Chapter 8.3, Problem 43PPS

a.

To determine

To determine the possible values of $x$ .

Answer to Problem 43PPS

  $x>4$

Explanation of Solution

Given:

  

Calculation:

From the given figures,

  

If $x=4$ the width of the rectangular sandbox would be zero and if x < 4 the width of the rectangular sandbox would be negative.

So, the possible values of x are $x>4$ .

Conclusion:

Therefore, the possible values of $x>4$ .

b.

To determine

To determine the shape having greater area.

Answer to Problem 43PPS

square

Explanation of Solution

Given:

  

Formula used:

Area of square $=sid{e}^2$

Area of rectangle $=\text{length×breadth}$

Calculation:

Consider an example,

Suppose the value of $x=5$ [ $\because x>4$ ]

Area of the square $={(3)}^2=9$ square units

[ $\because$ Area of square $=sid{e}^2$ ]

Area of the rectangle $=5\times 1=5$ square units

[ $\because$ Area of rectangle $=\text{length×breadth}$ ]

The shape having greater area is always square.

So, taking any values of $x>4$ , area of square will be always greater than that of rectangle.

Conclusion:

Therefore, the shape having greater area is always square.

c.

To determine

To determine the difference in area of the two.

Answer to Problem 43PPS

  $4{\text{ ft}}^2$

Explanation of Solution

Given:

  

Formula used:

Area of rectangle $=\text{length×breadth}$

  ${(a+b)}^2=a^2+b^2+2ab$

Calculation:

From the first figure, the side is $=x-2$ .

Area of square $={(x-2)}^2$

[ $\because$ Area of square $=sid{e}^2$ ]

Area of square $=x^2+4-4x$

  $\left[\because {(a+b)}^2=a^2+b^2+2ab\right]$

Now consider second figure.

Here, length $=x$ and breadth $=x-4$

Area of rectangle $=x(x-4)$

[ $\because$ Area of square $=sid{e}^2$ ]

Area of rectangle $=x^2-4x$$\left[\because {(a+b)}^2=a^2+b^2+2ab\right]$

Now the difference in their area $=$ Area of square $-$ Area of rectangle

  $=$ $x^2+4-4x$ $-$ $x^2-4x$ $=$ $4$ ${\text{ft}}^2$

Conclusion:

Therefore, the difference in area of the two is $4$ ${\text{feet}}^2.$