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Chapter 10: Problem 47

A rectangle has an area of 180 \(\mathrm{cm}^{2}\) and a perimeter of 54 \(\mathrm{cm} .\) What are its dimensions?

Short Answer

Expert verified
The rectangle's dimensions are 12 cm by 15 cm.

Step by step solution

01

Introduction to the Problem

We are given a rectangle with an area of 180 cm虏 and a perimeter of 54 cm. Our goal is to find its dimensions, which consist of the length (\(l\)) and width (\(w\)).
02

Understanding Formulas

We know two key formulas: the area of a rectangle is \(A = l \times w\) and the perimeter is \(P = 2(l + w)\). We will use these formulas to set up two equations based on the given values.
03

Setting Up Equations

Based on the given area, \(l \times w = 180\). For the perimeter, we have \(2(l + w) = 54\), which simplifies to \(l + w = 27\). Now we have two equations: \(l \times w = 180\) and \(l + w = 27\).
04

Express One Variable in Terms of the Other

From the second equation, \(l + w = 27\), we can express \(l\) in terms of \(w\) as \(l = 27 - w\).
05

Substitute into the Area Equation

Substitute \(l = 27 - w\) into the area equation \(l \times w = 180\). This gives us \((27 - w) \times w = 180\).
06

Simplify and Solve the Quadratic Equation

Expand the equation to get \(27w - w^2 = 180\), which can be rearranged to \(-w^2 + 27w - 180 = 0\). Multiply the entire equation by -1 to simplify: \(w^2 - 27w + 180 = 0\).
07

Factoring the Quadratic Equation

Factor the quadratic equation \(w^2 - 27w + 180 = 0\). The factors are \((w - 12)(w - 15) = 0\).
08

Finding the Width

Set each factor equal to zero to find potential solutions for \(w\): \(w - 12 = 0\) and \(w - 15 = 0\). Solving these gives \(w = 12\) and \(w = 15\).
09

Finding Corresponding Lengths

For \(w = 12\), using \(l = 27 - w\), we find \(l = 27 - 12 = 15\). For \(w = 15\), similarly, \(l = 12\). Thus, the dimensions of the rectangle are 12 cm by 15 cm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Area of a Rectangle
The area of a rectangle is a measure of the space contained within its boundaries. To calculate it, you multiply the length (\(l\) ) by the width (\(w\) ). The formula for the area (\(A\) ) is:
  • \(A = l \times w\)
This formula is useful when you know both dimensions of the rectangle. However, in cases where you only know the area and one dimension, you can rearrange the formula to find the missing dimension. For instance, if the area is 180 cm虏 and one side measures 12 cm, you can find the other side by dividing the area by the known dimension:
  • \(l = \frac{A}{w} = \frac{180}{12} = 15\) cm
Understanding this concept is crucial when tackling problems involving the dimensions of rectangles since it allows flexible manipulation of the formula as needed.
Perimeter of a Rectangle
The perimeter of a rectangle is the total length around the rectangle. It is calculated by adding the lengths of all four sides. Since a rectangle has two pairs of equal sides, the formula simplifies the calculation:
  • \(P = 2(l + w)\)
This formula helps you determine the perimeter when both the length and width are known. But if you only know one of the dimensions and the perimeter, it enables you to solve for the unknown dimension. For instance, if the perimeter is 54 cm, and you know one side, say 12 cm, you use the equation:\
  • \(2(12 + w) = 54\)
  • Solving for \(w\), you get:\(l + w = 27\), hence\(w = 15\) cm
Thus, the formula offers a means to find unknown lengths and widths quickly and efficiently.
Quadratic Equation
A quadratic equation is an important concept in algebra that appears in many geometry problems. It takes the form of \(ax^2 + bx + c = 0\). In our problem, we arrive at a quadratic equation when we substitute known values for the dimensions:
  • \(w^2 - 27w + 180 = 0\)
This particular form allows us to find variables like width or length by simplifying and factoring the expression. To solve the quadratic equation, you can factor it, use the quadratic formula, or complete the square. Factoring, as seen in our solution, breaks down to simpler expressions:
  • \((w - 12)(w - 15) = 0\)
The solutions to this factorized equation are the potential dimensions of the rectangle. Understanding quadratic equations is key to solving problems involving unknown measures, particularly when they don't fit into simple linear equations.

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Most popular questions from this chapter

19鈥�40 Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned} y & \geq x^{3} \\ y & \leq 2 x+4 \\ x+y & \geq 0 \end{aligned}\right.$$

\(29-44\) Use Cramer's Rule to solve the system. $$ \left\\{\begin{aligned} 3 y+5 z &=4 \\ 2 x &-z=10 \\ 4 x+7 y &=0 \end{aligned}\right. $$

Find the partial fraction decomposition of the rational function. \(\frac{2 x+1}{x^{2}+x-2}\)

19鈥�40 Graph the solution of the system of inequalities. Find the coordinates of all vertices, and determine whether the solution set is bounded. $$\left\\{\begin{aligned} x & \geq 0 \\ y & \geq 0 \\ x & \leq 5 \\ x+y & \leq 7 \end{aligned}\right.$$

Find the partial fraction decomposition of the rational function. \(\frac{2 x^{3}+7 x+5}{\left(x^{2}+x+2\right)\left(x^{2}+1\right)}\)
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