Chapter 9.5, Problem 3CE

a.

To determine

To find: The sum of measures of the red and blue arcs.

Answer to Problem 3CE

The sum of measures of the red and blue arcs can be find out using the Theorem 9.5 that states when there is a diameter which is orthogonal to a chord, then the diameter will bisects the chord and its arc.

Explanation of Solution

Given:

If there is circle with center $\mathrm{O}$ and chord of the circle is represented by $A B$ and $A B$ is bisected by OX . Therefore, $\mathrm{A}\mathrm{X}=\mathrm{B}\mathrm{X}$

  

Now it is to be proved that $O X$ is perpendicular to $A B$.

From, $\mathrm{\Delta }\mathrm{A}\mathrm{O}\mathrm{X}$ and $△\mathrm{B}\mathrm{O}\mathrm{X},$ it can be observed that

  $\mathrm{O}\mathrm{A}=\mathrm{O}\mathrm{B}$ (As both are the radius of circle)

  $\mathrm{O}\mathrm{X}=\mathrm{O}\mathrm{X}(\text{Common})$

  $\mathrm{A}\mathrm{X}=\mathrm{B}\mathrm{X}$ (This is given)

Therefore, $\mathrm{\Delta }\mathrm{A}\mathrm{O}\mathrm{X}\cong \mathrm{\Delta }\mathrm{B}\mathrm{O}\mathrm{X}$ (SSS)

and $\angle \mathrm{A}\mathrm{X}\mathrm{O}=\angle \mathrm{B}\mathrm{X}\mathrm{O}(\text{CPCT})$

Therefore, it can be said that $\angle \mathrm{A}\mathrm{X}\mathrm{O}$ and $\angle \mathrm{B}\mathrm{X}\mathrm{O}$ are forming a linear pair. $\angle \mathrm{A}\mathrm{X}\mathrm{O}+\angle \mathrm{B}\mathrm{X}\mathrm{O}={180}^{°}$

  $\angle \mathrm{A}\mathrm{X}\mathrm{O}+\angle \mathrm{A}\mathrm{X}\mathrm{O}={180}^{°}$

  $2\angle \mathrm{A}\mathrm{X}\mathrm{O}={180}^{°}$

  $\angle \mathrm{A}\mathrm{X}\mathrm{O}={90}^{°}$

Therefore, from this, it can be concluded that $\angle \mathrm{A}\mathrm{X}\mathrm{O}=\angle \mathrm{B}\mathrm{X}\mathrm{O}={90}^{°}$ .

This implies that $O X$ is perpendicular to $A B$.

b.

To determine

To explain: How part (a) helps to state that $\mathrm{x}+\mathrm{y}=180$ .

Answer to Problem 3CE

It can be stated using Theorem 9.6 which states that when two chords are at equal distances from the center then both the chords will be congruent. Therefore, due to this reason,

  $\mathrm{x}+\mathrm{y}=180$ .

Explanation of Solution

In first part of the question (a), as it was seen that $\mathrm{x}$ and $\mathrm{y}$ are forming a linear pair. $\mathrm{x}+\mathrm{y}={180}^{°}$

  $\mathrm{x}+\mathrm{x}={180}^{°}$

  $2\mathrm{x}={180}^{°}$

  $\mathrm{x}={90}^{°}$

Therefore, from this, it can be concluded that $\mathrm{x}=\mathrm{y}=90$ .

  

c.

To determine

To state: The Theorem 9.7.

Answer to Problem 3CE

It can be stated when chords are at the same distance from the center of the circle, they will always be equal in length.

Explanation of Solution

Given:

A circle is given with center O.AB and CD are the chords of circle. Now OX is perpendicular to $A B$ and $O Y$ is perpendicular to CD . Also OX is equal to OY .

Here, it is needed to be proved that $\mathrm{A}\mathrm{B}=\mathrm{C}\mathrm{D}$ .

  

  $\mathrm{In<mn>\Delta </mn><mn>A</mn><mn>O</mn><mn>X</mn>}$ and $\mathrm{\Delta }\mathrm{D}\mathrm{O}\mathrm{Y}$

  $\angle \mathrm{O}\mathrm{X}\mathrm{A}=\angle \mathrm{O}\mathrm{Y}\mathrm{D}(90$ degrees $)$

  $\mathrm{O}\mathrm{A}=\mathrm{O}\mathrm{D}$ (Radius of the circle) $\mathrm{O}\mathrm{X}=\mathrm{O}\mathrm{Y}$ (Given) $\mathrm{\Delta }\mathrm{A}\mathrm{O}\mathrm{X}\cong \mathrm{\Delta }\mathrm{D}\mathrm{O}\mathrm{Y}$

  $\mathrm{A}\mathrm{X}=\mathrm{D}\mathrm{Y}(\text{CPCT})$

Now OX is perpendicular th $A B,$ therefore $\mathrm{X}$ is bisecting $A B$

Therefore ${}_{\mathrm{a}}\mathrm{A}\mathrm{B}=2\mathrm{A}\mathrm{X}$

Now $O Y$ is perpendicular $\mathrm{to<mn>C</mn><mn>D</mn><mo>,</mo>}$ therefore $\mathrm{Y}$ is bisecting $C D$

Therefore, $\mathrm{C}\mathrm{D}=2\mathrm{D}\mathrm{Y}$

Since $\mathrm{A}\mathrm{X}=\mathrm{D}\mathrm{Y}$

  $2\mathrm{A}\mathrm{Y}=2\mathrm{D}\mathrm{Y}$

  $\mathrm{A}\mathrm{B}=\mathrm{C}\mathrm{D}$