   Chapter 8.3, Problem 21E ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698

#### Solutions

Chapter
Section ### Elementary Geometry for College St...

6th Edition
Daniel C. Alexander + 1 other
ISBN: 9781285195698
Textbook Problem
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# In Exercises 17 to 30, use the formula A = 1 2 a P to find the area of the regular polygon described.Find the area of a regular hexagon whose sides have length 6 cm.

To determine

To find:

The area of a regular hexagon.

Explanation

1) Any radius of a regular polygon bisects the angle at the vertex to which it is drawn.

2) Any apothem of a regular polygon bisects the side of the polygon to which it is drawn.

3) The measure of any central angle of a regular polygon of n sides is given by c=360n

4) The perimeter of a regular polygon is given by P = ns, where n is the number of sides and s is the length of any side.

5) The area of a regular hexagon with apothem a and perimeter P is given by A=12(aP).

Calculation:

Consider a regular hexagon ABCDEF with QE as radius and QG as apothem.

It is given that the side of the regular hexagon ABCDEF is 6 cm i.e., AB = BC = CD = DE = EF = FA = 6 cm

Any apothem of a regular polygon bisects the side of the polygon to which it is drawn.

Therefore, GE= GD

Now, DF = GE + GD

6=GE+GD6=2GDGD=3cm

The regular hexagon has 6 sides.

Use the formula of central angle, c=360n

Substitute n = 6 in c=360n.

c=3606=60

Therefore, EQD=60

With EQD=60 and QE=QD, ΔQED is an equiangular and equilateral triangle.

Therefore, QE = QD = 6 cm

Any radius of a regular polygon bisects the angle at the vertex to which it is drawn

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