Math Problem Statement

The length of the longer leg of a right triangle is 20 ft more than twice the length of the shorter leg. The length of the hypotenuse is 22 ft more than twice the length of the shorter leg. Find the side lengths of the triangle.

Solution

Let's define the variables:

  • Let xx be the length of the shorter leg.
  • The length of the longer leg is given as 2x+202x + 20 feet.
  • The length of the hypotenuse is 2x+222x + 22 feet.

We can apply the Pythagorean theorem to solve for xx:

x2+(2x+20)2=(2x+22)2x^2 + (2x + 20)^2 = (2x + 22)^2

Expanding both sides:

  1. (2x+20)2=4x2+80x+400(2x + 20)^2 = 4x^2 + 80x + 400
  2. (2x+22)2=4x2+88x+484(2x + 22)^2 = 4x^2 + 88x + 484

Substituting these into the equation:

x2+4x2+80x+400=4x2+88x+484x^2 + 4x^2 + 80x + 400 = 4x^2 + 88x + 484

Simplifying:

5x2+80x+400=4x2+88x+4845x^2 + 80x + 400 = 4x^2 + 88x + 484

Move all terms to one side:

5x2+80x+4004x288x484=05x^2 + 80x + 400 - 4x^2 - 88x - 484 = 0

This simplifies to:

x28x84=0x^2 - 8x - 84 = 0

Solving this quadratic equation:

Using the quadratic formula:

x=(8)±(8)24(1)(84)2(1)x = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(1)(-84)}}{2(1)}

x=8±64+3362x = \frac{8 \pm \sqrt{64 + 336}}{2}

x=8±4002x = \frac{8 \pm \sqrt{400}}{2}

x=8±202x = \frac{8 \pm 20}{2}

Thus, the possible solutions are:

x=8+202=14orx=8202=6x = \frac{8 + 20}{2} = 14 \quad \text{or} \quad x = \frac{8 - 20}{2} = -6

Since the length cannot be negative, x=14x = 14.

Final side lengths:

  • Shorter leg =14= 14 feet.
  • Longer leg =2(14)+20=48= 2(14) + 20 = 48 feet.
  • Hypotenuse =2(14)+22=50= 2(14) + 22 = 50 feet.

Would you like further clarification or a detailed explanation of any step?

Here are 5 related questions to expand on this topic:

  1. How does the Pythagorean theorem apply to different types of triangles?
  2. What are the possible methods to solve quadratic equations?
  3. Can you derive a formula for the hypotenuse in any right triangle?
  4. What are some real-life applications of the Pythagorean theorem?
  5. How would the equation change if the triangle were isosceles?

Tip: Always check the discriminant (b24acb^2 - 4ac) when solving quadratic equations to see if real solutions exist.

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Pythagorean Theorem

Formulas

Pythagorean theorem: a^2 + b^2 = c^2
Quadratic equation formula: ax^2 + bx + c = 0
Quadratic formula: x = [-b ± sqrt(b^2 - 4ac)] / 2a

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 8-10