Chapter 1.5, Problem 8PSA

a.

To determine

To Evaluate: The value of $\overline{AB}$

Answer to Problem 8PSA

  $\overline{AB}=4\text{units}$

Explanation of Solution

Given:

  

B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

Concept Used:

Trisected points divides the line segment in three equal parts.

Calculation:

As here, B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

  $\begin{array}{l}\Rightarrow \overline{AB}=\overline{BC}=\overline{CD}=\frac{1}{3}\times \overline{AD}=\frac{1}{3}\times 12=4\\ \therefore \overline{AB}=4\text{units}\end{array}$

Conclusion:

  $\overline{AB}=4\text{units}$

b.

To determine

To Evaluate: The value of $\overline{AB}$

Answer to Problem 8PSA

  $\overline{AC}=8\text{units}$

Explanation of Solution

Given:

  

B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

Concept Used:

Trisected points divides the line segment in three equal parts.

Calculation:

As here, B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

  $\begin{array}{l}\Rightarrow \overline{AB}=\overline{BC}=\overline{CD}=\frac{1}{3}\times \overline{AD}=\frac{1}{3}\times 12=4\\ \Rightarrow \overline{AC}=\overline{AB}+\overline{BC}=4\text{units + }4\text{units = }8\text{units}\\ \therefore \overline{AC}=8\text{units}\end{array}$

Conclusion:

  $\overline{AC}=8\text{units}$

c.

To determine

To Find: The value of $x$

Answer to Problem 8PSA

  $x=1$

Explanation of Solution

Given:

  

B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$ and $\overline{AB}=x+3$

Concept Used:

Trisected points divides the line segment in three equal parts.

Calculation:

As here, B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

  $\begin{array}{l}\Rightarrow \overline{AB}=\overline{BC}=\overline{CD}=\frac{1}{3}\times \overline{AD}=\frac{1}{3}\times 12=4\\ \text{Also }\overline{AB}=x+3\\ \Rightarrow 4=x+3\\ \Rightarrow x=1\end{array}$

Conclusion:

  $x=1$

d.

To determine

To Find: The value of $\overline{AE}$ .

Answer to Problem 8PSA

  $\overline{AE}=9\text{units}$

Explanation of Solution

Given:

  

B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

  $\overline{AB}=x+3\text{and}\overline{AE}=3x+6$

Concept Used:

Trisected points divides the line segment in three equal parts.

Calculation:

As here, B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

  $\begin{array}{l}\Rightarrow \overline{AB}=\overline{BC}=\overline{CD}=\frac{1}{3}\times \overline{AD}=\frac{1}{3}\times 12=4\\ \text{Also }\overline{AB}=x+3\\ \Rightarrow 4=x+3\\ \Rightarrow x=1\\ \therefore \overline{AE}=3\times 1+6=9\text{units}\end{array}$

Conclusion:

  $\overline{AE}=9\text{units}$

e.

To determine

To Find: The segment for which the point C is the mid-point.

Answer to Problem 8PSA

  $BD$

Explanation of Solution

Given:

  

B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

Concept Used:

Trisected points divides the line segment in three equal parts.

Calculation:

As here, B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

  $\begin{array}{l}\Rightarrow \overline{AB}=\overline{BC}=\overline{CD}=\frac{1}{3}\times \overline{AD}=\frac{1}{3}\times 12=4\\ \because \overline{BC}=\overline{CD}=4\text{units}\\ \Rightarrow \text{C is the mid-point of the segment BD}\text{.}\end{array}$

Conclusion:

  $\text{C is the mid-point of the segment BD}\text{.}$

f.

To determine

To Find: $\text{Whether the }\overrightarrow{EB}\text{ and }\overrightarrow{EC}\text{trisect }\angle AED$ ?

Answer to Problem 8PSA

  $\text{Yes, the }\overrightarrow{EB}\text{ and }\overrightarrow{EC}\text{trisect }\angle AED.$

Explanation of Solution

Given:

  

B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

Concept Used:

Trisected points divides the line segment in three equal parts.

Calculation:

As here, B and C are trisection point of $\overline{AD}$ , where $\overline{AD}=12$

  $\begin{array}{l}\Rightarrow \overline{AB}=\overline{BC}=\overline{CD}=\frac{1}{3}\times \overline{AD}=\frac{1}{3}\times 12=4\text{units}\\ \Rightarrow \text{B is the mid-point of the segment AC}\text{.}\\ \Rightarrow \text{C is the mid-point of the segment BD}\text{.}\end{array}$

Line joining the points B and C trisect the angle AED

  $\text{Thus, the }\overrightarrow{EB}\text{ and }\overrightarrow{EC}\text{trisect }\angle AED.$

Conclusion:

  $\text{Thus, the }\overrightarrow{EB}\text{ and }\overrightarrow{EC}\text{trisect }\angle AED.$