Chapter 12.7, Problem 19E

a.

To determine

To draw and calculate the perimeter of triangle ABC.

Answer to Problem 19E

The triangle is given below:

  

The perimeter of the triangle is $\text{12 units}$ .

the area of the triangle is $6\text{ square units}$

Explanation of Solution

Given information:

The vertices of triangle ABC are $\text{A}\left(0,-3\right)\text{,B}\left(3,1\right)\text{ and C}\left(3,-3\right)$

Calculation :

The triangle is given below:

  

The length of the sides of the triangle are:

  $\begin{array}{l}\text{AC}=3\text{ units}\\ \text{BC}=\text{4 units}\\ {\text{AB}}^2={\text{AC}}^2+{\text{BC}}^2\\ \text{AB}=\sqrt{9+16}=\sqrt{25}=5\text{ units}\end{array}$

The perimeter can be calculated by:

  $\text{AC}+\text{BC}+\text{AB}=\left(3+4+5\right)\text{ units }=\text{12 units}$

Hence, the perimeter of the triangle is $\text{12 units}$ .

The area of the triangle can be calculated by:

  $\begin{array}{l}\text{Area}=\frac{1}{2}\text{bh}\\ \text{Area}=\frac{1}{2}\cdot 4\cdot 3\\ \text{Area}=6\text{ square units}\end{array}$

Hence, the area of the triangle is $6\text{ square units}$

b.

To determine

To dilate and draw the triangle and find its area and perimeter

Answer to Problem 19E

The coordinates of the vertices after dilation after dilation are:

  $\text{A'}\left(0,-9\right)\text{,B'}\left(9,3\right)\text{ and C'}\left(9,-9\right)$

  

The perimeter of the image of the triangle is $36\text{ units}$ .

The area of the image of the triangle is $54\text{ square units}$ .

Explanation of Solution

Given information:

The vertices of triangle ABC are $\text{A}\left(0,-3\right)\text{,B}\left(3,1\right)\text{ and C}\left(3,-3\right)$

The vertices are to be dilated by a scale factor of $3$

Calculation :

The coordinates of the vertices after dilation are:

  $\begin{array}{l}\text{A}\left(0,-3\right)\to \text{A'}\left(0,-9\right)\left[\text{multiply the coordinates by 3}\right]\\ \text{B}\left(3,1\right)\to \text{B}\text{'}\left(9,3\right)\\ \text{C}\left(3,-3\right)\to \text{C'}\left(9,-9\right)\end{array}$

The coordinates of the vertices after dilation after dilation are:

  $\text{A'}\left(0,-9\right)\text{,B'}\left(9,3\right)\text{ and C'}\left(9,-9\right)$

The image of the triangle is drawn below:

  

The length of the sides of the triangle are:

  $\begin{array}{l}\text{A'C'}=9\text{ units}\\ \text{B'C'}=12\text{ units}\\ {\text{A'B'}}^2={\text{A'C'}}^2+{\text{B'C'}}^2\\ \text{A'B'}=\sqrt{81+144}=\sqrt{225}=15\text{ units}\end{array}$

The perimeter can be calculated by:

  $\text{A'C'}+\text{B'C'}+\text{A'B'}=\left(9+12+15\right)\text{ units }=36\text{ units}$

Hence, the perimeter of the image of the triangle is $36\text{ units}$ .

The area of the triangle can be calculated by:

  $\begin{array}{l}\text{Area}=\frac{1}{2}\text{bh}\\ \text{Area}=\frac{1}{2}\cdot 12\cdot 9\\ \text{Area}=54\text{ square units}\end{array}$

Hence, the area of the image of the triangle is $54\text{ square units}$ .

c.

To determine

To compare the ratio of the perimeter and the areas of both the triangles.

Answer to Problem 19E

The ratio of perimeter is the same as scale factor.

The ratio of the area is the square of the scale factor.

Explanation of Solution

Given information:

The perimeter of the triangle is $\text{12 units}$

The area of the triangle is $6\text{ square units}$

The perimeter of the image of the triangle is $36\text{ units}$

The area of the image of the triangle is $54\text{ square units}$

The image was dilated by a scale factor of $3$

Calculation :

The ratio of the perimeter of both triangles can be calculated by:

  $\begin{array}{l}\frac{\text{Perimeter of image of triangle ABC}}{\text{Perimeter of triangle ABC}}\\ =\frac{36}{12}\\ =3\end{array}$

Hence, the perimeter of the image of triangle ABC is $3$ times greater than the original triangle ABC.

The ratio of the area of both triangles can be calculated by:

  $\begin{array}{l}\frac{\text{Area of image of triangle ABC}}{\text{Area of triangle ABC}}\\ =\frac{54}{6}\\ =9\end{array}$

Hence, the area of the image of triangle ABC is $9$ times greater than the area of original triangle ABC.

The ratio of perimeter is the same as scale factor.

The ratio of the area is the square of the scale factor.