Chapter 8.6, Problem 37PPE

a.

To determine

To calculate: Area of both rectangle is same or different.

Answer to Problem 37PPE

Area of the first rectangle is $A_1=\left(x+2\right)\left(2x+2\right)=2x^2+6x+4$

Area of the second rectangle is $A_2=\left(2x+4\right)\left(x+1\right)=2x^2+6x+4$

Explanation of Solution

Given:length and breadth of both the rectangle is given.

calculation: length and breadth of first rectangle is $\left(2x+2\right)$ and $\left(x+2\right)$ while the length and breadth of second rectangle is $\left(2x+4\right)$ and $\left(x+1\right)$

Area of the rectangle is given by area= length multiplied by breadth

Therefore, Area of the first rectangle is $A_1=\left(x+2\right)\left(2x+2\right)=2x^2+6x+4$

And Area of the second rectangle is $A_2=\left(2x+4\right)\left(x+1\right)=2x^2+6x+4$

b.

To determine

To show:howboth the rectangle is same

Explanation of Solution

Given:length and breadth of both the rectangle is given.

Verification: From the above calculation we can observe that the area of both the rectangle is same because after finding the area the expression is same and if we consider any value of x

the result will be same not mater the dimension is different.

Suppose $x=0$ ,

Therefore $\begin{array}{l}{A}_1=2x^2+6x+4=4\\ A_2=2x^2+6x+4=4\end{array}$

Hence Area of both rectangle is same.