Chapter 10.6, Problem 11PSA

a.

To determine

Find the value for the given segment using diagram.

Answer to Problem 11PSA

The length of the segment $AX=6$ cm.

Explanation of Solution

Given:

The given diagram is

  

$Z$ is the center of the circle and $ArcWX={120}^{°}$ . The given segment is $\overline{AX}$ .

Calculation:

Use the inscribed angle theorem $\to$ the measure of an inscribed angle is half the measure of its intercepted arc.

  $\begin{array}{l}m\angle WAX=\frac{1}{2}marcWX=\frac{1}{2}\cdot {120}^{°}={60}^{°}\\ m\angle WXA=\frac{1}{2}marcWA=\frac{1}{2}\cdot {180}^{°}={90}^{°}\end{array}$

  $m\angle AWX={90}^{°}-{60}^{°}={30}^{°}$

The sides for the right angle triangle $\left({30}^{°},{60}^{°},{90}^{°}\right)$ is $\left(a,a\sqrt{3},2a\right)$ .

Given $WZ=6$ cm

  $AW=2a=6\cdot 2=12$ cm

  $AX=a=6$ cm

  $WX=a\sqrt{3}=6\sqrt{3}$ cm

Hence the length of the segment $AX=6$ cm.

b.

To determine

Find the perimeter of the given triangle using diagram.

Answer to Problem 11PSA

The perimeter of the $\Delta WXA=6\left(3+\sqrt{3}\right)\text{cm}$ .

Explanation of Solution

Given:

The given diagram is

  

$Z$ is the center of the circle and $ArcWX={120}^{°}$ . The given segment is $\overline{AX}$ .

Calculation:

Use the inscribed angle theorem $\to$ the measure of an inscribed angle is half the measure of its intercepted arc.

  $\begin{array}{l}m\angle WAX=\frac{1}{2}marcWX=\frac{1}{2}\cdot {120}^{°}={60}^{°}\\ m\angle WXA=\frac{1}{2}marcWA=\frac{1}{2}\cdot {180}^{°}={90}^{°}\end{array}$

  $m\angle AWX={90}^{°}-{60}^{°}={30}^{°}$

The sides for the right angle triangle $\left({30}^{°},{60}^{°},{90}^{°}\right)$ is $\left(a,a\sqrt{3},2a\right)$ .

Given $WZ=6$ cm

  $AW=2a=6\cdot 2=12$ cm

  $AX=a=6$ cm

  $WX=a\sqrt{3}=6\sqrt{3}$ cm

Perimeter of the triangle $\Delta WXA=AW+WX+AX$

  $\begin{array}{l}=12+6\sqrt{3}+6\\ =18+6\sqrt{3}\\ =6\left(3+\sqrt{3}\right)\text{cm}\end{array}$

Hence the perimeter of the $\Delta WXA=6\left(3+\sqrt{3}\right)\text{cm}$ .