex]
[tex]\[ x = \frac{4 \pm 36}{4} \][/tex]
This gives us two solutions:
[tex]\[ x = \frac{4 + 36}{4} = 10 \quad \text{and} \quad x = \frac{4 - 36}{4} = -8 \][/tex]
Since a length cannot be negative, [tex]\( x = 10 \)[/tex] ft (shorter leg).

6. Calculate the Other Dimensions:
- Longer leg:
[tex]\[ x + 14 = 10 + 14 = 24 \text{ ft} \][/tex]
- Hypotenuse:
[tex]\[ 2x + 6 = 2(10) + 6 = 20 + 6 = 26 \text{ ft} \][/tex]

Therefore, the dimensions of the triangle are:
- Short leg = 10 ft
- Long leg = 24 ft
- Hypotenuse = 26 ft

The correct answer is b. Short leg = 10, long leg = 24, hypotenuse = 26.