Chapter 9, Problem 27CT

(a)

To determine

To calculate:

The dimension of the rectangle to the nearest foot

Answer to Problem 27CT

The estimated dimension of the rectangle to the nearest foot is $7$ ft by $8$ ft.

Explanation of Solution

Given information:

Area of rectangle: $60{\text{ft}}^2$

Dimension of the rectangle: $x$ and $x+1$

Calculation:

  $x$ and $x+1$ are two consecutive numbers.

Area of rectangle:

  $A=W\times L$

Where:

  $A$ : Area of rectangle

  $W$ : Width of rectangle

  $L$ : Length of rectangle

The dimensions of the rectangle, the two consecutive integers that product is closest to $60$ .

The integers $7$ and $8$ are two integers,

  $\begin{array}{l}=7\times 8\\ =56\end{array}$

  $56$ is near to $60$ , $7$ and $8$ are consecutive numbers.

The estimated dimension of the rectangle is $7$ ft by $8$ ft.

(b)

To determine

To calculate:

The dimension to the nearest hundredth of the foot

Answer to Problem 27CT

The quadratic equation that models the situation is $x^2+x-60=0$ .

Explanation of Solution

Given information:

Area of rectangle: $60{\text{ft}}^2$

Dimension of the rectangle: $x$ and $x+1$

Area of the rectangle:

  $A=W\times L$

Where:

  $A$ : Area of rectangle

  $W$ : Width of rectangle

  $L$ : Length of rectangle

Substitute $60$ for $A$ , $x$ for the width and $x+1$ for length of rectangle in the formula for the area:

  $\begin{array}{l}A=W\times L\\ 60=x\left(x+1\right)\end{array}$

Then, simplify and form quadratic equation:

  $\begin{array}{l}x\left(x+1\right)=60\\ x^2+x=60\\ x^2+x-60=0\end{array}$

The quadratic equation that models the situation is $x^2+x-60=0$ .