Chapter 7.2, Problem 49PPS

(a)

To determine

To Prove: The proof that the given triangles have the same ratio of the side and the areas.

Explanation of Solution

Given:

The given triangle is shown in Figure 1

  

Figure 1

Calculation:

Consider the ratio of the sides of the two right triangles is,

  $\begin{array}{c}\frac{GH}{YZ}=\frac{a}{3a}\\ =\frac{1}{3}\end{array}$

Also,

  $\begin{array}{c}\frac{HF}{ZX}=\frac{b}{3b}\\ =\frac{1}{3}\end{array}$

Also,

  $\begin{array}{c}\frac{GF}{YX}=\frac{c}{3c}\\ =\frac{1}{3}\end{array}$

Then, then ratio of the perimeter of the two triangles is,

  $\begin{array}{c}\frac{P_{GHF}}{P_{\Delta YZX}}=\frac{GF+HF+FG}{YZ+ZX+XY}\\ =\frac{a+b+c}{3a+3b+3c}\\ =\frac{1}{3}\end{array}$

Hence, proved.

(b)

To determine

To find: Whether the area of the new triangle is similar when the 6 units are added to both the sides.

Answer to Problem 49PPS

The new triangle is not similar as both the sides are not proportional.

Explanation of Solution

Consider the ratio of the sides of the two right triangles is,

  $\begin{array}{c}\frac{GH}{YZ}=\frac{a+6}{3a+6}\\ =\frac{a+6}{3\left(a+2\right)}\end{array}$

Also,

  $\begin{array}{c}\frac{HF}{ZX}=\frac{b+6}{3b+6}\\ =\frac{b+3}{2\left(b+2\right)}\end{array}$

Also,

  $\begin{array}{c}\frac{GF}{YX}=\frac{c+6}{3c+6}\\ =\frac{c+6}{3\left(c+2\right)}\end{array}$

Thus, the new triangle is not similar as both the sides are not proportional.