Problem 111 - College Algebra with Modeling and Visualization

Question

Equilateral Triangle The area of an equilateral triangle with sides of length \(s\) is given by $$ A(s)=\frac{\sqrt{3}}{4} s^{2} $$ (a) Find \(A(4 s)\) and interpret the result. (b) Find \(A(s+2)\) and interpret the result.

Step-by-Step Solution

Answer

(a) 16 times the original area; it scales by 16. (b) Area changes with a quadratic polynomial expression.

Step 1: Understand the Problem

We are given a formula for the area of an equilateral triangle with side length \( s \), which is \( A(s) = \frac{\sqrt{3}}{4} s^2 \). We need to find the area when side length is \( 4s \) and \( s+2 \).

Step 2: Substitute and Simplify for Part (a)

For the first part, substitute \( 4s \) into the original area formula: \[ A(4s) = \frac{\sqrt{3}}{4} (4s)^2 \] Simplify the expression: \[ A(4s) = \frac{\sqrt{3}}{4} \times 16s^2 = 4\sqrt{3}s^2 \]

Step 3: Interpretation of A(4s)

The area \( A(4s) = 4\sqrt{3}s^2 \) indicates that the area of the triangle increases by a factor of 16 when the side length is quadrupled. This is because, in general, the area scales with the square of the side length.

Step 4: Substitute and Simplify for Part (b)

Now, substitute \( s + 2 \) into the original formula: \[ A(s+2) = \frac{\sqrt{3}}{4} (s+2)^2 \] Expand \((s+2)^2\): \[ (s+2)^2 = s^2 + 4s + 4 \] Thus, \[ A(s+2) = \frac{\sqrt{3}}{4} (s^2 + 4s + 4) = \frac{\sqrt{3}}{4}s^2 + \sqrt{3}s + \sqrt{3} \]

Step 5: Interpretation of A(s+2)

The expression \( A(s+2) = \frac{\sqrt{3}}{4}s^2 + \sqrt{3}s + \sqrt{3} \) shows how the area changes when 2 is added to each side. It increases by a combination of terms related to both the original side length \( s \) and the constant 2, indicating a complex polynomial relationship.

Key Concepts

Algebraic Manipulation

Algebraic manipulation is an essential tool in solving mathematical problems, especially when working with formulas. It involves rearranging and simplifying expressions to make them easier to work with. In this exercise, the given formula for the area of an equilateral triangle, \( A(s) = \frac{\sqrt{3}}{4} s^2 \), is manipulated by substituting different expressions for \( s \).

When tasked with finding \( A(4s) \), we're essentially replacing \( s \) with \( 4s \) in the formula. This requires us to first substitute and then simplify the expression:

• Substitute: \( A(4s) = \frac{\sqrt{3}}{4} (4s)^2 \)
• Simplify: \( A(4s) = \frac{\sqrt{3}}{4} \times 16s^2 = 4\sqrt{3}s^2 \)

Being careful with algebraic steps is crucial to revealing how changes in dimensions influence other quantities, such as area. The simplification shows how dramatically the area expands with increased side lengths, emphasizing the power of algebraic manipulation in understanding mathematical relationships.

Polynomial Expansion

Polynomial expansion plays a key role when dealing with expressions like \( (s+2)^2 \), which appears in the problem. It is the process of expanding expressions raised to a power, transforming them into a sum of terms according to binomial theorem principles.

In the case of \( A(s+2) \), the expansion of \((s+2)^2\) is necessary:

• Start with \( (s+2)^2 \).
• Expand: \( (s+2)^2 = s^2 + 4s + 4 \).

Plugging this expanded form back into the area formula, \( A(s+2) = \frac{\sqrt{3}}{4} (s^2 + 4s + 4) \), allows further simplification by distributing \( \frac{\sqrt{3}}{4} \) across the sum:

• Resulting in: \( A(s+2) = \frac{\sqrt{3}}{4}s^2 + \sqrt{3}s + \sqrt{3} \).

Understanding polynomial expansion is critical not only in geometry but also in broader applications of algebra, as it breaks down complex expressions into manageable pieces. This process highlights how each component of a polynomial contributes to the overall shape and size of geometric figures.

Geometric Interpretation

Geometric interpretation is about visualizing mathematical expressions in terms of shape and space. It's crucial for comprehending how changes in a formula affect the actual geometry. In the context of this exercise, we interpret the area formulas for different side lengths.

For \( A(4s) = 4\sqrt{3}s^2 \), quadrupling the side length results in an increase of the area by a factor of 16, as the area formula depends on the square of the side lengths. This illustrates how dramatically area can grow with small increases in dimension.

For \( A(s+2) = \frac{\sqrt{3}}{4}s^2 + \sqrt{3}s + \sqrt{3} \), the interpretation is more nuanced. Adding 2 units to each side increases the area not only by squaring the sum but also introducing additional terms related to both the original \( s \) and the constant term.

• Visualize: It resembles a basic area with an "extra" forming from the added length.
• Intricacy: This extra consists of both linear and constant terms, enhancing the complexity of the area increment.

Geometric interpretation helps in visualizing how polynomial operations relate to physical shapes' properties, thereby bridging algebra with tangible concepts familiar from geometry.