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Chapter 1: Problem 27

What is the length of the side of a square if its area and perimeter are numerically equal?

Short Answer

Expert verified
The side length of the square is 4 units.

Step by step solution

01

Define the variables

Let the side length of the square be denoted as \( s \). The area of the square is then given by \( A = s^2 \) and the perimeter is given by \( P = 4s \).
02

Set up the equation

According to the problem, the area and perimeter are numerically equal. Therefore, set up the equation \( s^2 = 4s \).
03

Solve the equation

To solve \( s^2 = 4s \), first rewrite it as \( s^2 - 4s = 0 \). Factor out \( s \) from the equation to get \( s(s - 4) = 0 \).
04

Find possible solutions

Since \( s(s - 4) = 0 \), there are two possible solutions: \( s = 0 \) or \( s = 4 \).
05

Verify the solutions

Since a side length of a square cannot be zero, the only feasible solution is \( s = 4 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

square properties
A square is a special type of rectangle where all four sides are equal in length and all angles are right angles (90 degrees).

To find the side length of a square when given its area or perimeter, it's important to remember these properties:
  • Area: The area of a square is found by squaring the length of one side, given by the formula \(A = s^2\), where \(s\) is the side length of the square.
  • Perimeter: The perimeter of a square is the total distance around the square. Since all sides are equal, the perimeter is given by \(P = 4s\).
Understanding these fundamental properties is crucial for solving problems related to squares efficiently.
Let's put this into practice with the given problem. The task is to find the length of the side of a square where its area and perimeter are numerically equal.
equation solving
Solving equations involves finding the value(s) of the variables that make the equation true.

Let's break down the process to solve the given problem:
1. From the problem, we know the area and perimeter of the square are numerically equal. Thus, we set the area equation \(s^2\) equal to the perimeter equation \(4s\). This gives us the equation: \[s^2 = 4s\ \] 2. To solve this, we rewrite the equation by moving all terms to one side: \[s^2 - 4s = 0\ \] 3. This forms a quadratic equation \(As^2 + Bs + C = 0\), where A = 1, B = -4, and C = 0.
Solving this quadratic equation is straightforward and can be done by factoring.
factoring equations
Factoring is the process of rewriting expressions as the product of their factors.

For the equation from the problem: \[s^2 - 4s = 0\ \] we can factor out a common factor. Here both terms include \(s\), so we factor \(s\) out:
\[s(s - 4) = 0\ \]
This gives us two possible solutions: \(s = 0\) or \(s - 4 = 0\). Solving \(s - 4 = 0\) gives \(s = 4\).

However, a side length of a square cannot be zero (\(s = 0\) is not a practical solution for this context), so the only viable solution is \(s = 4\). Therefore, the side length of our square is 4 units.

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Most popular questions from this chapter

Solve each equation. $$625 x^{-4}-125 x^{-2}+4=0$$

Concept Check Match the inequality in each exercise in Column I with its equivalent interval notation in Column II. A. \((-2,6]\) B. \([-2,6)\) C. \((-\infty,-6]\) D. \([6, \infty)\) E. \((-\infty,-3) \cup(3, \infty)\) F. \((-\infty,-6)\) G. \((0,8)\) H. \((-\infty, \infty)\) I. \([-6, \infty)\) J. \(\quad(-\infty, 6]\) $$x \leq 6$$

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