Chapter 9.4, Problem 17WE

To determine

To Find : The measure of chord JK of the given circle.

Answer to Problem 17WE

The measure of chord JK is $10\sqrt{3}$ .

Explanation of Solution

Given information:

Consider the figure as below:

  

In above figure circle contain chords JK, point O is a center point of circle.

Radius $OJ=10$ ,

Measure of arc $JK=120°.$

To find measure of chord JK,

Firstly draw a perpendicular segment OM which bisects chord JK and arc JK into two equal parts as KM = JM and line segment OK= 10 which a radius as shown in above figure.

If measure of arc $JK=120°$ then angle $\angle KOJ=120$ because measure of minor arc and central angles are congruent and equal to each other.

To find the measure of KM.

Use trigonometric identity:

  $\sin \theta =\frac{\text{perpendicular}}{\text{Hypotenuse}}$ in ΔKMO here $\angle KOM=60$ °

Because perpendicular segment OM which bisects $\angle KOJ=120$ into two equal parts

  $\angle KOM=\angle JOM=60°$

Also let $perpendicularKM=x$ and $hypotenuseOK=10$ and $\theta =60°$

  $sin60°=\frac{x}{10}$

Measure of

  $\begin{array}{l}sin60°=\frac{\sqrt{3}}{2}\\ sin60°=\frac{x}{10}\\ \frac{\sqrt{3}}{2}=\frac{x}{10}\end{array}$

Multiply both sides of equation $\frac{\sqrt{3}}{2}=\frac{x}{10}$ by 10 and 2.

  $\begin{array}{l}\frac{\sqrt{3}}{\cancel{2}}\times 10\times \cancel{2}=\frac{x}{\cancel{10}}\times \cancel{10}\times 2\hfill \\ 10\sqrt{3}=2x\hfill \end{array}$

Divide both sides of equation $10\sqrt{3}=2x$ by 2.

  $\begin{array}{l}\frac{10\sqrt{3}}{2}=\frac{2x}{2}\hfill \\ 5\sqrt{3}=x\hfill \\ x=5\sqrt{3}\hfill \end{array}$

Hence, the value is $KM=5\sqrt{3}$

In circle M is the midpoint of the chord JK.

That means $JK=MK+MJ$ and $KM=JM$ .

Then if $KM=5\sqrt{3}$ then $JM=5\sqrt{3}$

Substitute these values in equation $JK=MK+MJ$

  $\begin{array}{l}JK=MK+MJ\hfill \\ JK=5\sqrt{3}+5\sqrt{3}\hfill \\ JK=10\sqrt{3}\hfill \end{array}$

Therefore, measure of chord JK is $10\sqrt{3}$ .