Chapter 8.3, Problem 39E

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

Chapter
Section

### Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085
Textbook Problem

# Consider a regular hexagon A B C D E F (not shown). By joining midpoints of consecutive sides, a smaller regular hexagon M N P Q R S is formed. Find the ratio of areas A M N P Q R S A A B C D E F

To determine

To find:

The ratio of areas AMNPQRSAABCDEF.

Explanation

1) 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.

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

Calculation:

Given,

A regular hexagon ABCDEF and a smaller regular hexagon MNPQRS by joining the midpoints of consecutive sides of a regular hexagon ABCDEF.

Calculation:

Let the length of the each side of regular hexagon ABCDEF is s.

Consider the triangle ROQ, where O is the center of the hexagon.

To find the apothem OQ of the hexagon ABCDEF

From the figure, EOQ is a right angle triangle.

and EQ=OE2=s2

by using Pythagoras theorem

OE2=EQ2+OQ2s2=(s2)2+OQ2OQ2=s2âˆ’s24=4s2âˆ’s24=3s24OQ=3s2

Let X be the midpoint of RQ. Now to find apothem and side length of the regular hexagon MNPQRS.

The triangle âˆ†Â OEQ is the form of 30âˆ˜âˆ’60âˆ˜âˆ’90âˆ˜.

Therefore by the properties of 30âˆ˜âˆ’60âˆ˜âˆ’90âˆ˜ triangle the hypotenuse is equal to twice the length of the shorter leg, which is the side across from the 30âˆ˜ angle. The longer leg, which is across from the 60âˆ˜ angle, is equal to multiplying the shorter leg by the 3.

i.e

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