Problem 30 - Prealgebra

Question

Draw and label the dimensions of each rectangle whose perimeter and area are given. $$P=16 \mathrm{m}, A=12 \mathrm{m}^{2}$$

Step-by-Step Solution

Answer

The rectangle dimensions are 6m by 2m (or 2m by 6m). Both satisfy the given perimeter and area.

Step 1: Understand the problem

You need to find the dimensions of a rectangle where the perimeter is 16 meters and the area is 12 square meters. Let's denote the length as \( l \) meters and the width as \( w \) meters.

Step 2: Setup perimeter equation

The formula for the perimeter \( P \) of a rectangle is \( P = 2l + 2w \). Substitute \( P = 16 \) to get:\[2l + 2w = 16 \]

Step 3: Setup area equation

The formula for the area \( A \) of a rectangle is \( A = l imes w \). Substitute \( A = 12 \) to get:\[l imes w = 12 \]

Step 4: Solve perimeter equation for one variable

Solve the perimeter equation for one variable. Here it’s convenient to solve for \( l \):\[2l + 2w = 16 \ l + w = 8 \ l = 8 - w\]

Step 5: Substitute into area equation

Substitute \( l = 8 - w \) into the area equation:\[(8 - w)w = 12 \ 8w - w^2 = 12\]

Step 6: Rearrange into a quadratic equation

Rearrange the equation into a standard quadratic form:\[-w^2 + 8w - 12 = 0 \ w^2 - 8w + 12 = 0\]

Step 7: Solve the quadratic equation

Solve the quadratic equation for \( w \). You can factor or use the quadratic formula:\[w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Here, \( a = 1, b = -8, c = 12 \):\[w = \frac{-(-8) \pm \sqrt{(-8)^2 - 4 \times 1 \times 12}}{2 \times 1} \ w = \frac{8 \pm \sqrt{64 - 48}}{2} \ w = \frac{8 \pm \sqrt{16}}{2} \ w = \frac{8 \pm 4}{2}\]Which gives solutions \( w = 6 \) and \( w = 2 \).

Step 8: Find corresponding lengths

Use the values of \( w \) to find \( l \) from \( l = 8 - w \).- If \( w = 6 \), then \( l = 8 - 6 = 2 \).- If \( w = 2 \), then \( l = 8 - 2 = 6 \).Thus, the possible dimensions are \( (l, w) = (6, 2) \) or \( (2, 6) \).

Step 9: Draw and label

Draw a rectangle and label the dimensions:- For \( l = 6 \) m and \( w = 2 \) m, or vice versa. Both configurations satisfy the conditions as the equations are symmetric with respect to \( l \) and \( w \).

Key Concepts

Perimeter

The perimeter of a rectangle is an important measurement that represents the total distance around the rectangle. It is calculated by adding up all the sides of the rectangle. Since a rectangle has two pairs of equal sides, the formula for the perimeter (P) is:

• \(P = 2l + 2w\)

This formula means you take the length (l) and width (w) of the rectangle, multiply each by two, and then add them together.
In our exercise, the problem states that the perimeter is 16 meters. This gives us the equation:

• \(2l + 2w = 16\)

With this equation, we can start finding the rectangle's dimensions.
Understanding perimeter is key in shaping how we set up the problem and start solving it.

Area

The area of a rectangle is a measure of the space it occupies on a plane. It is found by multiplying the length of the rectangle by its width. Thus, the area (A) can be calculated using the formula:

• \(A = l \times w\)

This formula means you multiply the length and the width. The area tells you how much surface the rectangle covers.
In our exercise where the area is defined as 12 square meters, the equation becomes:

• \(l \times w = 12\)

By using this equation in conjunction with the perimeter equation, we can find the dimensions that satisfy both conditions.
Understanding the area is crucial to solve the exercise as it directly influences one side of our equation.

Quadratic Equations

Quadratic equations appear when solving problems involving areas or perimeters under certain constraints. A standard quadratic equation takes the form:

• \(ax^2 + bx + c = 0\)

These equations can have two solutions, and in the context of rectangle geometry, they often help determine possible dimensions.
In our case, substituting the perimeter-derived expression for length into our area equation creates a quadratic equation:

• \(w^2 - 8w + 12 = 0\)

The solutions to this quadratic equation are values of width \(w\), which can be found using methods like factoring, completing the square, or using the quadratic formula:

• \(w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Solving this yields the width values; in this exercise, \(w = 2\) and \(w = 6\), giving us options for corresponding lengths.

Problem Solving Steps

When solving rectangle problems involving both perimeter and area, having a structured approach can be very helpful. Here's the general step-by-step process we used:

• Understand the problem, identifying given quantities like perimeter and area and what needs to be found.
• Set up the perimeter equation using \(P = 2l + 2w\), ensuring all sides' lengths contribute appropriately to the total perimeter.
• Set up the area equation via \(A = l \times w\). Both of these equations are necessary to proceed logically.
• Choose one of the equations to express one variable in terms of the other. Substitute it into the other equation. We solved for length using the perimeter equation.
• Along the way, solve any quadratic forms encountered to find possible dimension solutions.
• Draw and label the rectangle using the found dimensions, verifying each satisfies both the area and perimeter constraints.

Following these clear and structured steps helps systematically address the problem to find solutions reliably.