Geometry For Enjoyment And Challenge
Geometry For Enjoyment And Challenge
91st Edition
ISBN: 9780866099653
Author: Richard Rhoad, George Milauskas, Robert Whipple
Publisher: McDougal Littell
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Question
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Chapter 7.4, Problem 4PSA

a

To determine

The number of sides of the equiangular polygon has to be found

a

Expert Solution
Check Mark

Answer to Problem 4PSA

The number of sides is 10

Explanation of Solution

Given information : The value of each angle is A=1440

We know the sum of the exterior angles is always 3600 . Thus, the measure of one exterior angle will be

  3600n , where n will be our answer in this case.

Thus, the interior angle will be

  A=18003600n

Thus, the value of n will be 10

b

To determine

The number of sides of the equiangular polygon has to be found using the value of each angle given as mentioned

b

Expert Solution
Check Mark

Answer to Problem 4PSA

The number of sides is 6

Explanation of Solution

Given information : The value of each angle is A=1200

We know the sum of the exterior angles is always 3600 . Thus, the measure of one exterior angle will be

  3600n , where n will be our answer in this case.

Thus, the interior angle will be

  A=18003600n

Thus, the value of n will be 6

c

To determine

The number of sides of the equiangular polygon has to be found by using the information given

c

Expert Solution
Check Mark

Answer to Problem 4PSA

The number of sides is 15

Explanation of Solution

Given information : The value of each angle is A=1560

We know the sum of the exterior angles is always 3600 . Thus, the measure of one exterior angle will be

  3600n , where n will be our answer in this case.

Thus, the interior angle will be

  A=18003600n

Thus, the value of n will be 15

d

To determine

The number of sides of the equiangular polygon has to be found with the help of the given data

d

Expert Solution
Check Mark

Answer to Problem 4PSA

The number of sides is 20

Explanation of Solution

Given information : The value of each angle is A=1620

We know the sum of the exterior angles is always 3600 . Thus, the measure of one exterior angle will be

  3600n , where n will be our answer in this case.

Thus, the interior angle will be

  A=18003600n

Thus, the value of n will be 20

e

To determine

The number of sides of the equiangular polygon has to be found by using the relevant data

e

Expert Solution
Check Mark

Answer to Problem 4PSA

The number of sides is 50

Explanation of Solution

Given information : The value of each angle is A=172.80

We know the sum of the exterior angles is always 3600 . Thus, the measure of one exterior angle will be

  3600n , where n will be our answer in this case.

Thus, the interior angle will be

  A=18003600n

Thus, the value of n will be 50

Chapter 7 Solutions

Geometry For Enjoyment And Challenge

Ch. 7.1 - Prob. 11PSACh. 7.1 - Prob. 12PSBCh. 7.1 - Prob. 13PSBCh. 7.1 - Prob. 14PSBCh. 7.1 - Prob. 15PSBCh. 7.1 - Prob. 16PSBCh. 7.1 - Prob. 17PSBCh. 7.1 - Prob. 18PSBCh. 7.1 - Prob. 19PSCCh. 7.1 - Prob. 20PSCCh. 7.1 - Prob. 21PSCCh. 7.1 - Prob. 22PSCCh. 7.1 - Prob. 23PSCCh. 7.2 - Prob. 1PSACh. 7.2 - Prob. 2PSACh. 7.2 - Prob. 3PSACh. 7.2 - Prob. 4PSACh. 7.2 - Prob. 5PSACh. 7.2 - Prob. 6PSACh. 7.2 - Prob. 7PSACh. 7.2 - Prob. 8PSACh. 7.2 - Prob. 9PSACh. 7.2 - Prob. 10PSACh. 7.2 - Prob. 11PSBCh. 7.2 - Prob. 12PSBCh. 7.2 - Prob. 13PSBCh. 7.2 - Prob. 14PSBCh. 7.2 - Prob. 15PSBCh. 7.2 - Prob. 16PSBCh. 7.2 - Prob. 17PSCCh. 7.2 - Prob. 18PSCCh. 7.2 - Prob. 19PSCCh. 7.2 - Prob. 20PSDCh. 7.3 - Prob. 1PSACh. 7.3 - Prob. 2PSACh. 7.3 - Prob. 3PSACh. 7.3 - Prob. 4PSACh. 7.3 - Prob. 5PSACh. 7.3 - Prob. 6PSACh. 7.3 - Prob. 7PSACh. 7.3 - Prob. 8PSBCh. 7.3 - Prob. 9PSBCh. 7.3 - Prob. 10PSBCh. 7.3 - Prob. 11PSBCh. 7.3 - Prob. 12PSBCh. 7.3 - Prob. 13PSBCh. 7.3 - Prob. 14PSBCh. 7.3 - Prob. 15PSBCh. 7.3 - Prob. 16PSBCh. 7.3 - Prob. 17PSBCh. 7.3 - Prob. 18PSBCh. 7.3 - Prob. 19PSBCh. 7.3 - Prob. 20PSCCh. 7.3 - Prob. 21PSCCh. 7.3 - Prob. 22PSCCh. 7.3 - Prob. 23PSCCh. 7.3 - Prob. 24PSDCh. 7.4 - Prob. 1PSACh. 7.4 - Prob. 2PSACh. 7.4 - Prob. 3PSACh. 7.4 - Prob. 4PSACh. 7.4 - Prob. 5PSACh. 7.4 - Prob. 6PSACh. 7.4 - Prob. 7PSACh. 7.4 - Prob. 8PSBCh. 7.4 - Prob. 9PSBCh. 7.4 - Prob. 10PSBCh. 7.4 - Prob. 11PSBCh. 7.4 - Prob. 12PSBCh. 7.4 - Prob. 13PSBCh. 7.4 - Prob. 14PSCCh. 7.4 - Prob. 15PSCCh. 7.4 - Prob. 16PSCCh. 7.4 - Prob. 17PSCCh. 7 - Prob. 1RPCh. 7 - Prob. 2RPCh. 7 - Prob. 3RPCh. 7 - Prob. 4RPCh. 7 - Prob. 5RPCh. 7 - Prob. 6RPCh. 7 - Prob. 7RPCh. 7 - Prob. 8RPCh. 7 - Prob. 9RPCh. 7 - Prob. 10RPCh. 7 - Prob. 11RPCh. 7 - Prob. 12RPCh. 7 - Prob. 13RPCh. 7 - Prob. 14RPCh. 7 - Prob. 15RPCh. 7 - Prob. 16RPCh. 7 - Prob. 17RPCh. 7 - Prob. 18RPCh. 7 - Prob. 19RPCh. 7 - Prob. 20RPCh. 7 - Prob. 21RPCh. 7 - Prob. 22RPCh. 7 - Prob. 23RPCh. 7 - Prob. 24RPCh. 7 - Prob. 25RPCh. 7 - Prob. 26RPCh. 7 - Prob. 27RPCh. 7 - Prob. 28RPCh. 7 - Prob. 29RPCh. 7 - Prob. 30RP
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