Chapter 11.8, Problem 13PSC

a.

To determine

To calculate:Thecoordinates of point O .

Answer to Problem 13PSC

Thecoordinates of point Oare $(6,2.81)$ .

Explanation of Solution

Given information:

In circle with centerO ,

Coordinates of point C are $(3,8)$

Coordinates of point D are $(9,8)$

  $m\stackrel{⌢}{CD}={60}^{\circ }$

Calculation:

  

Coordinates of E , the midpoint of side CD , is $(6,8).$

ΔCEO is ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ with CE = $3$

In ΔCEO,

  $\begin{array}{l}\tan {30}^{\circ }=\frac{1}{\sqrt{3}}\\ \frac{CE}{EO}=\frac{1}{\sqrt{3}}\\ \frac{3}{EO}=\frac{1}{\sqrt{3}}\\ EO=3\sqrt{3}\end{array}$

The x coordinate of point O will be same as that of point E because they both lie on same x coordinate.

The y coordinate of point O will be difference between 8 and $3\sqrt{3}$ .

Coordinates of point O are $(6,8-3\sqrt{3})=(6,2.81)$

b.

To determine

To find: The circumference of circle with center O .

Answer to Problem 13PSC

The circumference of circle with center Ois $37.7$ .

Explanation of Solution

Given information:

In circle with center O ,

Coordinates of point C are $(3,8)$

Coordinates of point D are $(9,8)$

  $m\stackrel{⌢}{CD}={60}^{\circ }$

Formula used:

Circumference of circle: $C=2\pi r=\pi d$

r = radius of circle.

Calculation:

  

Coordinates of E , the midpoint of side CD , is $(6,8).$

ΔCEO is ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ with CE = $3$

In ΔCEO,

  $\begin{array}{l}\tan {30}^{\circ }=\frac{1}{\sqrt{3}}\\ \frac{CE}{EO}=\frac{1}{\sqrt{3}}\\ \frac{3}{EO}=\frac{1}{\sqrt{3}}\\ EO=3\sqrt{3}\end{array}$

The x coordinate of point O will be same as that of point E because they both lie on same x coordinate.

The y coordinate of point O will be difference between 8 and $3\sqrt{3}$ .

Coordinates of point O are $(6,8-3\sqrt{3})=(6,2.81)$

ΔCOD is equilateral triangle, i.e. ${60}^{\circ }{60}^{\circ }{60}^{\circ }\Delta$ with CO = $6$ .

CO is radius of circle with center O .

  $r=CO=6$

  $C=2\pi r=2\pi \ast 6=12\pi \approx 37.7$