Find the sum of the measures of the interior angles of each of the following convex polygons: a pentagon, an octagon, a dodecagon, a forty-sided polygon, a fifty-two sided polygon, a hundred sided polygon.

Respuesta :

Answer:

pentagon=540°

octagon= 1080°

decagon= = 1440°

forty-sided polygon= 6840°

fifty-two-sided polygon=9000°

hundred-sided polygon=17640°

Step-by-step explanation:

To find the measure of the sum of the interior angle of a polygon, we will simple use the formula;

sum of interior angle = (n-2)180

where n is the number of sides of the polygon

We are ask to find the sum of the interior angle of a pentagon, octagon, decagon, a forty-sided polygon, a fifty-two-sided polygon and a hundred-sided polygon.

First we need to define some terminology;

A penatgon is a five(5)-sided polygon

An Octagon is an eight(8)-sided polygon

A Decagon is a ten(10)-sided polygon

We can now proceed to solve

sum of interior angle = (n-2)180

Sum of interior angle of a pentagon = (5-2)180

                                                             = 3 × 180

                                                             =540°

Sum of interior angle of a pentagon = 540°

Sum of interior angle of an octagon  = (8-2)180

                                                             = 6 × 180

                                                             =1080°

Sum of interior angle of an octagon = 1080°

Sum of interior angle of a decagon  = (10-2)180

                                                             = 8 × 180

                                                             =1440°

Sum of interior angle of a decagon = 1440°

Sum of interior angle of a forty-sided polygon  = (40-2)180

                                                                               = 38 × 180

                                                                                =6840°

Sum of interior angle of a forty-sided polygon = 6840°

Sum of interior angle of a fifty-two-sided polygon  = (52-2)180

                                                                                    = 50 × 180

                                                                                     =9000°

Sum of interior angle of a fifty-two-sided polygon = 9000°

Sum of interior angle of a hundred-sided polygon  = (100-2)180

                                                                                     = 98 × 180

                                                                                      =17640°

Sum of interior angle of a hundred-sided polygon = 17640°

The sum of the measure of the interior angles of each of the following convex polygons with hundred sided polygon is 17,640°.

What is the sum of the interior angles of a polygon?

The sum of the interior angles of n sided polygon can be written as            (n-2)180°.

Given to us

  • A pentagon,
  • An octagon,
  • A dodecagon,
  • A forty-sided polygon,
  • A fifty-two sided polygon,
  • A hundred-sided polygon.

We know the sum of the interior angles of n sided polygon can be written as (n-2)180°. therefore,

The sum of the interior angles of a pentagon

We know that a pentagon has 5 number of sides, therefore, the value of n is 5.

The sum of the interior angles of a pentagon = (n-2)180°

                                                                            = (5-2)180°

                                                                            = 540°

The sum of the interior angles of an Octagon

We know that an octagon has 8 sides, therefore, the value of n is 8.

The sum of the interior angles of an octagon = (n-2)180°

                                                                         = (8-2)180°

                                                                         = 1,080°

The sum of the interior angles of a dodecagon

We know that a dodecagon has 12 sides, therefore, the value of n is 12.

The sum of the interior angles of a dodecagon = (n-2)180°

                                                                               = (12-2)180°

                                                                               = 1,800°

The sum of the interior angles of a forty-sided polygon

The sum of the interior angles of a forty-sided polygon = (n-2)180°

                                                                                            = (40-2)180°

                                                                                            = 6,840°

The sum of the interior angles of a fifty-sided polygon

The sum of the interior angles of a fifty-sided polygon = (n-2)180°

                                                                                          = (52-2)180°

                                                                                          = 9,000°

The sum of the interior angles of a hundred-sided polygon

The sum of the interior angles of a hundred-sided polygon = (n-2)180°

                                                                                                  = (100-2)180°

                                                                                                  = 17,640°

Hence, the sum of the measure of the interior angles of each of the following convex polygons with hundred sided polygon is 17,640°.

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