Problem 16 - Precalculus Student Solutions Manual 5th

Question

The lengths of the sides of a right triangle are consecutive positive integers. Find these lengths. (Hint: Use the Pythagorean theorem.)

Step-by-Step Solution

Answer

The side lengths are 3, 4, and 5.

Step 1: Identify Variables

Let's denote the lengths of the sides of the right triangle as consecutive positive integers, i.e., let the sides be represented by \( a, a+1, a+2 \) where \( a \) is a positive integer.

Step 2: Apply the Pythagorean Theorem

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. In our case, \( a \), \( a+1 \), and \( a+2 \) are the sides, and \( a+2 \) must be the hypotenuse. So, we use the Pythagorean theorem: \[ a^2 + (a+1)^2 = (a+2)^2 \].

Step 3: Expand and Simplify the Equation

Expand the equation: \[ a^2 + (a+1)^2 = (a+2)^2 \]\[ a^2 + a^2 + 2a + 1 = a^2 + 4a + 4 \]Combine like terms: \[ 2a^2 + 2a + 1 = a^2 + 4a + 4 \].

Step 4: Solve the Quadratic Equation

Move all terms to one side to form a quadratic equation: \[ 2a^2 + 2a + 1 - a^2 - 4a - 4 = 0 \]Simplify: \[ a^2 - 2a - 3 = 0 \].Factor the quadratic equation: \[ (a-3)(a+1) = 0 \].Set each factor to zero: \( a - 3 = 0 \) or \( a + 1 = 0 \).Solve for \( a \): \( a = 3 \) or \( a = -1 \). Since \( a \) must be positive, we have \( a = 3 \).

Step 5: Determine Side Lengths

Substitute \( a = 3 \) back into the original variables to find the side lengths. The sides are: \( a = 3 \), \( a+1 = 4 \), \( a+2 = 5 \). Thus, the side lengths are 3, 4, and 5.

Key Concepts

Pythagorean theorem

The Pythagorean theorem is a fundamental principle in geometry. It states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. Mathematically, it is written as:

a^2 + b^2 = c^2.

In the given problem, the sides of the right triangle are consecutive integers, specifically 3, 4, and 5. We use the Pythagorean theorem to determine the relationship between these sides. For this problem, we have:

• The smaller side is denoted as a.
• The middle side is a + 1.
• The hypotenuse, which is the largest side, is a + 2.

Substituting these into the formula, we get:

a^2 + (a+1)^2 = (a+2)^2.

This equation helps us set up the problem to solve for the sides of the triangle.

Quadratic equations

Quadratic equations are polynomials of degree 2, generally represented as:

ax^2 + bx + c = 0.

In our problem, to solve for the sides of the triangle, we first expanded the equation:

a^2 + (a+1)^2 = (a+2)^2,

and simplified it step by step until we reached:

a^2 - 2a - 3 = 0.

This is a quadratic equation where:

• a is the coefficient of the quadratic term: 1.
• b is the coefficient of the linear term: -2.
• c is the constant term: -3.

To solve this quadratic equation, we used factoring. The equation factors into:

a^2 - 2a - 3 = (a - 3)(a + 1) = 0.

We then solved for the variable a by setting each factor equal to zero and finding that a = 3 or -1. Since a must be positive, we concluded a = 3.

Solving equations

Solving equations involves finding the values of the variables that satisfy the equation. In this problem, we solved a quadratic equation by performing the following steps:

• First, we expanded and simplified the equation.
• Next, we combined like terms to form a standard quadratic equation.
• We then moved all terms to one side to set the equation to zero.

From our equation:

a^2 - 2a - 3 = 0,

we proceeded by factoring it to:

a^2 - 2a - 3 = (a - 3)(a + 1) = 0.

Setting each factor equal to zero gave us:

a - 3 = 0 or a + 1 = 0,

leading to solutions a = 3 or a = -1.

Since a must be positive, we chose a = 3. Finally, substituting a into the original variables, we found the side lengths of the triangle to be 3, 4, and 5.