# The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 5...

## Question:

The length of a rectangle is 1 cm longer than its width. If the diagonal of the rectangle is 5 cm, what are the dimensions of the rectangle in centimeters?

## Dimensions of a Rectangle:

Lengths and widths make a rectangle. Sometimes a length is partly a measurement of unknown constant width. In this case, we can express the length as a measurement of width. With this, we can compute for different roots of the width once we relate it to the area.

## Answer and Explanation:

The length of a rectangle is 1 cm longer than its width.

The length of the rectangle can be expressed as $$Rect_{length}=Rect_{width}+1 \ \text{cm} \quad \left [ \text{length of the rectangle} \right ] $$ and the width of the rectangle is $$Rect_{width} \quad \left [ \text{width of the rectangle} \right ] $$.

The diagonal of the rectangle is 5 cm which can be expressed as the hypotenuse of the rectangle.

Therefore, we can apply the Pythagorean theorem:

$$\begin{align} \left (Rect_{hypotenuse} \right )^{2}&=\left (Rect_{width} \right )^{2}+\left (Rect_{length} \right )^{2} && \left [ \text{Pythagorean Theorem} \right ]\\[0.2cm] \left (5 \ \text{cm} \right )^{2}&=\left (Rect_{width} \right )^{2}+\left (Rect_{width}+1 \ \text{cm} \right )^{2} && \left [ \text{Substitute the equivalence of the length of rectangle to simplify} \right ]\\[0.2cm] 25\ \text{cm}^{2}&=\left (Rect_{width} \right )^{2}+\left (Rect_{width} \right )^{2}+2 \ \text{cm}\left (Rect_{width} \right )+1\ \text{cm}^{2}\\[0.2cm] 0&=2 \ \text{cm}\left (Rect_{width} \right )^{2}+2 \ \text{cm}\left (Rect_{width} \right )-25\ \text{cm}^{2} && \left [ \text{Quadratic expression 1} \right ]\\[0.2cm] \end{align} $$

To solve for the roots of the quadratic equation:

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

Where the main quadratic equation formula is:

$$\displaystyle a\left ( Rect_{width} \right )^{2} +b \ \text{cm}\left ( Rect_{width} \right )+c \ \text{cm}^2=0 $$

Therefore, we have:

$$\begin{align} \displaystyle Roots_{Rect_{width}}&=\frac{-b\pm \sqrt{b^{2}-4ac} }{2a} \ \text{cm}\\[0.2cm] \displaystyle Roots_{Rect_{width}}&=\frac{-2\pm \sqrt{\left (2 \right )^{2}-4\left (2 \right )\left (-25 \right )} }{2\left (2 \right )} \ \text{cm} && \left [ \text{Substitute the values for a, b, and c from the quadratic expression 1} \right ]\\[0.2cm] \displaystyle Roots_{Rect_{width}}&=\frac{-2\pm \sqrt{4+200}}{4} \ \text{cm}\\[0.2cm] \displaystyle Roots_{Rect_{width}}&=\frac{-2\pm \sqrt{204}}{4} \ \text{cm}\\[0.2cm] \displaystyle Roots_{+Rect_{width}}&=\frac{-2+ \sqrt{204}}{4} \ \text{cm} && \left [ \text{Addition Condition} \right ]\\[0.2cm] \displaystyle Roots_{+Rect_{width}}&=\frac{-2+ 2\sqrt{51}}{4}\ \text{cm}\\[0.2cm] \displaystyle Roots_{+Rect_{width}}&=\frac{-1+ \sqrt{51}}{2}\ \text{cm}\\[0.2cm] \displaystyle Roots_{+Rect_{width}}&\approx3.1\ \text{cm}\\[0.2cm] \displaystyle Roots_{-Rect_{width}}&=\frac{-2- \sqrt{204}}{4} \ \text{cm} && \left [ \text{Minus Condition} \right ]\\[0.2cm] \displaystyle Roots_{-Rect_{width}}&=\frac{-2- 2\sqrt{51}}{4}\ \text{cm}\\[0.2cm] \displaystyle Roots_{-Rect_{width}}&=\frac{-1- \sqrt{51}}{2}\ \text{cm}\\[0.2cm] \displaystyle Roots_{-Rect_{width}}&\approx-4.1\ \text{cm} \end{align} $$

Select the addition (+) condition since it produces a positive value for the rectangle of: {eq}\displaystyle Roots_{+Rect_{width}}=\frac{-1+ \sqrt{51}}{2}\ \text{cm}\approx3.1 \ \text{cm} {/eq}. Thus, the length of a rectangle is 1 cm longer than its width is expressed as {eq}\displaystyle Rect_{length}=Rect_{width}+1 \ \text{cm}=\frac{-1+ \sqrt{51}}{2}\ \text{cm}=1 \ \text{cm}\approx4.1 \ \text{cm} {/eq}.

Therefore, the dimensions of the rectangle are given by {eq}\approx3.1 \ \text{cm} \ \text{and} \ \approx4.1 \ \text{cm} {/eq}, for the width and length, respectively.

#### Learn more about this topic:

from Geometry: High School

Chapter 8 / Lesson 7