Chapter 13.6, Problem 38PPSE

Solution

a.

The area of the patio in square feet.

The area of the patio is 140.4 ft².

Given:

A patio in the shape of an equilateral triangle with the base 18 ft and height 15.6 ft.

Concept used:

Area of the equilateral triangle is $A=\frac{1}{2}bh$

Where, b is the side and h is the height of the triangle.

Calculation:

Area of an equilateral triangle is $A=\frac{1}{2}bh$

On substituting the values of b and h in the formula of area,

  $\begin{array}{c}A=\frac{1}{2}\times 18\times 15.6\\ A=140.4{\text{ ft}}^{\text{2}}\end{array}$

Thus, the area of the patio is 140.4 ft².

Conclusion:

The area of the patio is 140.4 ft².

b.

The graph of the function $y=3\tan \left(\theta \right)$ and the height of the tile.

The graph of the function $y=3\tan \left(\theta \right)$ is drawn.

When $\theta =30°$ , the height of the tile is 1.732 in and when $\theta =60°$ , the height of the tile is 5.2 in.

Given:

The tile in the shape of isosceles triangle with the base 6 in.

The height of the triangular tile is:

  $y=3\tan \left(\theta \right)$

Calculation:

The graph of the function $y=3\tan \left(\theta \right)$ on a graphing calculator is shown below:

  

As per the given problem,

When $\theta =30°$

Substitute the value of $\theta$ in the given function of height $y=3\tan \left(\theta \right)$ , we get

  $\begin{array}{c}y=3\tan \left(30°\right)\\ =\frac{3}{\sqrt{3}}\\ y=\sqrt{3}\\ y\simeq 1.732\end{array}$

Thus, the height of the tile is 1.732 in when $\theta =30°$ .

When $\theta =60°$

Substitute the value of $\theta$ in the given function of height $y=3\tan \left(\theta \right)$ , we get

  $\begin{array}{c}y=3\tan \left(60°\right)\\ =3\sqrt{3}\\ y=3\times 1.732\\ y\simeq 5.196\end{array}$

Thus, the height of the tile is 5.2 in when $\theta =60°$ .

Conclusion:

When $\theta =30°$ , the height of the tile is 1.732 in and when $\theta =60°$ , the height of the tile is 5.2 in.

c.

The area of one tile in square inches.

When $\theta =30°$ , the area of the tile is 5.1 square inches and when $\theta =60°$ , the area of the tile is 15.6 square inches.

Given:

The base of the tile is 6 in.

From part (b), when $\theta =30°$ , the height of the tile is 1.7 in and when $\theta =60°$ , the height of the tile is 5.2 in.

Concept used:

Area of the isosceles triangle is $A=\frac{1}{2}bh$

Where, b is the side and h is the height of the triangle.

Calculation:

Area of an isosceles triangle is $A=\frac{1}{2}bh$

When $\theta =30°$ , $b=6\text{ in and }h=1.7\text{ in}$

On substituting the values of b and h in the formula of area,

A $\begin{array}{c}A=\frac{1}{2}\times 6\times 1.7\\ A=5.1{\text{ in}}^{\text{2}}\end{array}$

Thus, the area of the tile is 5.1 in².

When $\theta =60°$ , $b=6\text{ in and }h=5.2\text{ in}$

On substituting the values of b and h in the formula of area,

  $\begin{array}{c}A=\frac{1}{2}\times 6\times 5.2\\ A=15.6{\text{ in}}^{\text{2}}\end{array}$

Thus, the area of the tile is 15.6 in².

Conclusion:

When $\theta =30°$ , the area of the tile is 5.1 square inches and when $\theta =60°$ , the area of the tile is 15.6 square inches.

d.

The number of tiles in the patio.

When $\theta =30°$ , the number of tiles required 3964 tiles and when $\theta =60°$ , the number of tiles required 1296 tiles.

Given:

The base of the tile is 6 in.

From part (b), when $\theta =30°$ , the height of the tile is 1.7 in and when $\theta =60°$ , the height of the tile is 5.2 in.

Concept used:

Conversion of ft to in, $1\text{ ft}=12\text{ in}$

To get the number of tiles, the following formula will be used:

  $\text{Number of tiles}=\frac{\text{Area of patio}}{\text{Area of one tile}}$

Calculation:

From part (a), the area of the patio is 140.4 ft². Now, converting the area in square inches.

Since $1\text{ ft}=12\text{ in}$ , then

140.4 ft² will be $(140.4\times 144){\text{ in}}^2$

Thus, the area of the patio is 20217.6 in².

From part (c), the area of a tile when $\theta =30°$ is 5.1 in²

So, the number of tiles will be:

  $\begin{array}{l}\frac{20217.6}{5.1}\\ \simeq 3964\text{ tiles}\end{array}$

Again, from part (c), the area of a tile when $\theta =60°$ is 15.6 in²

So, the number of tiles will be:

  $\begin{array}{l}\frac{20217.6}{15.6}\\ \simeq 1296\text{ tiles}\end{array}$

Conclusion:

When $\theta =30°$ , the number of tiles required 3964 and when $\theta =60°$ , the number of tiles required 1296.