Chapter 12, Problem 6CR

a.

To determine

To calculate: The ratio of sides AB and DE .

Answer to Problem 6CR

The ratio is $3:5$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\Delta ABC\sim \Delta DEF\\ A\left(\Delta ABC\right)=9,\\ A\left(\Delta DEF\right)=25.\end{array}$

Formula used:

The below theorem is used:

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Calculation:

  

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

  $\begin{array}{l}\frac{A_1}{A_2}={\left(\frac{s_1}{s_2}\right)}^2\\ \frac{9}{25}={\left(\frac{s_1}{s_2}\right)}^2\\ \left(\frac{s_1}{s_2}\right)=\sqrt{\frac{9}{25}}\\ \left(\frac{s_1}{s_2}\right)=\frac{3}{5}\end{array}$

b.

To determine

To find: The perimeter ratio of triangle ABC and triangle DEF .

Answer to Problem 6CR

The ratio is $3:5$ .

Explanation of Solution

Given information:

  $\begin{array}{l}\Delta ABC\sim \Delta DEF\\ A\left(\Delta ABC\right)=9,\\ A\left(\Delta DEF\right)=25.\end{array}$

Formula used:

The below theorem is used:

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Calculation:

  

If two triangles are similar, then the ratio of the area of both triangles is proportional to the square of the ratio of their corresponding sides.

Perimeter is sum of all sides. Thus, the ratio of the area of both triangles is also proportional to their perimeters.

  $\begin{array}{l}\frac{A_1}{A_2}={\left(\frac{P_1}{P_2}\right)}^2\\ \frac{9}{25}={\left(\frac{P_1}{P_2}\right)}^2\\ \left(\frac{P_1}{P_2}\right)=\sqrt{\frac{9}{25}}\\ \left(\frac{P_1}{P_2}\right)=\frac{3}{5}\end{array}$