Problem 47 - Algebra 2 and Trigonometry

Question

The area of a rectangle is 24 square inches. The length of the rectangle is \(\sqrt{5}+1\) inches. Express the width of the rectangle in simplest form.

Step-by-Step Solution

Answer

The width of the rectangle is \( 6\sqrt{5} - 6 \) inches.

Step 1: Understand the area formula for rectangles

The area of a rectangle is calculated by multiplying its length by its width. The formula for the area is \( \text{Area} = \text{Length} \times \text{Width} \).

Step 2: Substitute the known values into the area formula

We know the area of the rectangle is 24 square inches and the length is \( \sqrt{5} + 1 \) inches. Substitute these values into the formula: \( 24 = (\sqrt{5} + 1) \times \text{Width} \).

Step 3: Solve for the width

To find the width, divide both sides of the equation by \( \sqrt{5} + 1 \). This yields \( \text{Width} = \frac{24}{\sqrt{5} + 1} \).

Step 4: Simplify the expression

Simplify the expression \( \frac{24}{\sqrt{5} + 1} \) by rationalizing the denominator. Multiply the numerator and the denominator by the conjugate of the denominator, \( \sqrt{5} - 1 \), resulting in \( \frac{24(\sqrt{5} - 1)}{(\sqrt{5} + 1)(\sqrt{5} - 1)} \).

Step 5: Calculate the simplified denominator

Calculate the denominator: \( (\sqrt{5} + 1)(\sqrt{5} - 1) = \sqrt{5}^2 - 1^2 = 5 - 1 = 4 \).

Step 6: Finalize the expression for the width

This gives us the expression \( \frac{24(\sqrt{5} - 1)}{4} \). Simplify by dividing 24 by 4, resulting in \( 6(\sqrt{5} - 1) \).

Step 7: Simplify further if possible

Since \( 6(\sqrt{5} - 1) \) is already in its simplest form, the width of the rectangle is \( 6\sqrt{5} - 6 \) inches.

Key Concepts

Rationalizing the Denominator

When dealing with expressions that have a square root in the denominator, like \( \frac{24}{\sqrt{5}+1} \), rationalizing the denominator helps us to convert the expression into a more manageable form without radicals in the denominator.

This process involves multiplying both the numerator and the denominator by the conjugate of the denominator. The conjugate of \( \sqrt{5} + 1 \) is \( \sqrt{5} - 1 \). By doing so, we create a denominator without a square root, making calculations easier.

The multiplication yields:

• Numerator: \( 24(\sqrt{5}-1) \)
• Denominator: \( (\sqrt{5}+1)(\sqrt{5}-1) = \sqrt{5}^2-1^2 = 5-1= 4 \)

This simplification is crucial as it allows us to work with whole numbers in the denominator, providing clarity and simplicity in further computations.

Simplifying Expressions

Expressions can often appear complicated, but by understanding and applying various mathematical techniques, we can simplify them into more straightforward and useful forms.

Simplifying expressions involves combining like terms, using the distributive property, and canceling common factors. In the exercise, we ended up with the expression \( \frac{24(\sqrt{5} - 1)}{4} \). Since 24 and 4 have a common factor of 4, we can simplify by dividing both:

• Divide the numerator: \( 24 \div 4 = 6 \)
• The expression simplifies to: \( 6(\sqrt{5} - 1) \)

Simplification is about making expressions easier to understand and solving problems efficiently. Here, after simplification, the width of the rectangle becomes \( 6\sqrt{5} - 6 \), a much neater expression.

Rectangles

Rectangles are fundamental geometric shapes characterized by having opposite sides that are equal and four right angles. Calculating the area of a rectangle is straightforward with the formula \( \text{Area} = \text{Length} \times \text{Width} \).

In the exercise, we used this formula to find the width when the area and length were known. Given an area of 24 square inches and a length of \( \sqrt{5}+1 \) inches, substituting into the area formula allows us to solve for the unknown width.

Rectangles are vital in both pure and applied mathematics and appear in numerous practical contexts like floor planning and architecture. Understanding how to manipulate rectangle-related equations is a core skill in algebra, helping to solve many real-world problems involving areas and dimensions.