Q28. - Algebra 1

Question

A piece of wire 42 centimeters long is bent into the shape of a rectangle with a width that is twice its length. Find the dimensions of the rectangle.

A. 5cm, 12cm

B. 7cm, 14cm

C. 9cm, 16cm

D. 11cm, 18cm

Step-by-Step Solution

Answer

Option B is correct. The dimensions of the rectangle are 7cm, 14cm.

Step 1. Write the formula for perimeter of rectangle.

Perimeter of a rectangle is given by:

$\text{Perimeter}=2\left(l+w\right)$

where $l$ is the length and $w$ is the width of a rectangle.

Step 2. Form the equation .

Let length be $x$ cm.

The width of the rectangle is twice its length.

Therefore, width $=2x$ cm

The length of the wire is 42 cm.

Therefore, perimeter cm 

The equation can be written as:

$$\begin{gathered} 2\left(l+w\right)=\text{Perimeter} \\ 2\left(l+w\right)=42 \end{gathered}$$

Step 3. Solve the equation.

The equation can be solved as follows:

$\begin{array}{c} 2\left(l+w\right)=42\\ 2\left(x+2x\right)=42\\ 2x+4x=42\end{array}$

        $\begin{array}{c}6x=42\\ \frac{6x}{6}=\frac{42}{6}\\ x=7\end{array}$

Therefore, the dimensions are $x=7cm\text{ and }2x=14cm.$