Math Problem Statement

The area of a rectangle is 28 yd², and the length of the rectangle is 5 yd less than three times the width. Find the dimensions of the rectangle.

Solution

In this problem, we are told that the area of a rectangle is 28 square yards, and the length is 5 yards less than three times the width. We need to find the dimensions of the rectangle.

Let:

  • ww represent the width of the rectangle.
  • ll represent the length of the rectangle.

We are given the following information:

  1. The length is 5 yards less than three times the width, so: l=3w5l = 3w - 5
  2. The area of the rectangle is 28 square yards, and the area is calculated as Area=l×w\text{Area} = l \times w. Thus: l×w=28l \times w = 28

Substitute the expression for ll from the first equation into the second equation: (3w5)×w=28(3w - 5) \times w = 28 Expand and simplify: 3w25w=283w^2 - 5w = 28 3w25w28=03w^2 - 5w - 28 = 0

This is a quadratic equation. We can solve it using the quadratic formula: w=(5)±(5)24(3)(28)2(3)w = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(3)(-28)}}{2(3)} Simplify: w=5±25+3366w = \frac{5 \pm \sqrt{25 + 336}}{6} w=5±3616w = \frac{5 \pm \sqrt{361}}{6} w=5±196w = \frac{5 \pm 19}{6}

So, ww can be: w=5+196=246=4w = \frac{5 + 19}{6} = \frac{24}{6} = 4 or [ w = \frac{5 -

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Math Problem Analysis

Mathematical Concepts

Algebra
Geometry
Quadratic Equations

Formulas

Area of a rectangle: Area = length × width
Quadratic equation: ax^2 + bx + c = 0

Theorems

Quadratic formula

Suitable Grade Level

Grades 8-10