Problem 229 - Elementary Algebra

Question

In the following exercises, solve using triangle properties. One side of a triangle is twice the shortest side. The third side is five feet more than the shortest side. The perimeter is 17 feet. Find the lengths of all three sides.

Step-by-Step Solution

Answer

The sides are 3 feet, 6 feet, and 8 feet.

Step 1 - Define Variables

Let the shortest side of the triangle be denoted as \(x\).

Step 2 - Express Other Sides in Terms of Variables

The side that is twice the shortest side can be expressed as \(2x\). The third side, which is five feet more than the shortest side, can be expressed as \(x + 5\).

Step 3 - Write the Perimeter Equation

The perimeter of the triangle is given as 17 feet. Using the perimeter formula \(\text{Perimeter} = a + b + c\), we can write the equation \(x + 2x + (x + 5) = 17\).

Step 4 - Simplify the Equation

Combine like terms to simplify the equation: \(x + 2x + x + 5 = 17\) simplifies to \(4x + 5 = 17\).

Step 5 - Solve for x

Subtract 5 from both sides of the equation: \(4x + 5 - 5 = 17 - 5\) simplifies to \(4x = 12\). Divide both sides by 4 to solve for \(x\): \(x = 3\).

Step 6 - Find the Lengths of All Sides

The shortest side is \(x = 3\) feet. The side that is twice the shortest side is \(2x = 2 \times 3 = 6\) feet. The third side is \(x + 5 = 3 + 5 = 8\) feet.

Key Concepts

Solving Equations

Solving equations is a fundamental skill in algebra. It involves finding the value of unknown variables that make the equation true. In this exercise, we use a linear equation to find the lengths of the triangle's sides, given the sum of their lengths (the perimeter). We start by writing an equation based on the given information and then solving it step-by-step.
To solve an equation, we usually:

• Combine like terms.
• Isolate the variable on one side of the equation.
• Perform operations (addition, subtraction, multiplication, division) to find the variable's value.

Practicing this method helps in understanding how variables interact in mathematical expressions and prepares one for more complex problems.

Defining Variables

Defining variables means assigning symbols to represent unknown quantities. This is an essential step in solving problems involving algebraic expressions and equations.
In this exercise, we define the shortest side of the triangle as the variable \(x\). From this, we express the other sides based on relationships described in the problem:

• Shortest side: \(x\)
• Twice the shortest side: \(2x\)
• Third side, five feet more than the shortest side: \(x + 5\)

By clearly defining variables, we can set up equations that are easier to manipulate and solve. This approach helps us make sense of the problem and find the correct values for each quantity.

Perimeter of Triangle

The perimeter of a triangle is the total length around the triangle. It is found by adding the lengths of all three sides together. Understanding this concept is crucial for solving problems involving triangles.
For this exercise, the perimeter is given as 17 feet. Using the variables defined earlier, we can set up the perimeter equation as:
\[ x + 2x + (x + 5) = 17 \]
This equation comes from adding the length of each side:

• Shortest side: \(x\)
• Twice the shortest side: \(2x\)
• Third side: \(x + 5\)

Summing these sides gives us a straightforward way to use the perimeter to solve for \(x\) and then find the lengths of all sides.

Algebraic Expressions

Algebraic expressions are combinations of variables, numbers, and operations (such as addition or multiplication). They are used to represent relationships and can be manipulated to solve problems.
In the context of this exercise, we use algebraic expressions to represent the sides of the triangle.
The expressions are:

• Shortest side: \(x\)
• Twice the shortest side: \(2x\)
• Third side: \(x + 5\)

These expressions are then combined in the perimeter equation:
\[ x + 2x + (x + 5) = 17 \]
We simplify this to:
\[ 4x + 5 = 17 \]
By solving this equation, we find \(x\), which gives us the lengths of all sides. This deepens our understanding of how algebraic expressions work and how they can be used to solve real-world problems.