Question

The perimeter of a rectangle is 16 inches and its area is 15 square inches. What are its dimensions?

Short answer

The dimensions of the rectangle are 5 inches and 3 inches.

Step-by-step solution

Understand the Problem

The perimeter of the rectangle is 16 inches, and its area is 15 square inches. Let the length be denoted by \( l \) and the width by \( w \). We need to find the values of \( l \) and \( w \).

Write Down the Formulas

The perimeter of a rectangle is given by 2 times the sum of its length and width:\[ P = 2(l + w) \]The area of a rectangle is given by the product of its length and width:\[ A = l \times w \]

Set Up the Equations

Using the given values, write down the equations:\[ 2(l + w) = 16 \]\[ l \times w = 15 \]

Solve the Perimeter Equation

Divide both sides of the perimeter equation by 2:\[ l + w = 8 \]Now solve for one of the variables, say \( w \):\[ w = 8 - l \]

Substitute into the Area Equation

Substitute \( w = 8 - l \) into the area equation:\[ l(8 - l) = 15 \]Simplify and solve the quadratic equation:\[ l^2 - 8l + 15 = 0 \]

Solve the Quadratic Equation

Solve the quadratic equation using the quadratic formula \( l = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):\[ l = \frac{8 \pm \sqrt{64 - 60}}{2} = \frac{8 \pm 2}{2} \]\[ l = 5 \text{ or } l = 3 \]

Find the Width

Using the solutions for \( l \):If \( l = 5 \), then \( w = 8 - 5 = 3 \)If \( l = 3 \), then \( w = 8 - 3 = 5 \)Thus, the dimensions are either (5,3) or (3,5).

Verify the Solutions

Check the perimeter and area with both sets of solutions:For (5, 3):Perimeter: \( 2(5 + 3) = 16 \)Area: \( 5 \times 3 = 15 \)For (3, 5):Perimeter: \( 2(3 + 5) = 16 \)Area: \( 3 \times 5 = 15 \)Both sets of dimensions satisfy the given problem.

Key concepts

rectangle perimeter

Let's start by understanding the concept of a rectangle's perimeter. The perimeter is the total distance around the outside of the rectangle. It is calculated by adding together the lengths of all four sides. In a rectangle, opposite sides are of equal length. Therefore, the formula to find the perimeter is: \[ P = 2(l + w) \] Here, P is the perimeter, l is the length, and w is the width. If we know the perimeter and one dimension (either length or width), we can find the other dimension by rearranging this equation.

quadratic equation

Quadratic equations are a key part of solving rectangle problems. A quadratic equation is any equation that can be written in the form: \[ ax^2 + bx + c = 0 \] In our problem, we derived the quadratic equation: \[ l^2 - 8l + 15 = 0 \] To solve this, we used the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, a, b, and c are coefficients from the quadratic equation. By substituting the values into this formula, we can find the solutions for l, which are the possible lengths of the rectangle.

area calculation

The area of a rectangle is the amount of space inside its boundary and is calculated by multiplying its length and width: \[ A = l \times w \] In our problem, we were given that the area is 15 square inches. By substituting this into the area formula, we could use algebra to find the dimensions of the rectangle once one variable was isolated from the perimeter equation.

algebraic substitution

Algebraic substitution is a method we used to solve the equations. After simplifying the perimeter equation to find an expression for w in terms of l: \[ w = 8 - l \] we substituted this into the area equation: \[ l(8 - l) = 15 \] This step transformed the original problem into a single variable quadratic equation, which we then solved using the quadratic formula. By substituting back, we found both the length and the width of the rectangle that satisfy both given conditions (perimeter and area).