Chapter 9.10, Problem 8PSB

a.

To determine

To calculate: The measure of angle E in a pentagon with center O .

Answer to Problem 8PSB

The measure of angle E is ${108}^{\circ }$ .

Explanation of Solution

Given information:

A pentagon with center O.

Side EN = 12 cm.

Formula used:

Exterior angle at E is $\frac{{360}^{\circ }}{n}$ .

n = number of sides.

Calculation:

  

m∠ EON or exterior angle at E is $\frac{{360}^{\circ }}{n}=\frac{{360}^{\circ }}{5}={72}^{\circ }$ .

  $m\angle NEP=180^o-72^o=108^o$

b.

To determine

To find: The measure of angle NOM if side EN is 12 cm.

Answer to Problem 8PSB

The measure of angle NOM is ${36}^{\circ }$ .

Explanation of Solution

Given information:

A pentagon with center O.

Side EN = 12 cm.

Formula used:

Exterior angle at T is $\frac{{360}^{\circ }}{n}$ .

n = number of sides.

Calculation:

  

m∠ NOT $=\frac{{360}^{\circ }}{n}=\frac{{360}^{\circ }}{5}={72}^{\circ }$ .

  $\overline{OM}$ bisects ∠ NOT.

  $m\angle NOM=\frac{72^o}{2}=36^o$

c.

To determine

To calculate: The length of side $\overline{OM}$ .

Answer to Problem 8PSB

The length of side $\overline{OM}$ is $8.26cm$ .

Explanation of Solution

Given information:

A pentagon with center O.

Side EN = 12 cm.

Formula used:

  $\tan \theta =\frac{oppositeside}{adjacentside}$

Calculation:

  

m∠ NOT $=\frac{{360}^{\circ }}{n}=\frac{{360}^{\circ }}{5}={72}^{\circ }$ .

  $\overline{OM}$ bisects ∠ NOT.

  $m\angle NOM=\frac{72^o}{2}=36^o$

In ΔNOM,

  $\begin{array}{l}tan\angle NOM=\frac{NM}{OM}\\ OM=\frac{NM}{tan\angle NOM}\approx \frac{6}{0.7265}\approx 8.26cm\end{array}$

d.

To determine

To find: The area of triangle NOT .

Answer to Problem 8PSB

The area of triangle NOT is $49.55c{m}^2$ .

Explanation of Solution

Given information:

A pentagon with center O.

Side EN = 12 cm.

Formula used:

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

Calculation:

  

m∠NOT $=\frac{{360}^{\circ }}{n}=\frac{{360}^{\circ }}{5}={72}^{\circ }$ .

  $\overline{OM}$ bisects ∠ NOT.

  $m\angle NOM=\frac{72^o}{2}=36^o$

In ΔNOM,

  $\begin{array}{l}tan\angle NOM=\frac{NM}{OM}\\ OM=\frac{NM}{tan\angle NOM}\approx \frac{6}{0.7265}\approx 8.26cm\end{array}$

NT = 12 cm

Area of triangle NOT $=\frac{1}{2}\times NT\times OM$

Area of triangle NOT $=\frac{1}{2}\times 12\times 8.26$

Area of triangle NOT $\approx 49.55c{m}^2$

e.

To determine

To find: The area of pentagon.

Answer to Problem 8PSB

The area of pentagon can be calculated by multiplying area of triangle NOT by 5.

Explanation of Solution

Given information:

A pentagon with center O.

Side EN = 12 cm.

Formula used:

Area of triangle: $A=\frac{1}{2}\times b\times h$

b = base of triangle

h = height of triangle

  

m∠ NOT $=\frac{{360}^{\circ }}{n}=\frac{{360}^{\circ }}{5}={72}^{\circ }$ .

  $\overline{OM}$ bisects ∠ NOT.

  $m\angle NOM=\frac{72^o}{2}=36^o$

In ΔNOM,

  $\begin{array}{l}tan\angle NOM=\frac{NM}{OM}\\ OM=\frac{NM}{tan\angle NOM}\approx \frac{6}{0.7265}\approx 8.26cm\end{array}$

NT = 12 cm

Area of triangle NOT $=\frac{1}{2}\times NT\times OM$

Area of triangle NOT $=\frac{1}{2}\times 12\times 8.26$

Area of triangle NOT $\approx 49.55c{m}^2$

Area of pentagon is sum of areas of triangles NOT , TOA , AOP , POE and EON.

All five triangles are identical and have same area as triangle NOT.

Area of Pentagon $\approx 49.55\times 5=247.75c{m}^2$