   Chapter 8.3, Problem 23E ### 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 an equilateral triangle whose radius measures 10 in.

To determine

To find:

The area of the equilateral triangle.

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 area of the equilateral triangle is given by, A=34s2 where s is side of the equilateral triangle.

5) 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.

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

Calculation:

Consider an equilateral triangle DEF with QG as apothem.

It is given that the length of the radius of the equilateral triangle DEF is 10 in. i.e., QD = 10 in.

Use the formula of central angle, c=360n

Substitute n = 3 in c=360n.

c=3603=120

Therefore, EQD=120

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

Therefore, GQD=60

Now, ΔDQG is a 30°-60°-90° triangle.

By properties of 30°-60°-90° triangle,

DG=sin60DQ=32(10)=53in.

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

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