The sides of a rectangle were 5cm { \times } 9cm. When both dimensions were increased by an...

Question:

The sides of a rectangle were 5cm {eq}\times {/eq} 9cm. When both dimensions were increased by an equal amounts the area of the rectangle increased by {eq}120 cm^2 {/eq} find the dimensions of the new rectangle.

Change in Dimensions of a Rectangle:

The new dimensions of the rectangle can be computed using the concept of area and factoring. With this, we should remember that the factors can yield both positive and negative numbers. Nevertheless, since we cannot consider everything since we are dealing with real-life applications.

Answer and Explanation:

Given:

The sides of a rectangle were {eq}Rectangle_{area}=\left (5 \ \text{cm} \right ) \left (9 \ \text{cm} \right ) {/eq}. Since the width is usually shorter than the length of a rectangle, the measurement can be expressed as:

$$\begin{align} Rectangle_{area}&=\left (R_W \right ) \left (R_L \right )\\[0.2cm] Rectangle_{area}&=\left (5 \ \text{cm} \right ) \left (9 \ \text{cm} \right ) \end{align} $$

Where {eq}R_W \ \text{and} \ R_L {/eq} represents the width and length of the rectangle, respectively.

When both dimensions were increased by an equal amount the area of the rectangle increased by {eq}120 cm^2 {/eq}.

This can be presented wherein the new width is valued at: {eq}R_{new-W}=R_W + x {/eq} and the new length valued at: {eq}R_{new-L}=R_L + x {/eq}

Therefore:

$$\begin{align} Rectangle_{newarea}&=\left (R_{new-W} \right ) \left (R_{new-L} \right )\\[0.2cm] Rectangle_{newarea}&=\left (R_W + x \right ) \left (R_L + x \right )\\[0.2cm] 120 \ \text{cm}^2&=\left (5 \ \text{cm}+x \right ) \left (9 \ \text{cm}+x \right ) \\[0.2cm] 120 \ \text{cm}^2&=45 \ \text{cm}^2+ \left (14\ \text{cm} \right )x + x^2\\[0.2cm] x^2+ \left (14\ \text{cm} \right )x- 75 \ \text{cm}^2&=0 && \left [ Quadratic Expression 1 \right ] \end{align} $$

To solve for the roots of the quadratic equation:

$$\displaystyle Roots_{x}=\frac{-b\pm \sqrt{b^{2}-4ac} }{2a} $$

The main quadratic equation formula is:

$$\displaystyle a\left ( x \right )^{2} +b\left ( x \right )+c=0 $$

Comparing this to the quadratic expression 1, we have: {eq}a=1, \ b=14, \ \text{and} \ c=-75 {/eq}

Therefore, we have:

$$\begin{align} \displaystyle Roots_{x}&=\frac{-b\pm \sqrt{b^{2}-4ac} }{2a} \ \text{cm}\\[0.2cm] \displaystyle Roots_{x}&=\frac{-\left (14 \right )\pm \sqrt{\left (14 \right )^{2}-4\left (1 \right )\left (-75 \right )} }{2\left (1 \right )} \ \text{cm}&& \left [ \text{Substitute the values for a, b, and c} \right ]\\[0.2cm] \displaystyle Roots_{x}&=\frac{-14\pm \sqrt{196+300}}{2} \ \text{cm}\\[0.2cm] \displaystyle Roots_{x}&=\frac{-14\pm 4\sqrt{31}}{2} \ \text{cm}\\[0.2cm] \displaystyle Roots_{x}&=-7\pm 2\sqrt{31} \ \text{cm}\\[0.2cm] \displaystyle Roots_{+x}&=-7+ 2\sqrt{31} \ \text{cm} && \left [ \text{Addition Condition} \right ]\\[0.2cm] \displaystyle Roots_{+x}&\approx4.14\ \text{cm}\\[0.2cm] \displaystyle Roots_{-x}&=-7- 2\sqrt{31} \ \text{cm}&& \left [ \text{Minus Condition} \right ]\\[0.2cm] \displaystyle Roots_{-x}&\approx-18.14\ \text{cm}\\[0.2cm] \end{align} $$

We will only select the addition (+) condition of: {eq}\displaystyle Roots_{+x}=-7+ 2\sqrt{31} \ \text{cm}\approx4.14\ \text{cm} {/eq}.

The new dimensions for the width:

$$\begin{align} R_{new-W}&=R_W + x\\ R_{new-W}&=5 \ \text{cm} +\left (-7+ 2\sqrt{31} \right ) \ \text{cm}\approx9.14 \ \text{cm} \end{align} $$

The new dimensions for the length:

$$\begin{align} R_{new-L}&=R_L + x\\ R_{new-L}&=9 \ \text{cm} +\left (-7+ 2\sqrt{31} \right ) \ \text{cm}\approx13.14 \ \text{cm} \end{align} $$

Therefore, the new dimensions for the width and length of the rectangle are given by {eq}\approx13.14 \ \text{cm} \ \text{and} \ \approx9.14 \ \text{cm} {/eq}, respectively.


Learn more about this topic:

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Measuring the Area of a Rectangle: Formula & Examples

from Geometry: High School

Chapter 8 / Lesson 7
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