Chapter 8.7, Problem 31E

a.

To determine

For thenth rectangle in the pattern, what are the dimensions in terms of n ?

Answer to Problem 31E

The dimensions of nth rectangle in the pattern would be $n\text{and }n+1$ .

Explanation of Solution

Given:

The first four rectangles in a pattern are shown below:

  

Calculation:

Let n be the length of rectangle.

We can see that width of rectangle is one more than length, so width of nth rectangle in terms of n would be $n+1$ .

Therefore, the dimensions of nth rectangle in the pattern would be $n\text{and }n+1$ .

b.

To determine

To write:

A function for the area $A(n)$ of the nth rectangle.

Answer to Problem 31E

The function for the area $A(n)$ of the nth rectangle would be $A(n)=n^2+n$ .

Explanation of Solution

Given:

The first four rectangles in a pattern are shown below:

  

Calculation:

We know that area of rectangle is width times length.

Upon substituting the dimensions of nth rectangle (in terms of n ) in area formula, we will get:

  $\begin{array}{l}A(n)=n(n+1)\\ A(n)=n^2+n\end{array}$

Therefore, the function for the area of the nth rectangle would be $A(n)=n^2+n$ .

c.

To determine

To write:

A function for the perimeter $P(n)$ of the nth rectangle.

Answer to Problem 31E

The function for the perimeter $P(n)$ of the nth rectangle would be $P(n)=4n+2$ .

Explanation of Solution

Given:

The first four rectangles in a pattern are shown below:

  

Calculation:

We know that perimeter of rectangle is 2 times the sum of width and length of rectangle.

Upon substituting the dimensions of nth rectangle (in terms of n ) in perimeter formula, we will get:

  $\begin{array}{l}P(n)=2(n+n+1)\\ P(n)=2(2n+1)\\ P(n)=4n+2\end{array}$

Therefore, the perimeter $P(n)$ of the nth rectangle would be $P(n)=4n+2$ .

d.

To determine

To find:

The area and perimeter of the50th rectangle.

Answer to Problem 31E

The area of the 50th rectangle would be 2550 square units.

The perimeter of the 50th rectangle would be 202 units.

Explanation of Solution

Given:

The first four rectangles in a pattern are shown below:

  

Calculation:

To find the area and perimeter of the50th rectangle, we will substitute $n=50$ in area and perimeter functions as:

  $\begin{array}{l}A(n)=n^2+n\\ A(50)={50}^2+50\\ A(50)=2500+50\\ (50)=2550\end{array}$

Therefore, the areaof the 50th rectangle would be 2550 square units.

  $\begin{array}{l}P(n)=4n+2\\ P(50)=4(50)+2\\ P(50)=200+2\\ P(50)=202\end{array}$

Therefore, the perimeterof the 50th rectangle would be 202 units.