Chapter 12, Problem 5CR

a.

To determine

To calculate: The value x of circle’s chord.

Answer to Problem 5CR

The value of x is $8$ .

Explanation of Solution

Given information:

A circle with two chords.

Formula used:

The below theorem is used:

Chord-Chord Power theorem states that if two chords of a circle intersect, then the product of the measures of the parts of one chord is equal to the product of the measures of the parts of the other chord.

Calculation:

  

By chord- chord power theorem, we get

  $\begin{array}{l}3x=12\times 2\\ 3x=24\\ x=\frac{24}{3}=8\end{array}$

b.

To determine

To find: The value x of triangle’s side.

Answer to Problem 5CR

The value of x is $3.2$ .

Explanation of Solution

Given information:

Side AB = 16,

Side BC = x ,

Side CD = 3 .

Side AD = 15.

Formula used:

The below theorem is used:

Corresponding sides of similar triangles are congruent.

Calculation:

  

△ABC and △ACD are similar triangles.

Corresponding sides of similar triangles are congruent.

  $\begin{array}{l}\frac{AD}{CD}=\frac{AB}{BC}\\ \frac{15}{3}=\frac{16}{x}\\ 15x=48\\ x=\frac{48}{15}=3.2\end{array}$

c.

To determine

To find the measure of arc x .

Answer to Problem 5CR

The measure of arc x is $44°$ .

Explanation of Solution

Given information:

  $\begin{array}{l}m\stackrel{⌢}{BD}=x,\\ m\stackrel{⌢}{AC}=20°,\\ m\angle BPD={12}^{\circ }.\end{array}$

Formula used:

The measure of secant-secant angle is one half the difference of its two intercepted arcs.

Calculation:

  

  $\begin{array}{l}m\angle BPD=\frac{m\stackrel{⌢}{BD}-m\stackrel{⌢}{AC}}{2}\\ {12}^{\circ }=\frac{x-{20}^{\circ }}{2}\\ {24}^{\circ }=x-{20}^{\circ }\\ x={44}^{\circ }\end{array}$

d.

To determine

To find: The value of side x if one of the side AB is 4.

Answer to Problem 5CR

The value of x is $12$ .

Explanation of Solution

Given information:

Side AB = 4,

Side CB = 3,

Side AD = 13.

Formula used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  

In right angle triangle,

  $a^2+b^2\text{= }{c}^2$

Proof:

  

Side AC can be calculated by applying Pythagoras Theorem.

In right angled triangle △ABC , we get

  $\begin{array}{l}{\left(AC\right)}^2={\left(AB\right)}^2+{\left(BC\right)}^2\\ {\left(AC\right)}^2={\left(4\right)}^2+{\left(3\right)}^2\\ {\left(AC\right)}^2=16+9\\ {\left(AC\right)}^2=25\\ AC=\sqrt{25}=5\end{array}$

Side DC can be calculated by applying Pythagoras Theorem.

In right angled triangle △ACD , we get

  $\begin{array}{l}{\left(AD\right)}^2={\left(AC\right)}^2+{\left(CD\right)}^2\\ {\left(5\right)}^2={\left(13\right)}^2+{\left(x\right)}^2\\ {\left(x\right)}^2=169-25\\ {\left(x\right)}^2=144\\ x=\sqrt{144}=12\end{array}$