The length of the rectangle is 5cm less than thrice its width. Find the dimensions of the rectangle if its area is 112cm²?

Sep 6, 2015

Length: $\text{16 cm}$
Width: $\text{7 cm}$

Explanation:

First, start by writing the formula for the area of a rectangle of width $w$ and length $l$

$\textcolor{b l u e}{A = l \cdot w}$

Now, you know that if you triple the rectangle's width and subtract 5 cm from the result, you get the rectangle's length.

This means that you can write

$l = 3 \cdot w - 5$

Since you know that the area of the rectangle is equal to ${\text{112 cm}}^{3}$, you can write a second equation using $l$ and $w$

$\left(3 w - 5\right) \cdot w = 112$

$3 {w}^{2} - 5 w = 112$

$3 {w}^{2} - 5 w - 112 = 0$

Use the quadratic formula to find the two solutions to this quadratic equation

${w}_{1 , 2} = \left(\left(- 5\right)\right) \pm \frac{\sqrt{{\left(- 5\right)}^{2} - 4 \cdot 3 \cdot \left(- 112\right)}}{2 \cdot 3}$

${w}_{1 , 2} = \frac{5 \pm \sqrt{1369}}{6}$

${w}_{1 , 2} = \frac{5 \pm 37}{6}$

Since $w$ represents the width of the rectangle, the negative solution will have no physical significance. This means that the only valid solution to this quadratic is

$w = \frac{5 + 37}{6} = \frac{42}{6} = \textcolor{g r e e n}{\text{7 cm}}$

The length of the rectangle will be

$3 \cdot 7 - 5 = 21 - 5 = \textcolor{g r e e n}{\text{16 cm}}$