Problem 36 - Algebra 2

Question

DESIGN. For Exercises \(36-38,\) use the following information. Jill is designing a picture frame for an art project. She plans to have a square piece of glass in the center and surround it with a decorated ceramic frame, which will also be a square. The dimensions of the glass and frame are shown in the diagram at the right. jill determines that she needs 27 square inches of material for the frame. Write a polynomial equation that models the area of the frame.

Step-by-Step Solution

Answer

The polynomial equation is \( f^2 - g^2 = 27 \).

Step 1: Understand Problem Setup

Jill's picture frame consists of a square piece of glass surrounded by a square ceramic frame. Both the glass and the outside of the frame are square-shaped. We need an equation for the area of the frame (the outer square minus the inner square) that equals 27 square inches.

Step 2: Represent Variables

Let the side length of the inner square (glass) be denoted as \( g \), and the side length of the outer square (ceramic frame) as \( f \).

Step 3: Express Outer and Inner Areas

Calculate the area of the outer square (frame) as \( f^2 \) and the area of the inner square (glass) as \( g^2 \).

Step 4: Model Frame Area

The area of the frame material is the area of the outer square minus the area of the inner square: \( f^2 - g^2 = 27 \).

Step 5: Write Polynomial Equation

To model the area of the frame in terms of the side lengths, use the equation of the frame area: \( (f - g)(f + g) = 27 \). This equation can also be seen as \( f^2 - g^2 = 27 \), showing the difference of squares.

Key Concepts

Area of a Square

In this exercise, we're dealing with squares, which are essential in understanding geometric problems like Jill's picture frame. A square is a simple four-sided figure where all sides are equal in length. Thus, the most important property is calculating its area.

To find the area of a square, you square the length of one of its sides. For example, if the side of a square is 3 units, the area will be computed as:

• Area = side × side = 3 × 3 = 9 square units

This property makes squares easy to work with in algebra.

In Jill's case, we have two squares: the glass piece, and the larger outer square which includes both the glass and the frame. By understanding the formula for the area of a square, we can model more complex situations involving composite figures such as frames.

Difference of Squares

The difference of squares is crucial to solving Jill's problem. Recognizing when and how to use it can simplify solving polynomial equations. It is a specific algebraic pattern where you subtract one squared term from another.

The formula for the difference of squares is:

• \[ a^2 - b^2 = (a - b)(a + b)\]

Here, "a" and "b" are any algebraic terms. This identity is valuable because it allows you to factor and simplify expressions easily.

In the problem, the frame's area is represented by this principle. Jill's frame's area can be modeled as:

• \[ f^2 - g^2 = (f - g)(f + g) = 27\]

This indicates the frame's material's area is the difference between the areas of two squares: the larger outer square and the smaller inner square.

Algebraic Modeling

Algebraic modeling is a method of representing real-world situations through algebraic expressions or equations. In this exercise, it helps us translate Jill's picture frame problem into a solvable mathematical equation.

The goal is to construct an equation that represents the conditions given in the problem. For Jill, that condition is needing 27 square inches of material for the frame. With this information, we use algebra to describe how the dimensions of the glass and frame relate to the total area.

By defining variables for the sides of the squares:

• Let \( f \) be the side length of the entire outer frame.
• Let \( g \) be the side length of the glass in the middle.

We can express the problem's requirements with the difference of squares formula. This leads us to model the area as:

• \[(f - g)(f + g) = 27\]

This algebraic equation efficiently encapsulates the physical layout of the picture frame and guides us to the solution.