Problem 34 - Elementary and Intermediate Algebra

Question

Pythagorean Theorem Problems Car Repairs. To create some space to work under the front end of a car, a mechanic drives it up steel ramps. A ramp is 1 foot longer than the back, and the base is 2 feet longer than the back of the ramp. Find the length of each side of the ramp. (IMAGE CANNOT COPY)

Step-by-Step Solution

Answer

The lengths of the ramp sides are 1 foot, 2 feet, and 3 feet.

Step 1: Understanding the Problem

We need to find the lengths of the three sides of a ramp. Let's denote the lengths: let \( b \) be the back of the ramp, \( b+1 \) be the ramp (hypotenuse), and \( b+2 \) be the base.

Step 2: Apply the Pythagorean Theorem

According to the Pythagorean Theorem, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This gives us the equation: \[ (b+1)^2 = b^2 + (b+2)^2 \]

Step 3: Expanding the Equation

Expand the equation to simplify it: 1. Square \( b+1 \): \[ (b+1)^2 = b^2 + 2b + 1 \] 2. Square \( b+2 \): \[ (b+2)^2 = b^2 + 4b + 4 \]

Step 4: Substitute and Simplify

Substitute the expressions from the expansions into the Pythagorean equation: \[ b^2 + 2b + 1 = b^2 + b^2 + 4b + 4 \] Combine and simplify the terms: \[ b^2 + 2b + 1 = 2b^2 + 4b + 4 \]

Step 5: Rearrange and Solve

Rearrange the terms to form a quadratic equation: \[ 0 = 2b^2 + 4b + 4 - b^2 - 2b - 1 \] Simplify: \[ 0 = b^2 + 2b + 3 \] Subtract \( 3 \) from both sides: \[ b^2 + 2b - 3 = 0 \] This quadratic equation can be factored: \[ (b + 3)(b - 1) = 0 \] Solve for \( b \): the solutions are \( b = -3 \) or \( b = 1 \). Since length cannot be negative, \( b = 1 \).

Step 6: Calculate Other Side Lengths

Using \( b = 1 \), compute the other side lengths: 1. Hypotenuse (ramp): \( b + 1 = 2 \). 2. Base: \( b + 2 = 3 \).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental concept in algebra, and they often show up in problems related to geometry, like the one involving the mechanics' ramp. A quadratic equation is usually presented in the form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( a \) must be non-zero.
In this ramp problem, we derived the quadratic equation:\[ b^2 + 2b - 3 = 0 \]To solve it, we factor the equation into two binomial expressions: \[ (b + 3)(b - 1) = 0 \] Setting each factor equal to zero gives the potential solutions for \( b \).

• \( b + 3 = 0 \) leads to \( b = -3 \)
• \( b - 1 = 0 \) leads to \( b = 1 \)

Since length cannot be negative, we accept \( b = 1 \). This simple problem elegantly illustrates the power of quadratic equations in solving geometric dimensions.

Algebraic Expressions

Algebraic expressions are combinations of numbers, variables, and arithmetic operations. In our ramp problem, algebraic expressions helped us describe the relationships between the different sides of the ramp. We defined:

• \( b \) as the back of the ramp
• \( b+1 \) as the hypotenuse
• \( b+2 \) as the base

Expanding these expressions using the formula for perfect squares is crucial. For example, the expression \((b+1)^2\) is expanded as:\[ (b+1)^2 = b^2 + 2b + 1 \] And similarly, \((b+2)^2\) becomes:\[ (b+2)^2 = b^2 + 4b + 4 \] These expansions allowed us to substitute back into the Pythagorean equation and simplify the problem into a manageable form, leading to a quadratic equation.

Geometry Applications

The Pythagorean Theorem is a crucial part of geometry, especially when dealing with right triangles. It 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: \[ c^2 = a^2 + b^2 \] In the mechanic's ramp problem, the components were a practical application of geometry.

• Here, the hypotenuse (ramp) is \( b+1 \).
• The base is \( b+2 \).
• The back of the ramp is \( b \).

Using these relationships, applying the Pythagorean Theorem becomes straightforward: \[ (b+1)^2 = b^2 + (b+2)^2 \] This equation helped find the dimensions of the ramp by mathematically representing a physical problem. Geometry applications like this one demonstrate how mathematical principles can solve real-world engineering challenges, providing clear guidance in designing and analyzing structures.