Chapter 7.7, Problem 46E

a.

To determine

The regions representing $a^2-b^2$ .

Answer to Problem 46E

Required figures and results are shown below.

Explanation of Solution

Given Information:Figure shows a large square with area $a^2$ and a smaller square with area $b^2$.

Calculation:

  $a^2-b^2=$ Area of larger square-Area of smaller square

  $\begin{array}{lllll}z\hfill & \hfill & \hfill & u\hfill & \hfill \\ y\hfill & \hfill & \hfill & x\hfill & \hfill \end{array}$

The area labeled by $x,u,z$ (excluding $y$ ) shows the region $a^2-b^2$ .

  $\begin{array}{l}\begin{array}{lll}z\hfill & \hfill & u\hfill \end{array}\\ \begin{array}{c}\begin{array}{}x\hfill \end{array}\end{array}\end{array}$  $zux$

Now, if the region $x$ is shifted to the right of region $u$ , then the new region is:

Clearly, area of this region is $(a+b)(a-b)$ .

Thus, $a^2-b^2=(a+b)(a-b)$

b.

To determine

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

Answer to Problem 46E

Required figures and results are shown below.

Explanation of Solution

Given Information:Figure shows a large square with area $a^2$ and a smaller square with area $b^2$.

Calculation:

  ${(a-b)}^2$ =Area of region $u$

  $a^2-2ab+b^2=a^2-ab-ab+b^2$

=Area of region $(u+x+y+z)$ -Area of region $(x+y)$ -Area of region $(y+z)$ +Area of region $y$

=Area of region $u$

Hence proved.