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Determine the perimeter of a regular octagon that has a distance from its center to any of its vertices of 10
inches. Round your answer to the nearest tenth of an inch.

Determine The Perimeter Of A Regular Octagon That Has A Distance From Its Center To Any Of Its Vertices Of 10 Inches Round Your Answer To The Nearest Tenth Of A class=

Sagot :

Answer:

61.2 in

Step-by-step explanation:

Draw a segment from O perpendicular to AB forming a right triangle.

Call the point of intersection J.

m∠HAB = (8 - 2)180/8 = 135

m∠BAO = 135/2 = 67.5

AJ/10 = cos 67.5

AJ = 10 cos 67.5

AB = 2(10 cos 67.5)

Perimeter = 8(AB) = 8(2)(10 cos 67.5) = 61.2 in.

The perimeter of a regular octagon that has a distance from its center to any of its vertices of 10 inches would be 61.2 inches.

What is a polygon?

A polygon is a closed figure made up of three or more line segments connected end to end.

First, we draw a segment from O perpendicular to AB forming a right triangle.

Let say the point of intersection J.

∠HAB = (8 - 2)180/8

          = 135

∠BAO = 135/2

           = 67.5

AJ/10 = cos 67.5

AJ = 10 cos 67.5

AB = 2(10 cos 67.5)

Perimeter = 8(AB)

               = 8(2)(10 cos 67.5)

               = 61.2 in.

Learn more about polygon;

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