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Chapter 13: Problem 58

Find the length and width of a rectangle whose perimeter is 40 feet and whose area is 96 square feet.

Short Answer

Expert verified
The dimensions of the rectangle are 8 feet and 12 feet.

Step by step solution

01

Formulate the Equations

Based on the problem, we have two formulas for a rectangle. The perimeter formula is \(P = 2l + 2w\) and the area formula is \(A = lw\). Since we know the perimeter is 40 feet and area 96 square feet, we can write down the equations as: \(2l + 2w = 40\) and \(lw = 96\).
02

Solve One of the Equations

To make this easier close, let's solve for one variable in the perimeter equation. For instance, solving for width, \(w\), we get \(w = 20 - l\).
03

Substitute in the Second Equation

Take the equation we just solved for \(w\) and substitute it into our area equation to get \(l(20 - l) = 96\). This simplifies to a quadratic equation \(l^2 - 20l + 96 = 0\).
04

Solve Quadratic Equation

By factoring the quadratic equation, we get \(l - 8\)(\(l -12\)= 0. Thus the solutions for \(l\) are 8 feet and 12 feet.
05

Find the Width

Substitute the solutions of the length into the equation \(w = 20 - l\). When \(l = 8\), then \(w = 12\). When \(l = 12\), then \(w = 8\). Hence, the length and width of the rectangle are 8 feet and 12 feet respectively.

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