Question

1. a rectangle has dimensions as shown in the graphic on the right.

algebraically determine the dimensions of the rectangle if
its perimeter is 24 cm.

Explanation:

Step1: Recall the formula for the perimeter of a rectangle

The perimeter \(P\) of a rectangle is given by \(P = 2(l + w)\), where \(l\) is the length and \(w\) is the width. Here, \(l=\frac{2}{x}\) and \(w = \frac{1 - x}{x^{2}}\), and \(P = 24\). So, \(24=2(\frac{2}{x}+\frac{1 - x}{x^{2}})\).

Step2: Simplify the equation

Divide both sides of the equation \(24 = 2(\frac{2}{x}+\frac{1 - x}{x^{2}})\) by \(2\). We get \(12=\frac{2}{x}+\frac{1 - x}{x^{2}}\).
Multiply through by \(x^{2}\) (since \(x
eq0\)) to clear the fractions: \(12x^{2}=2x+(1 - x)\).

Step3: Expand and solve the quadratic equation

Expand the right - hand side: \(12x^{2}=2x + 1-x\), which simplifies to \(12x^{2}-x - 1=0\).
Use the quadratic formula \(x=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for a quadratic equation \(ax^{2}+bx + c = 0\). Here, \(a = 12\), \(b=-1\), and \(c=-1\).
First, calculate the discriminant \(\Delta=b^{2}-4ac=(-1)^{2}-4\times12\times(-1)=1 + 48=49\).
Then \(x=\frac{1\pm\sqrt{49}}{2\times12}=\frac{1\pm7}{24}\).
We have two solutions for \(x\):

• \(x_{1}=\frac{1 + 7}{24}=\frac{8}{24}=\frac{1}{3}\)
• \(x_{2}=\frac{1-7}{24}=\frac{-6}{24}=-\frac{1}{4}\)

Since \(x\) represents a dimension - related value in the context of the rectangle (denominator in the expressions for length and width), and if \(x =-\frac{1}{4}\), the length \(\frac{2}{x}=-8\) (negative length is not valid for a rectangle's dimension). So we take \(x=\frac{1}{3}\).

Step4: Find the dimensions of the rectangle

Substitute \(x = \frac{1}{3}\) into the expressions for length and width.
Length \(l=\frac{2}{x}\), substituting \(x=\frac{1}{3}\), we get \(l = 6\) cm.
Width \(w=\frac{1 - x}{x^{2}}\), substituting \(x=\frac{1}{3}\), we have \(w=\frac{1-\frac{1}{3}}{(\frac{1}{3})^{2}}=\frac{\frac{2}{3}}{\frac{1}{9}}=6\) cm.

Answer:

The length and the width of the rectangle are both \(6\) cm.