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Title

Apothem and Area of a hexagon

Problem Statement

Determine an equation of the area of a regular hexagon in terms of its apothemand perimeter.

The apothem is the shortest distance from the center of the polygon to one of the sides.

Will this equation be true for other regular polygons? Explain.

Problem setup

Can a hexagons area be calculated using the perimeter and the apothem? f so! "hat is theequation and does it hold true for other polygons?

Plans to Solve/Investigate the Problem

"ill construct a hexagon using #$% s&etchpad. plan to sol'e the problem by using the

constructed hexagon to determine the role of the apothem in the hexagon. predicate that theapothem can be considered as the height of a triangle! therefore! the area can be calculated using

triangles found "ithin the hexagon.

m NO = 2.02 cm

m OP = 2.02 cm

m PM = 2.02 cm

m MJ = 2.02 cm

m KJ = 2.02 cm

m KN = 2.02 cm

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measure the length of the sides of the hexagon to ma&e sure they are all equal. (ext! constructa line segment from point ) to *. Then construct the midpoint on the line and measure the

distance from the center of the hexagon to segment %+. The distance "as ,.- cm.

BC

AD

EF

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Distance Q to PM = 1.75 cm

m NO = 2.02 cm

m OP = 2.02 cm

m PM = 2.02 cm

m MJ = 2.02 cm

m KJ = 2.02 cm

m KN = 2.02 cm

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"ill then create a segment from point / to midpoint 0 on line segment %+. 1ine 0 "ill ser'es

as the apothem of the hexagon.

m NO = 2.02 cm

m OP = 2.02 cm

m PM = 2.02 cm

m MJ = 2.02 cm

m KJ = 2.02 cm

m KN = 2.02 cm

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(ext! construct a triangle using points %! /! and +. ma&e line segments from points % and +

to point /.

m NO = 2.02 cm

m OP = 2.02 cm

m PM = 2.02 cm

m MJ = 2.02 cm

m KJ = 2.02 cm

m KN = 2.02 cm

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(ext! di'ide the hexagon into the equal triangles.

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m NO = 2.02 cm

m OP = 2.02 cm

m PM = 2.02 cm

m MJ = 2.02 cm

m KJ = 2.02 cm

m KN = 2.02 cm

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$ince created the segment 0/ from the midpoint on %+! can use the segment 0/ as the

height of triangle /+%! "hich the height of the triangle is ,.- cm. (o" that ha'e the height of

triangle /+%! can use the formula for the area of the triangle to determine the area of triangle

/+%. The apothem "ill be substituted for the height of one of the equilateral triangles. Theformula "ill be 2 base3%+4 5apothem3height4 "hich "ould gi'e me ,.-6 cm! "hich is the area

of the triangle /+%. used the calculate function on #$% to ma&e sure that my calculations "ere

correct.

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2( )m PM Distance Q to PM( )= 1.7 cm2

The area of triangle /+% is 7.58.785,.-9 ,.-6 cm. (o" that ha'e the area of this triangle can use this area to determine the area of the remaining six triangles in the hexagon. Therefore!

based on this obser'ation! &no" that can find the area of the hexagon by multiplying 2 base5

h5 number of triangles in the hexagon. The equation "ill be as follo"s: 258.785,.-569 ,7.cm. The area of the hexagon "ould be ,7. cm. To ma&e sure that my calculations "ere correct!

used the calculate function of #$% and my calculations "ere correct.

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2( )m PM Distance Q to PM( ) = 10.55 cm2 also used the interior of the hexagon to chec& the area of

the hexagon and it "as ,7. cm.

A!ea MPONKJ = 10.55 cm2

(ext! find the perimeter of the hexagon! and from my prior &no"ledge! &no" that theperimeter of an ob;ect is found by adding the sides of an ob;ect. Therefore to find the perimeterof this hexagon "ill ta&e the length of the sides "hich "ere 8.78 cm and add it up 6 times

"hich "ould ma&e the perimeter ,8.7< cm.

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m NO = 2.02 cm

m OP = 2.02 cm

m PM = 2.02 cm

m MJ = 2.02 cm

m KJ = 2.02 cm

m KN = 2.02 cm

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Pe!imete! MPONKJ = 12.0" cm

To ma&e sure that the calculations "ere correct! used the calculate function in #$% and added

up the length of sides six times.

m KN#m KJ#m MJ#m PM#m OP#m NO = 12.0" cm

Conclusion:

As stated earlier line 0 "ill ser'e as the apothem. reali=ed that the area can be calculated by

e'aluating the area of a series of triangles "ithin the hexagon. Therefore! the area for that

triangle "ill be 2 base times the height. 3,>8 b5h4. Also! as noticed in my constructions! thereare also fi'e other triangles that "ere created by ;oining the 'ertices "ith the center point. (o"

"e can replace base "ith side and h "ith apothem. The equation "ill then be 25s5a. $ince

there are six triangles that ma&e up the area of the regular hexagon then the 25s5a "ill ha'e tobe added six times to get the total area.

The equation "ould then be A9,>8as ,>8as ,>8as ,>8as ,>8as ,>8as "here a is theapothem and s the length of the sides. The total area adds up to one half the base "hich is the

side times the height "hich is the apothem and this occurs six times in the entire hexagon

236as4. Therefore! the equation can be "ritten to isolate the perimeter of the hexagon as 236s4

3a4. This form of the equation allo"s the perimeter to be isolated in terms of 6s. Thus! theequation of the area of a regular pentagon in terms of the apothem and the perimeter is one@half

the perimeter times the apothem. $tated as 23%5a4 9 Area. %erimeter is equal to the length of

sides added six times for the number of sides.

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Extensions of the Problem

0a'e students create a quadrilateral that is a square and find the apothem of the square and find a

formula for area and perimeter of the square.

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% H

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D C

A B

1ine * is the apothem of the square "hich is a side of the square #*D. This square #*D is ,.8