Chapter A.3, Problem 106E

(a)

To determine

Describe the region that represents $a^2-b^2$

Answer to Problem 106E

The region is shown in the figure below

Explanation of Solution

Given:

  

The area of the shaded region $a^2-b^2$ can be expressed as the sum of 3 rectangles from the above figure. That is

  $\begin{array}{l}{a}^2-b^2={(a-b)}^2+2(a-b)b\hfill \\ a^2-b^2=(a-b)(a-b+2b)\hfill \\ a^2-b^2=(a-b)(a+b)\hfill \end{array}$

The area of the inner square with sides $a-b$ is ${(a-b)}^2$ .This area is equal to

  $\begin{array}{l}{(a-b)}^2=a^2-2b(a-b)-b^2\hfill \\ {(a-b)}^2=a^2-2ab+2b^2-b^2\hfill \\ {(a-b)}^2=a^2-2ab+b^2\hfill \end{array}$

The proof is complete.

(b)

To determine

Use the figure to show that ${(a-b)}^2=a^2-2ab+b^2$

Answer to Problem 106E

The prove is given below.

Explanation of Solution

Given:

  

Area of the rectangle on the upper left sides ${(a-b)}^2$ , area of smaller square is $b^2$ and the area of the rectangle on the lower right sides is ${(a-b)}^2$

Therefore, from the above figure, note that after subtracting all the areas from the area of the larger square to get the desired result.

  $\begin{array}{cc}{(a-b)}^2& =a^2-b^2-(a-b)b-(a-b)b\\ & =a^2-b^2-ab+b^2-ab+b^2\\ & =a^2+b^2-2ab\end{array}$

The proof is complete.

(a)

To determine

Use the figure to show that ${(a+b)}^2=a^2+2ab+b^2$

Answer to Problem 106E

The prove is given below.

Explanation of Solution

Given:

  

Total area of the above square ${(a+b)}^2$ , is the sum of one 2 squares and 2 rectangles.

  $\begin{array}{l}{(a+b)}^2=a^2+b^2+ab+ab\hfill \\ {(a+b)}^2=a^2+2ab+b^2\hfill \end{array}$

The proof is complete.