Chapter 13.2, Problem 27WE

(a)

To determine

To find: that $\Delta OAB\cong \Delta ORS$ .

Answer to Problem 27WE

The $\Delta OAB$ is congruent to $\Delta ORS$ .

Explanation of Solution

Given information:

The given figure is

  

Figure $\left(1\right)$ .

Calculation:

Consider the figure $\left(1\right)$ ,

In $\Delta OAB$ and $\Delta ORS$

  $\begin{array}{c}\overline{OA}=\overline{OR}\\ \overline{AB}=\overline{RS}\\ \angle A=\angle R\\ \Delta OAB\cong \Delta ORS\end{array}$

Therefore the $\Delta OAB$ is congruent to $\Delta ORS$ .

(b)

To determine

To find: That $\overline{OB}\perp \overline{OS}$ .

Answer to Problem 27WE

The line $\overline{OB}$ is perpendicular to the line $\overline{OS}$ .

Explanation of Solution

Given information:

The given figure is

  

Figure $\left(1\right)$ .

Calculation:

Consider the given figure,

Assume that

  $\angle BOA={\text{n}}^{\text{o}}$

Then, $\angle ROB$

  $\angle ROB={90}^{\text{o}}-{\text{n}}^{\text{o}}$

The $\Delta OAB$ and $\Delta ORS$ are congruent.

So,

  $\angle BOE=\angle FNE\left(=n^{\text{o}}\right)$

Calculate the angle $\angle SOB$ ,

  $\begin{array}{l}\angle SOB=\angle ROS+\angle ROB\\ \angle SOB=n^{\text{o}}+{90}^{\text{o}}-n^{\text{o}}\\ \angle SOB={90}^{\text{o}}\end{array}$

Therefore, the line $\overline{OB}$ is perpendicular to the line $\overline{OS}$ .

(c)

To determine

To find: The product of the slopes of $\overline{OB}$ and $\overline{OS}$ .

Answer to Problem 27WE

The product of the slopes of $\overline{OB}$ and $\overline{OS}$ . is $1$ .

Explanation of Solution

Given information:

The given figure is

  

Figure $\left(1\right)$ .

Calculation:

Consider the figure $\left(1\right)$

Given coordinates of $\overline{OB}$ are $\left(0,0\right)$ and $\left(5,3\right)$ .

Calculate slope of the line

  $\left(m\right)=\frac{y_2-y_1}{x_2-x_1}$

Substitute the values for the slope of $\overline{OB}$ .

  $\begin{array}{l}m=\frac{3-0}{5-0}\\ m=\frac{3}{5}\end{array}$

The coordinates of $\overline{OS}$ are $\left(0,0\right)$ and $\left(3,5\right)$ .

Slope of the $\overline{OS}$

  $\begin{array}{l}m=\frac{5-0}{3-0}\\ m=\frac{5}{3}\end{array}$

Calculate the product of the slopes.

  $\begin{array}{l}P=\frac{3}{5}\times \frac{5}{3}\\ P=1\end{array}$

Therefore the product of the slopes of $\overline{OB}$ and $\overline{OS}$ . is $1$ .