Chapter 8, Problem 15STP

(a)

To determine

Find x .

Answer to Problem 15STP

10

Explanation of Solution

Given :

Calculation:

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

$$

$\begin{array}{l}\angle CAB\cong \angle CDA\\ \angle ACD\cong \angle ACB\end{array}$

So, by AA Similarity Postulate, $\Delta CAB\sim \Delta CDA$ .

Hence,

$\begin{array}{l}\frac{AD}{AB}=\frac{DC}{AC}=\frac{AC}{BC}\\ \Rightarrow \frac{x}{y}=\frac{12.5}{z}=\frac{z}{20.5}\\ \Rightarrow \frac{12.5}{z}=\frac{z}{20.5}\\ \Rightarrow z^2=(12.5)(20.5)\\ \Rightarrow z^2=256.25\\ \Rightarrow z\approx 16\end{array}$

Similarly , $\Delta BAC\sim \Delta BDA$

Hence,

$\begin{array}{l}\frac{AD}{CA}=\frac{BD}{BA}=\frac{BA}{BC}\\ \Rightarrow \frac{x}{z}=\frac{8}{y}=\frac{y}{20.5}\\ \Rightarrow \frac{8}{y}=\frac{y}{20.5}\\ \Rightarrow \frac{8}{y}=\frac{y}{20.5}\\ \Rightarrow y^2=8(20.5)\\ \Rightarrow y^2=164\\ \Rightarrow y\approx 12.8\end{array}$

Since ,

$\begin{array}{l}\frac{x}{z}=\frac{8}{y}\\ \Rightarrow x=z\frac{8}{y}\\ \Rightarrow x\approx 16\frac{8}{12.8}\\ \Rightarrow x=10\end{array}$

(b)

To determine

Find y .

Answer to Problem 15STP

12.8

Explanation of Solution

Given :

Calculation:

In $\Delta BAC\text{ and }\Delta BDA$ ,

$$

$\begin{array}{l}\angle BAC\cong \angle BDA\\ \angle B\cong \angle B\end{array}$

So, by AA Similarity Postulate, $\Delta BAC\sim \Delta BDA$ .

Hence,

$\begin{array}{l}\frac{AD}{CA}=\frac{BD}{BA}=\frac{BA}{BC}\\ \Rightarrow \frac{x}{z}=\frac{8}{y}=\frac{y}{20.5}\\ \Rightarrow \frac{8}{y}=\frac{y}{20.5}\\ \Rightarrow \frac{8}{y}=\frac{y}{20.5}\\ \Rightarrow y^2=8(20.5)\\ \Rightarrow y^2=164\\ \Rightarrow y\approx 12.8\end{array}$

(c)

To determine

Find z .

Answer to Problem 15STP

16

Explanation of Solution

Given :

Calculation:

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

$$

$\begin{array}{l}\angle CAB\cong \angle CDA\\ \angle ACD\cong \angle ACB\end{array}$

So, by AA Similarity Postulate, $\Delta CAB\sim \Delta CDA$ .

Hence,

$\begin{array}{l}\frac{AD}{AB}=\frac{DC}{AC}=\frac{AC}{BC}\\ \Rightarrow \frac{x}{y}=\frac{12.5}{z}=\frac{z}{20.5}\\ \Rightarrow \frac{12.5}{z}=\frac{z}{20.5}\\ \Rightarrow z^2=(12.5)(20.5)\\ \Rightarrow z^2=256.25\\ \Rightarrow z\approx 16\end{array}$