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Elementary Geometry For College St...

7th Edition
Alexander + 2 others
ISBN: 9781337614085

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BuyFindarrow_forward

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=s2s24=4s2s24=3s24OQ=3s2

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

The triangle  OEQ is the form of 306090.

Therefore by the properties of 306090 triangle the hypotenuse is equal to twice the length of the shorter leg, which is the side across from the 30 angle. The longer leg, which is across from the 60 angle, is equal to multiplying the shorter leg by the 3.

i.e

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