Chapter 7.4, Problem 14WE

(a)

To determine

To Name: two triangles that are similar to $\Delta ABC$ in the figure shown.

Answer to Problem 14WE

  $\Delta ABC\sim \Delta ADC$ and $\Delta ABC\sim \Delta BDC$ .

Explanation of Solution

Given:

Figure is shown as.

  

Calculation:

In $\Delta ABC\text{and }\Delta ADC$ .

  $\begin{array}{l}\angle A=\angle A\therefore \left(\text{Common}\right)\\ \angle ACB=\angle ADC\therefore \left(\text{Right angle}\right)\end{array}$

Therefore, by AA postulate $\Delta ABC\sim \Delta ADC$.

In $\Delta ABC\text{and }\Delta BDC$ .

  $\begin{array}{l}\angle B=\angle B\therefore \left(\text{Common}\right)\\ \angle ABC=\angle BDC\therefore \left(\text{Right angle}\right)\end{array}$

Therefore, by AA postulate $\Delta ABC\sim \Delta BDC$.

therefore, the answer is $\Delta ABC\sim \Delta ADC$ and $\Delta ABC\sim \Delta BDC$ .

(c)

To determine

To Find: the values of $x\text{ and }y$ in the figure shown.

Answer to Problem 14WE

  $33.33$ and $9$ .

Explanation of Solution

Given:

Figure is shown as.

  

Calculation:

In $\Delta ABC\text{and }\Delta ADC$ .

  $\begin{array}{l}\angle A=\angle A\therefore \left(\text{Common}\right)\\ \angle ACB=\angle ADC\therefore \left(\text{Right angle}\right)\end{array}$

Therefore, by AA postulate $\Delta ABC\sim \Delta ADC$.

In a two similar triangle, the ratio of their corresponding sides is equal.

  $\begin{array}{l}\frac{AB}{BD}=\frac{BC}{DC}\\ \frac{16+y}{y}=\frac{x}{12}\\ \frac{16+9}{9}=\frac{x}{12}\\ \frac{25}{9}=\frac{x}{12}\\ x=\frac{25\times 12}{9}\\ x=33.33\end{array}$

  $\begin{array}{l}\frac{AB}{AC}=\frac{AC}{AD}\\ \frac{16+y}{20}=\frac{20}{16}\\ \frac{16+y}{20}=\frac{5}{4}\\ 16+y=25\\ y=9\end{array}$

In $\Delta ABC\text{and }\Delta BDC$ .

  $\begin{array}{l}\angle B=\angle B\therefore \left(\text{Common}\right)\\ \angle ABC=\angle BDC\therefore \left(\text{Right angle}\right)\end{array}$

Therefore, by AA postulate $\Delta ABC\sim \Delta BDC$.

In a two similar triangle, the ratio of their corresponding sides is equal.

  $\begin{array}{l}\frac{AB}{BD}=\frac{BC}{DC}\\ \frac{16+y}{y}=\frac{x}{12}\\ \frac{16+9}{9}=\frac{x}{12}\\ \frac{25}{9}=\frac{x}{12}\\ x=\frac{25\times 12}{9}\\ x=33.33\end{array}$

therefore, the answer is $33.33$ and $9$ .