Chapter 10.7, Problem 6PSA

a.

To determine

Find $m\stackrel{⌢}{PQ}$ .

Answer to Problem 6PSA

  $m\stackrel{⌢}{PQ}=72°$

Explanation of Solution

Given information:

  

  $PQRST$ is regular pentagon and $ABCDEF$ is regular hexagon.

Calculation:

  $PQRST$ is regular pentagon

All the sides of pentagon are congruent and the chords of circumscribed circles are congruent and circle arc is $360°$ .

  $m\stackrel{⌢}{PQ}=\frac{360°}{5}=72°$

Hence, $m\stackrel{⌢}{PQ}=72°$

b.

To determine

Find $m\stackrel{⌢}{RT}$ .

Answer to Problem 6PSA

  $m\stackrel{⌢}{RT}=144°$

Explanation of Solution

Given information:

  

  $PQRST$ is regular pentagon and $ABCDEF$ is regular hexagon.

Calculation:

  $PQRST$ is regular pentagon

All the sides of pentagon are congruent and the chords of circumscribed circles are congruent and circle arc is $360°$ .

  $m\stackrel{⌢}{PQ}=\frac{360°}{5}=72°$

  $\begin{array}{l}m\stackrel{⌢}{RT}=m\stackrel{⌢}{RS}+m\stackrel{⌢}{ST}\\ m\stackrel{⌢}{RT}=2\left(72°\right)=144°\end{array}$

Hence, $m\stackrel{⌢}{RT}=144°$

c.

To determine

Find $m\stackrel{⌢}{AB}$ .

Answer to Problem 6PSA

  $m\stackrel{⌢}{AB}=60°$

Explanation of Solution

Given information:

  

  $PQRST$ is regular pentagon and $ABCDEF$ is regular hexagon.

Calculation:

  $PQRST$ is regular pentagon

All the sides of polygon are congruent and the chords of circumscribed circles are congruent and circle arc is $360°$ .

  $m\stackrel{⌢}{AB}=\frac{360°}{6}=60°$

Hence, $m\stackrel{⌢}{AB}=60°$ .

d.

To determine

Find $m\stackrel{⌢}{BD}$ .

Answer to Problem 6PSA

  $m\stackrel{⌢}{BD}=120°$

Explanation of Solution

Given information:

  

  $PQRST$ is regular pentagon and $ABCDEF$ is regular hexagon.

Calculation:

  $PQRST$ is regular pentagon

All the sides of polygon are congruent and the chords of circumscribed circles are congruent and circle arc is $360°$ .

  $m\stackrel{⌢}{AB}=\frac{360°}{6}=60°$

Arc $\stackrel{⌢}{BD}$ is a sum of two arcs,

  $\begin{array}{l}m\stackrel{⌢}{BD}=m\stackrel{⌢}{BC}+m\stackrel{⌢}{CD}\\ =\left(2\right)60°\\ m\stackrel{⌢}{BD}=120°\end{array}$

Hence, $m\stackrel{⌢}{BD}=120°$ .

e.

To determine

Find $m\stackrel{⌢}{DE}A$ .

Answer to Problem 6PSA

  $m\stackrel{⌢}{DE}A=180°$

Explanation of Solution

Given information:

  

  $PQRST$ is regular pentagon and $ABCDEF$ is regular hexagon.

Calculation:

  $PQRST$ is regular pentagon

All the sides of polygon are congruent and the chords of circumscribed circles are congruent and circle arc is $360°$ .

  $m\stackrel{⌢}{AB}=\frac{360°}{6}=60°$

Arc $m\stackrel{⌢}{DE}A$ is a sum of three arcs,

  $\begin{array}{l}m\stackrel{⌢}{DEA}=m\stackrel{⌢}{DE}+m\stackrel{⌢}{EF}+m\stackrel{⌢}{FA}\\ =\left(3\right)60°\\ m\stackrel{⌢}{DE}A=180°\end{array}$

Hence, $m\stackrel{⌢}{DE}A=180°$ .