Chapter 1.3, Problem 70PS

Solution

To find: New dimension of rectangular deck.

The new dimension of deck is $15\text{ feet long by 14 feet wide}\text{.}$

Given:

A rectangular deck for recreation measures 21 feet long by 20 feet wide.

Concept Used:

Factoring method to find the roots of the quadratic equation.

Calculation:

Length of rectangle $=21\text{feet}$ .

Width of rectangle $=20\text{feet}$ .

  $\text{Area of rectangle}=\text{length }\times \text{ width}$ .

  $\begin{array}{l}\text{Area of rectangle}=21\times 20\\ =420{\text{feet}}^2\end{array}$

Now, half the area $=420\times \frac{1}{2}{\text{feet}}^2=210{\text{feet}}^2$

After subtracting the same distance of $x$ to the length and the width:

New length of rectangle is $\left(21-x\right)\text{feet}$ .

New width of rectangle is $\left(20-x\right)\text{feet}$ .

Here, Applying the area of triangle formula,

  $\therefore \left(21-x\right)\left(20-x\right)=210$

Solving this equation:

  $\begin{array}{c}\left(21-x\right)\left(20-x\right)=210\\ 21\left(20-x\right)-x\left(20-x\right)=210\\ 420-21x-20x+x^2=210\\ x^2-41x+420-210=0\\ x^2-41x+210=0\end{array}$

Applying the factoring method of quadratic equation:

  $\begin{array}{c}{x}^2-41x+210=0\\ x^2-35x-6x+210=0\\ x\left(x-35\right)-6\left(x-35\right)=0\\ \left(x-35\right)\left(x-6\right)=0\end{array}$

Using Zero Product property:

  $\begin{array}{c}x-35=0\text{or}x-6=0\\ x=35\text{or}x=6\end{array}$

Here, Neglecting the value of $x=35$ As putting in the value of length and width it makes them negative and it is known that distance cannot be negative.

Therefore, The value of $x\text{ is 6}\text{.}$

Deck’s new Dimension:

  $\begin{array}{c}\text{length is }21-x=21-6\\ =15\text{ feet}\\ \text{width is }20-x=20-6\\ =14\text{ feet}\end{array}$

Conclusion:

The new dimension of deck is $15\text{ feet long by 14 feet wide}\text{.}$