Q. 27 - Geometry

Question

 a. Show that$\Delta OAB\cong \Delta ORS$ .

b. Why is $\overline{OB}\perp \overline{OS}$?

c. Find the product of the slopes of $\overline{OB}$ and $\overline{OS}$.

Step-by-Step Solution

Answer

1. We have proved that$\Delta OAB\cong \Delta ORS$ .
2. $\Delta OCD\cong \Delta NEF$by SAS rule.

      3. $\overline{OD}$and$\overline{NF}$  have the same angle with the$x\text{-axis}$ .

Part a. Step-1 – Given

Given figure is:

Step-2 – To determine

We have to show that $\Delta OAB\cong \Delta ORS$.

Step-3 – Calculation

In the given figure:

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

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

$So, \Delta OAB\cong \Delta ORS$

Part b. Step-1 – Given

Given figure is:

Step-2 – To determine

We have to find why$\overline{OB}\perp \overline{OS}$ .

Step-3 – Calculation

In the given figure:

Let us assume that $\angle BOA={\text{n}}^{\text{o}}$

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

From part a,$\Delta OAB\cong \Delta ORS$ .

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

Then we 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, $\overline{OB}\perp \overline{OS}$

Part c. Step-1 – Given

Given figure is:

Step-2 – To determine

We have to find the product of the slopes of $\overline{OB}$ and $\overline{OS}$.

Step-3 – Calculation

In the given figure:

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

Then we calculate the slope of the line

$m=\frac{y_2-y_1}{x_2-x_1}$

Substituting 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 are (0, 0) and (3, 5).

Slope of the $\overline{OS}$

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

Then we calculate the product of the slopes.

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

So, the product of the slopes of  and  is .