# The dimensions of a rectangle are such that it's length is 9 inches more than its width. If the...

## Question:

The dimensions of a rectangle are such that it's length is 9 inches more than its width. If the length were doubled and if the width were decreased by 4 inches, the area would be increased by 110 in{eq}^2 {/eq}. What are the length and width of the rectangle?

## Dimensions of a Rectangle:

When relating different areas of a rectangle which at the same time related to each other through the changes of their dimensions, we can use this expression to derive the original length and width measurements.

The dimensions of a rectangle are such that it's length is 9 inches more than its width. This can be written as

\begin{align} RecL&=RecW+ 9\ \text{in}\\[0.2cm] RecW&=RecW\\[0.2cm] RecA&=RecW \cdot RecL \end{align}.

Where {eq}RecL, \ RecW, \text{and} \ RecA {/eq} is the length, width, and area of the rectangle.

If the length were doubled and if the width were decreased by 4 inches, the area would be increased by {eq}110 \ \text{in}^2 {/eq}. This can be written as:

\begin{align} RecL_N&=2\left (RecW+ 9\ \text{in} \right ) && \left [ \text{New Length of the Rectangle} \right ]\\[0.2cm] RecW_N&=RecW- 4\ \text{in}&& \left [ \text{New Width of the Rectangle} \right ]\\[0.2cm] RecA_N&=RecW_N \cdot RecL_N\\[0.2cm] RecA_N&=RecA+110 \ \text{in}^2\\[0.2cm] RecA_N&=RecW \cdot RecL+110 \ \text{in}^2 && \left [ \text{New Area of the Rectangle} \right ] \end{align}

Where {eq}RecL_N, \ RecW_N, \text{and} \ RecA_N {/eq} is the new length and width of the rectangle.

Solving for the length and width dimensions, we have:

\begin{align} RecA_N&=RecW \cdot RecL+110 \ \text{in}^2 && \left [ \text{New Area of the Rectangle} \right ]\\[0.2cm] RecW_N \cdot RecL_N&=RecW \cdot RecL+110 \ \text{in}^2\\[0.2cm] \left (RecW- 4\ \text{in} \right )\cdot \left ( 2\left (RecW+ 9\ \text{in} \right ) \right )&=\left ( RecW \right )\cdot \left ( RecW+ 9\ \text{in} \right )+110 \ \text{in}^2\\[0.2cm] 2\left ( \left ( RecW \right )^2+ \left (5\ \text{in} \right )RecW-36\ \text{in}^2 \right )&=\left ( RecW \right )^2+\left (9\ \text{in} \right )RecW+110 \ \text{in}^2\\[0.2cm] 2\left ( RecW \right )^2+ \left (10\ \text{in} \right )RecW-72\ \text{in}^2&=\left ( RecW \right )^2+\left (9\ \text{in} \right )RecW+110 \ \text{in}^2\\[0.2cm] \left ( RecW \right )^2+\left (1\ \text{in} \right )RecW-182\ \text{in}^2&=0 && \left [ \text{Apply Factoring Method} \right ]\\[0.2cm] \left (RecW + 14 \ \text{in} \right )\left (RecW - 13 \ \text{in} \right )&=0\\[0.2cm] \end{align}

The roots are then:

\begin{align} RecW&=-14 \ \text{in} && \left [ \text{There should be no negative value for inches} \right ]\\[0.2cm] RecW&=13 \ \text{in} && \left [ \text{Use this since this is the only positive root} \right ] \end{align}

Therefore the length and width of the rectangle are the following:

\begin{align} RecL&=13 \ \text{in}+ 9\ \text{in}\\[0.2cm] RecL&=22 \ \text{in} && \left [ \text{Length of the rectangle} \right ]\\[0.2cm] RecW&=13 \ \text{in} && \left [ \text{Width of the rectangle} \right ]\\[0.2cm] \end{align}.