Chapter 1.6, Problem 16P

To determine

To translate: each of the two given problem into an equation and also to find the dimensions of the rectangle given its area.

Answer to Problem 16P

The dimensions of the rectangle are $l=6\text{units},w=5\text{units}$

Explanation of Solution

Given Information:

The statements

(1) Length of a rectangle is one unit more than its width

(2) Area of rectangle is 30 square units.

The given statements express the relation between length and width of a rectangle. Also, the area of the rectangle is given.

Let the length of the rectangle be denoted by l and width be denoted by w . The area is given by the expression,

  $A=l\times w$

1. The first statement relates the length and breadth which can be expressed as,

  $l=w+1$

2. The second statement gives the area of the rectangle as,

  $l\times w=30$

It is expressed in the following figure.

  

Now, substituting for length of the rectangle in terms of width , in equation under 2., the new equation becomes,

  $\begin{array}{l}\left(w+1\right)\times w=30\\ w^2+w=30\\ w^2+w-30=0\end{array}$

Solving the above quadratic equation leads to width of the rectangle as,

  $w=-6,5$

The valid solution is $w=5\text{ units}$

Using this value, the length of the rectangle can be obtained as,

  $\begin{array}{c}l=w+1\\ =5+1\text{ units}\\ \text{= 6 units}\end{array}$