Chapter 8.3, Problem 51PFA

a.

To determine

To write: An expression for the area of each square.

Answer to Problem 51PFA

Area of the square of side length $x$ is given by $x^2$

Area of square of side length $x+3$ is given by $x^2+6x+9$

Explanation of Solution

Given:

A square of side length $x$ and another square of side length $x+3$

Calculation:

The area of side is given by ${\left(\text{side}\right)}^2$

Hence, area of the square of side length $x$ is given by $x^2$

And the area of square of side length $x+3$ is given by

  $\begin{array}{l}{\left(x+3\right)}^2\\ =x^2+6x+9\end{array}$

b.

To determine

To find: The value of x

Answer to Problem 51PFA

  $x=1$

Explanation of Solution

Given:

The area of larger square is 15 square units more than the area of the smaller square.

Calculation:

From part (a)

Area of larger square is $x^2+6x+9$ and that smaller square is $x^2$

Therefore,

  $\begin{array}{l}{x}^2+6x+9=15+x\\ x^2+5x-6=0\\ \left(x+6\right)\left(x-1\right)=0\\ x=-6,1\end{array}$

Since, side length cannot ne negative. Hence, discard the value $x=-6$

Thus, $x=1$

c.

To determine

To find: The area of each square using the value of x .

Answer to Problem 51PFA

Area of the larger square is given by $16\text{ sq}\text{. units}$

Area of the smaller square is given by $1\text{sq}\text{. unit}$

Explanation of Solution

Given:

The value of x is 1

Calculation:

From part (a)

Area of larger square is $x^2+6x+9$ and that smaller square is $x^2$

Substituting $x=1$ in both of these expressions to find the area of each square.

Area of the larger square is given by

  $\begin{array}{l}{\left(1\right)}^2+6\left(1\right)+9\\ =16\text{ sq}\text{. units}\end{array}$

Area of the smaller square is given by

  $\begin{array}{l}{\left(1\right)}^2\\ =1\text{ sq}\text{. unit}\end{array}$