Math Problem Statement

The area of a rectangular painting is 35 square feet. The perimeter is 24 feet. What are the dimensions of the painting?

Solution

To solve for the dimensions of the rectangular painting, we can set up a system of equations using the area and perimeter formulas.

Let the length of the rectangle be ll and the width be ww.

Step 1: Write the area equation

The area AA of a rectangle is given by:

A=l×wA = l \times w

We are told that the area is 35 square feet:

l \times w = 35 \tag{1}

Step 2: Write the perimeter equation

The perimeter PP of a rectangle is given by:

P=2l+2wP = 2l + 2w

We are told that the perimeter is 24 feet:

2l+2w=242l + 2w = 24

Simplify the perimeter equation:

l + w = 12 \tag{2}

Step 3: Solve the system of equations

We now have the system of two equations:

  1. l×w=35l \times w = 35
  2. l+w=12l + w = 12

From equation (2), solve for ll:

l=12wl = 12 - w

Substitute this into equation (1):

(12w)×w=35(12 - w) \times w = 35

Expand the equation:

12ww2=3512w - w^2 = 35

Rearrange the terms:

w212w+35=0w^2 - 12w + 35 = 0

Step 4: Solve the quadratic equation

Solve the quadratic equation w212w+35=0w^2 - 12w + 35 = 0 using the quadratic formula:

w=(12)±(12)24(1)(35)2(1)w = \frac{-(-12) \pm \sqrt{(-12)^2 - 4(1)(35)}}{2(1)}

w=12±1441402w = \frac{12 \pm \sqrt{144 - 140}}{2}

w=12±42w = \frac{12 \pm \sqrt{4}}{2}

w=12±22w = \frac{12 \pm 2}{2}

This gives two solutions:

w=12+22=7orw=1222=5w = \frac{12 + 2}{2} = 7 \quad \text{or} \quad w = \frac{12 - 2}{2} = 5

Step 5: Find the corresponding lengths

For w=7w = 7, from equation (2), l=127=5l = 12 - 7 = 5.

For w=5w = 5, from equation (2), l=125=7l = 12 - 5 = 7.

Thus, the dimensions of the rectangle are:

5 feet×7 feet\boxed{5 \text{ feet} \times 7 \text{ feet}}

These are the possible dimensions of the painting.


Would you like further details or clarifications on any of the steps? Here are some related questions:

  1. How can we use the quadratic formula to solve other real-world problems?
  2. What would happen if the perimeter was increased to 30 feet?
  3. How can we solve similar problems involving the area and perimeter of triangles?
  4. What are some applications of quadratic equations in real-life scenarios?
  5. How do you derive the quadratic formula?

Tip: When solving problems involving area and perimeter, always make sure to use the correct formulas and keep track of your units!

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Math Problem Analysis

Mathematical Concepts

Algebra
Quadratic Equations
Geometry

Formulas

Area of a rectangle: A = l * w
Perimeter of a rectangle: P = 2l + 2w

Theorems

Quadratic Formula

Suitable Grade Level

Grades 7-9