Problem 53 - Algebra 2 and Trigonometry

Question

The difference in the lengths of the sides of two squares is 1 meter. The difference in the areas of the squares is 13 square meters. What are the lengths of the sides of the squares?

Step-by-Step Solution

Answer

The lengths of the sides of the squares are 7 meters and 6 meters.

Step 1: Define Variables

Let the side of the larger square be denoted as \( x \) and the side of the smaller square as \( y \). According to the problem, the difference in their side lengths is \( x - y = 1 \).

Step 2: Express Areas in Terms of Variables

The area of the larger square is \( x^2 \) and the area of the smaller square is \( y^2 \). The difference in their areas is given as \( x^2 - y^2 = 13 \).

Step 3: Use Difference of Squares Formula

The formula for the difference of squares states that \( x^2 - y^2 = (x-y)(x+y) \). Given \( x - y = 1 \), substitute into the formula to get \( 1 \times (x + y) = 13 \).

Step 4: Solve for Sum of Variables

From \( 1 \times (x + y) = 13 \), we have \( x + y = 13 \).

Step 5: Create and Solve System of Equations

We now have two equations: \( x - y = 1 \) and \( x + y = 13 \). Add these two equations:\[ (x - y) + (x + y) = 1 + 13 \]This simplifies to \( 2x = 14 \), which gives \( x = 7 \).

Step 6: Solve for Smaller Side

Using \( x = 7 \) in the equation \( x - y = 1 \), we find:\[ 7 - y = 1 \]Therefore, \( y = 6 \).

Step 7: Verify Solution

To ensure correctness, check the conditions: The side lengths are \( x = 7 \) and \( y = 6 \), so their difference is 1 (as given). Their areas are \( 49 \) and \( 36 \), and their difference is \( 13 \), which matches the problem statement.

Key Concepts

Systems of Equations

Systems of equations are a powerful mathematical tool that involves solving two or more equations simultaneously to find the values of the unknown variables. In our problem, we have two unknowns: the side lengths of two squares, which we've denoted as \( x \) for the larger square and \( y \) for the smaller square. These equations together form a system:

• First equation: \( x - y = 1 \)
• Second equation: \( x + y = 13 \)

Understanding these equations allows us to find two key relationships: the difference in their side lengths and their sum. By adding the two equations, we effectively eliminate one variable, simplifying the system to a single equation. This technique of adding or subtracting equations is known as the elimination method. It enables us to find that \( x = 7 \), and subsequently, \( y = 6 \). By finding these values, we solve the system of equations and provide the solution to the problem. This method is essential in algebra for efficiently solving problems involving multiple unknowns.

Difference of Squares

The difference of squares is a special algebraic technique used to simplify expressions of the form \( x^2 - y^2 \). This expression can be rewritten using the identity \( x^2 - y^2 = (x-y)(x+y) \), where \( x \) and \( y \) are any real numbers. In our problem, this identity is particularly useful because it simplifies our equation regarding the areas of the squares.

Given that \( x - y = 1 \), we can directly substitute into the formula to find that \( 1 \times (x + y) = 13 \). This substitution is key as it reduces the complexity of the problem, turning an equation of quadratic terms into a linear equation, which is much easier to solve.

Ultimately, the difference of squares concept allows us to manipulate and simplify the given conditions easily while providing insights into the geometric nature of the squares. This algebraic tool not only simplifies calculations but also deepens understanding of relationships between different elements in mathematical problems.

Area of Squares

In geometry, the area of a square is determined by squaring the length of its side, or expressed algebraically as \( x^2 \) for a square with side length \( x \). Understanding this concept is vital when solving problems that involve the measurable properties of shapes.

For our specific exercise, we deal with two squares where the difference in their side lengths is 1 meter, and their area difference is 13 square meters. This quantitative relationship is heavily reliant on finding the precise area, which provides context for the problem’s equations. As we defined earlier, the area of the larger square is \( x^2 \) while the smaller square has an area of \( y^2 \).

By knowing that the difference in these areas is \( x^2 - y^2 = 13 \), we connect this geometric property to algebraic equations using the difference of squares principle and solve for the unknown side lengths. Thus, understanding the area of squares translates geometrical shapes into algebraic terms, offering a bridge between visual and numerical problem-solving.