1Cademy - Finding the Sides of a Right Triangle Deck with a 17-Foot Hypotenuse

Learn Before

Solving a Quadratic Equation by Factoring

Example

Finding the Sides of a Right Triangle Deck with a 17-Foot Hypotenuse

Apply the seven-step problem-solving strategy to a right triangle word problem where the side lengths are described relative to each other, producing a quadratic equation that must be solved by factoring.

Problem: A right-triangle-shaped deck has a hypotenuse of $17$ feet. One leg is $7$ feet shorter than the other. Find the lengths of the two legs.

Read: A right triangle deck has a hypotenuse of $17$ ft and one side that is $7$ ft less than the other side.
Identify: The lengths of the two legs of the deck.
Name: Let $x$ = the length of one leg. Then $x - 7$ = the length of the other leg.
Translate: Apply the Pythagorean Theorem and substitute:

$a^2 + b^2 = c^2$

$x^2 + (x - 7)^2 = 17^2$

Solve: Expand the squared binomial: $x^2 + x^2 - 14x + 49 = 289$ . Combine like terms: $2x^2 - 14x + 49 = 289$ . Subtract $289$ from both sides to obtain standard form: $2x^2 - 14x - 240 = 0$ . Factor out the GCF of $2$ : $2(x^2 - 7x - 120) = 0$ . Factor the trinomial by finding two numbers whose product is $-120$ and whose sum is $-7$ : the pair $8$ and $-15$ works. So $2(x - 15)(x + 8) = 0$ . Since $2 \neq 0$ , apply the Zero Product Property to the variable factors:

$x - 15 = 0 \quad \text{or} \quad x + 8 = 0$

$x = 15 \quad \text{or} \quad x = -8$

Because $x$ represents a physical length, $x = -8$ does not make sense and is discarded. So $x = 15$ , and the other leg is $15 - 7 = 8$ .

Check: $8^2 + 15^2 = 64 + 225 = 289$ and $17^2 = 289$ , so $289 = 289$ ✓
Answer: The sides of the deck are $8$ , $15$ , and $17$ feet.

This example differs from simpler Pythagorean Theorem problems in two important ways. First, because one leg is expressed in terms of the other ( $x$ and $x - 7$ ), substituting into $a^2 + b^2 = c^2$ produces a quadratic equation rather than a simple square-root problem. Second, solving the quadratic by factoring yields two algebraic solutions, but the negative value must be discarded because a side length cannot be negative. The problem thus combines the Pythagorean Theorem, squaring a binomial, GCF extraction, trinomial factoring, and the rejection of unrealistic solutions — all within a single application.

Updated 2026-04-21

Contributors are:

Who are from:

References

OpenStax Elementary Algebra

Learn Before

Related

Learn After