/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 4 You have 16 miles of fence that ... [FREE SOLUTION] | 91Ó°ÊÓ

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You have 16 miles of fence that you will use to enclose a rectangular field. a. Draw a picture to show that you can arrange the 16 miles of fence into a rectangle of width 3 miles and length 5 miles. What is the area of this rectangle? b. Draw a picture to show that you can arrange the 16 miles of fence into a rectangle of width 2 miles and length 6 miles. What is the area of this rectangle?

Short Answer

Expert verified
a. Area = 15 square miles; b. Area = 12 square miles.

Step by step solution

01

Understanding the Perimeter Constraint

For any rectangle, the perimeter is calculated as \( 2 imes ( \text{length} + \text{width} ) \). In this problem, the perimeter of the field, which is the length of the fence, is given as 16 miles. We'll use this to verify the dimensions in parts (a) and (b).
02

Arranging the Fence for Part (a)

For a rectangle with a width of 3 miles and a length of 5 miles, calculate the perimeter: \( 2 \times (5 + 3) = 2 \times 8 = 16 \) miles. This matches the given fence length, confirming that such a rectangle is possible. The area is calculated as \( \text{length} \times \text{width} = 5 \times 3 = 15 \) square miles.
03

Arranging the Fence for Part (b)

For a rectangle with a width of 2 miles and a length of 6 miles, calculate the perimeter: \( 2 \times (6 + 2) = 2 \times 8 = 16 \) miles. This also matches the given fence length, confirming that such a rectangle is achievable. The area is \( \text{length} \times \text{width} = 6 \times 2 = 12 \) square miles.

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Key Concepts

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

Rectangles and Perimeters
A rectangle is a four-sided shape where opposite sides are equal in length and every angle is a right angle (90 degrees). When working with rectangles, understanding the concept of perimeter is crucial.
For a rectangle, the perimeter can be calculated using the formula:
  • Perimeter = 2 × (Length + Width)
This formula helps determine the total distance around the rectangle. In our original exercise, we used 16 miles of fencing, which means that the sum of all four sides of the rectangle must equal 16 miles.
This knowledge allows us to find different combinations of length and width that fit the requirement, and it’s important to understand that several rectangles with different dimensions can have the same perimeter.
Understanding Area Calculation in Rectangles
Area refers to the amount of space contained within a two-dimensional shape. For rectangles, the area is calculated using the formula:
  • Area = Length × Width
In our exercise, by rearranging the dimensions of a rectangle while keeping the perimeter constant, we noticed how the area changes. For example:
  • A rectangle with dimensions 5 miles by 3 miles yields an area of \(5 \times 3 = 15\) square miles.
  • Another rectangle with dimensions 6 miles by 2 miles results in an area of \(6 \times 2 = 12\) square miles.
These variations demonstrate an important concept: while the perimeter might remain unchanged, the area can vary significantly based on the ratio of length to width. It's vital to understand this relationship, especially when optimizing layouts in real-world applications, such as allocating land.
Solving Algebraic Problems with Perimeter and Area
Algebraic problems involving perimeter and area can seem challenging at first, but breaking them down into smaller steps makes them manageable. In exercises like these, the initial step is usually to define the given values and determine the unknowns.
Let's say we are given a fixed perimeter, like in our 16-mile example, and we need to establish possible dimensions of the rectangle. This requires setting up an equation:
  • Given: Perimeter is 16 miles: \(2 \times (L+W) = 16\)
  • Solve for one variable, for instance, \(L\) or \(W\).
By manipulating these equations, we can explore the various possible combinations of dimensions that satisfy the perimeter condition. Additionally, we can further calculate and compare the areas to see how different dimensions affect the space enclosed by the rectangle.
Algebraic skills are essential here as they allow us to systemically solve real-world problems where balancing different dimensions for the best design is key.

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Most popular questions from this chapter

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