Problem 11 - A Graphical Approach to College Algebra

Question

$$\begin{aligned} &\text {Solve each problem analytically, and support your solution}\\\ &\text {graphically.} \end{aligned}$$ Dimensions of a Square If the length of a side of a square is increased by 3 centimeters, the perimeter of the new square is 40 centimeters more than twice the length of the side of the original square. Find the length of the side of the original square.

Step-by-Step Solution

Answer

The length of the side of the original square is 14 cm.

Step 1: Understanding the Problem

We have a square whose side length, let's say, is \( s \). When we increase the side length by 3 cm, the perimeter of the new square becomes 40 cm more than twice the length of the side of the original square.

Step 2: Setting up the Equation for the New Square

Increasing the side of the square by 3 cm makes the new side length \( s + 3 \). The perimeter of a square is calculated as 4 times the side length. Therefore, the perimeter of the new square is \( 4(s + 3) \).

Step 3: Relating Perimeters to the Original Side Length

According to the problem, the new perimeter equals 40 cm more than twice the side of the original square, which can be formulated as: \( 4(s + 3) = 40 + 2s \).

Step 4: Solving the Equation

Simplify the equation \( 4(s + 3) = 40 + 2s \) to find \( s \). Begin by expanding the left side: \( 4s + 12 = 40 + 2s \). Then, solve for \( s \) by bringing like terms to the same side: \( 4s - 2s = 40 - 12 \). This simplifies to \( 2s = 28 \). Dividing by 2 gives \( s = 14 \).

Step 5: Verifying the Solution

To verify, calculate the new perimeter with \( s = 14 \). Then, the original side plus 3 is 17, and the new perimeter is \( 4 \times 17 = 68 \). Verify \( 68 = 40 + 2 \times 14 \), which holds true, confirming \( s = 14 \) cm is correct.

Key Concepts

Perimeter

Understanding the concept of perimeter is crucial when dealing with squares and other geometric figures. The perimeter of a shape is the total length around it. For squares, this is simple: since all sides are of equal length, the perimeter is four times the length of one side.
In our exercise, the calculation involved increasing the side length of the square and determining how this affects the perimeter. Originally, the perimeter of a square with side length \( s \) is \( 4s \). When the side length is increased by 3 cm, the new side becomes \( s + 3 \). Thus, the perimeter of the new square is \( 4(s + 3) \).
Understanding this fundamentally helps you link the perimeter change to the problem requirements, such as how the new perimeter compares to expressions involving the original side length.

Equation Solving

Solving equations is an essential skill in algebra to find unknown values. In this problem, we set up an equation based on the condition given in the problem. We know that the new perimeter \( 4(s + 3) \) is 40 cm more than twice the original side length \( 2s \). This is expressed as:

• \( 4(s + 3) = 40 + 2s \)

To solve this equation, we start by expanding the left-hand side:

• \( 4(s + 3) = 4s + 12 \)

Then, we bring like terms together:

• \( 4s + 12 = 40 + 2s \)

Simplifying, we get:

• \( 2s = 28 \)

Finally, divide each side by 2 to isolate \( s \):

• \( s = 14 \)

Equation solving involves manipulation of terms to isolate the variable of interest, here \( s \). This skill allows us to translate word problems into mathematical expressions, which can then be systematically solved step by step.

Square Dimensions

Understanding square dimensions involves recognizing that a square is a special type of rectangle where all sides are equal. When working with squares, knowing the side length is enough to calculate other properties.
In estimation or problem-solving, like in this exercise, knowing how changing one side affects perimeter or area is crucial. For example, increasing each side length by a certain measure (3 cm in this problem) and seeing how that impacts the perimeter demonstrates how tightly linked square dimensions are to formulas for perimeter and area.
Visualizing helps:

• Draw a small square for the original side, then add another layer around it for the increase. This helps solidify the understanding of how all sides change simultaneously.


• Because these dimensions are simple multiples of the side length, calculations can be straightforward.

Understanding square dimensions makes it easier to apply the appropriate algebraic operations to solve problems involving changes in such dimensions.