Tower Project

Thanvir Ahmed

Mr. Acre

9A GAT

5-31-13

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Tower Project

Part One: Introduction

A multimillionaire by the name of Marie Copeland is trying to build a dog house.

Mrs. Copeland’s wants a dog house for her Golden Retriever, called Scot. Mrs. Copeland is a

math teacher and she loves polygons. She wants Scot’s doghouse to consist of a decagon base

(10 sides). Scot is addicted to water and finds it fascinating, so Mrs. Copeland decided she

wanted him to have an aquarium to keep him busy. Mrs. Copeland lives in the City of Warren,

and the City has requested her to build Scot’s doghouse 3’ in from each side because Scot is

crazy and attacks people if they walk by the sidewalk. Mrs. Copeland’s front yard is a 20’ X 20’

square piece of land. Mrs. Copeland wants the doghouse to resemble a tower, but she is too old

to build it herself. Mrs. Copeland noticed that Scot has a crush on the neighbor’s dog, Beyoncé,

but Scot is too shy to approach her. Mrs. Copeland want’s the doghouse to have windows so Scot

can admire from a distance. I was hired by Mrs. Copeland to build the tower doghouse. Being an

excellent mathematician and architect (The Amazing Thanvir) it was no problem for me. Scot is

a really big dog, so Mrs. Copeland wants the doghouse to be as big as possible without going

outside of the borders requested by the city.

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Part Two: Polygonal Base

Figure 1. Maximized Decagon and Calculations

The figure above shows the original 20’ X 20’ plot of land. Three feet are subtracted

from both sides due to the boundaries. The new smaller piece of land is 14’ X 14’. The

decahedron polygonal base is maximized by putting the points on the square plot of land.

This big decahedron is known as polygon 1. I did not choose sides because if sides were put on

the square plot of land, then some parts of the decagon leave the boundaries.

20’

20’

14’

14’

20 -3 -3 = 14’20-3 -3 = 14’

Polygon 1

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Figure 2. Polygon One Area

The figure above shows the height of one triangular piece of polygon 1. I came up with

the 6.66 ft. for the height by using Sine. The formula I used was Sin(Ø) = (Opposite/Hypotenuse)

Sin(72°) = y/7 I used to find the height. The 72° was the angle measure, y was the height I was

trying to solve for, and 7 was the hypotenuse because 14/2 is 7. Sin= opposite/hypotenuse the

opposite side of the angle in this equation is the height (y), and the hypotenuse is the side

opposite the 90° angle (7). I was able to determine that the angle was 72° because the base is a

decagon, a decagon and a decagon has 10 sides. A circle measure 360°, so 360/10=36 which is a

central angle measurement. I divided 36 by 2 to get 18 because I needed a right triangle in order

to use Sine. A triangles angle measures add up to make 180°. 180 - (90+18) = 72. Sin(72°) = y/7

I solved this to get 6.66 ft. The height of one triangle making up the decagon, or apothem is 6.66

ft. for polygon 1. I found that one side of the decagon was 4.33 ft. this was done by using Sine.

Sin(Ø) = (Opposite/Hypotenuse) Sin(18)=x/7 The formula I used to find the length of one side of

the decagon. The angle measure was 18°, x was the length that I was trying to solve for, and 7 is

the hypotenuse. I was able to determine that the angle measure is 18° because 36° is the central

14’

14’4.33’

6.66’

20’

20’

72°

18°

7’

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angle and in order to use Sine the triangle must be a right triangle. So I divided 36 by 2 and get

18. Sin(18)=x/7 I solved this and got x=2.16 ft. This number is only ½ the length of the side of

the decagon because ½ the triangle was used to find the length of the side. So I multiplied 2.16

by 2 and get 4.33 ft (2.16*2 = 4.33). The length of the side for polygon 1 is 4.33 ft. The Area of

polygon 1 is 144.01 ft2. The area is found by multiplying the apothem (height of triangular piece

of polygon) and the length of one side of the polygon together this is then divided by 2 because

the polygon is produced by triangles. Then that number was multiplied by 10 because there are

10 triangles in a decagon. A stands for area A = (B*H/2)*10 this is the formula I used to find the

area. (6.66*4.33/2)*10 = 144.01 the area of polygon 1 was 144.01 ft2.

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Figure 3. Polygon Two Area

The figure above shows the height of one triangular piece of polygon 2. The height of

one triangle making up polygon 2, or apothem, was 5.66 ft. In order to get 5.66 ft. I took the

height of one triangle from polygon 1 and subtracted 1 foot 6.66 - 1 = 5.66. One side length of

polygon 2 was 3.68 ft. to get this I used Tangent. Tan(Ø)=Opposite/Adjacent this was the

formula I used to find one side of the base. Tangent = opposite/adjacent Tan(18)=x/5.66 18° was

the angle measure, x was what I was trying to solve for, and 5.66 was the adjacent.

Tan(18)=x/5.66 I solved this and got that x = 1.84 ft. this number was only ½ of the length of the

base so it was multiplied by 2. (1.84*2 = 3.68) the length of one side of polygon 2 was 3.68 ft.

The area of polygon 2 was 103.99 ft2. I found the area of the polygon by multiplying the

apothem (height of triangle) and the length of the base, and then I divided that by 2 because a

decagon consists of triangles. Then I multiplied the area of one triangle by 10 to get the area of

polygon 2 because a decagon is made up of 10 triangles. A = (B*H/2)*10 this is the formula I

14’

14’1’

5.66’

20’

20’

72°

18°

3.68’

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used to find the area. A= (3.68*5.66/2) *10 I solved this to get that A = 103.99. The area of

polygon 2 was 103.99 ft2.

Figure 4. Polygon Three Area

The figure above shows the height of one triangular piece of polygon 3. The height of

one triangle making up polygon 3, or the apothem, was 4.66 ft. To find 4.66 ft. as the apothem I

took the height of one triangle from polygon 1 and subtracted 2 feet 6.66 – 2 = 4.66. One side

length of polygon 3 was 3.03ft. to get this I used Tangent. Tan(Ø)=Opposite/Adjacent was the

equation I used to find the side length of polygon 3. Tan(18) = x/4.66 18° was the angle measure,

x was what I what I was trying to solve for, and 4.66 was the adjacent. Tan(18) = x/4.66 I solved

this and got that x = 1.51 ft. this number was only ½ of the length of one side of polygon 3. I

multiplied 1.51 by 2 in order to get the side length of polygon 3 (1.51*2 = 3.03) the length of one

side of polygon 3 was 3.03 ft. The area of polygon 3 was 70.48 ft2. I found the area of polygon 3

by multiplying the apothem (height of triangular piece of decagon) and the length of the base,

14’

14’

2’

4.66’ 20’

20’

72°

18°

3.03’

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and then I divided that by 2 because decagon consists of triangles. Then I multiplied the area of

that one triangle by 10 to get the area of polygon 3 because a decagon is made up of 10 triangles.

A = (B*H/2)*10 this is the formula I used to find the area. A= (3.03*4.66/2)*10 I solved this to

get that A = 70.48. The area of polygon 3 was 70.48 ft2.

Figure 5. Polygon Four Area

The figure above shows the height of one triangular piece of polygon 5. The height of

one triangle making up polygon 4, or the apothem, was 3.66 ft. To find 3.66 ft. as the apothem I

took the height of one triangle from polygon 1 and subtracted 3 feet 6.66 – 3 = 3.66. One side

length of polygon 4 was 2.38ft. to get this I used Tangent. Tan(Ø)=Opposite/Adjacent was the

equation I used to find the side length of polygon 3. Tan(18) = x/3.66 18° was the angle measure,

x was what I what I was trying to solve for, and 3.66 was the adjacent. Tan(18) = x/3.66 I solved

this and got that x = 1.19 ft. this number was only ½ of the length of one side of polygon 4. I

multiplied 1.19 by 2 in order to get the side length of polygon 4 (1.19*2 = 2.38) the length of one

side of polygon 4 was 2,38 ft. The area of polygon 4 was 43.46 ft2. I found the area of polygon 4

14’

14’

3’

3.66’ 20’

20’

72°

18°

2.38’

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by multiplying the apothem (height of triangular piece of decagon) and the length of the base,

and then I divided that by 2 because decagon consists of triangles. Then I multiplied the area of

that one triangle by 10 to get the area of polygon 4 because a decagon is made up of 10 triangles.

A = (B*H/2)*10 this is the formula I used to find the area. A= (2.83*3.66/2)*10 I solved this to

get that A = 43.46. The area of polygon 4 was 43.46 ft2.

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Part Three: Volume of Footing, Floor, and Aquarium

Figure 6. Volume and of Concrete Footing and Aquarium

The figure above shows the concrete footing. The footing extended from polygon 1 to

polygon 4. The footing was 3.5 ft. deep and 3 ft. wide. I found that the volume of the concrete

footing was 351.91 ft3. I found the volume of the footing by first finding the volume of the entire

piece of concrete including the hollow center, and then I found the volume of the hollow center. I

then subtracted the volume of the hollow center from the volume of the whole piece of concrete

in order to get the volume of the footing. The volume of the entire piece of concrete was 504.03

ft3. I found 504.03 ft3 by multiplying the area of polygon 1 by the depth of the footing. The area

of polygon 1 was 144.01 ft2 (Figure 2) the depth of the footing was 3.5 ft. The formula I used to

find V, or the volume, of the entire piece of concrete with the hollow center was

(V= Area of polygon 1 * height) V = 144.01*3.5 I solved this and got that V = 504.03. The

volume of the entire piece of concrete including the hollow center was 504.03 ft3. The volume of

the hollow center within the concrete was 152.12 ft3. I found 152.12 ft3 by multiplying the area

3.5’

3’

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of polygon 4 by the depth of the footing. The area of polygon 4 was 43.46 ft2 (Figure 5) the

depth of the footing was 3.5 ft. the formula I used to find the volume of the hollow center was

(V = Area of polygon 4 * height) V = 43.46 *3.5 I solved this and found that V = 152.12. The

volume of the hollow center was 152.12 ft3. I then subtracted the volume of the hollow center

from the volume of the entire piece of concrete including the hollow center. The formula I used

to find the volume of the footing without the hollow center was (V= volume of whole concrete –

volume of hollow center) V = 504.03 – 152.12 I solved this and got that V = 351.19. The volume

of the concrete footing was 351.19 ft3. The cost of the concrete was $1,610. The concrete

company installs the concrete for $115 for 27 ft3. So to find the cost of the concrete footing I

divided the volume of the footing by 27, and then rounded up because I cannot buy less than a

bag of concrete. The formula I used to find C, or the cost, of the concrete was C = (volume of

concrete/27*115) C = 351.19/27*115 I solved this and got that C = 1,610. The cost of the

concrete to make the footing was $1,610. The volume of the water in the aquarium was 114.09

ft3. The hollow center was for an aquarium the aquarium was 75% filled with water the aquarium

is on polygon 4. To find the volume of the water I multiplied the volume of the hollow center

by .75. The formula I used to find the volume of the water within the aquarium was (V = Volume

of aquarium*.75) V = 152.12*.75 I solved and found that V = 114.09. The volume of water in

the aquarium was 114.09 ft3.

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Figure 7. Volume of Plexiglas

The figure above shows the Plexiglas floor. The Plexiglas floor covers the aquarium and

polygon 4. The volume of the Plexiglas was the area of polygon 4 multiplied by 4 inches. The

Plexiglas company installs Plexiglas for $1,100 per 32 ft2 X 4”. The area of polygon 4 was 43.46

ft2 (Figure 5). I divided the area of polygon 4 by 32 ft2 and rounded up because I could not buy

less than 1 sheet of Plexiglas. The formula I used to find the cost of the Plexiglas needed to make

the floor was ( C = area of polygon 4/32*1,100) C = (43.46 /32)*1,100 I solved this to get that V

= 2,200. There was no need to multiply by 4 in the equation because the thickness of the

Plexiglas was 4”. The cost of the Plexiglas was $2,200.

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Part Four: One Lateral Face of the Outer Prism

Figure 8. Base and Door

The figure above shows the side length of the base of one lateral face of the tower. I used

the 3.68’ the side length for polygon 2(Figure 3) as the base of one lateral face because the prism

was built on polygon 2. The height of the prism was 7.35 ft. I got 7.35 ft. because the height of

the prism was 2 times the side length for polygon 2. H = (2*3.68) this was the equation I used to

find the height of the prism. I solved the equation and got that H = 7.35 so the height of the prism

was 7.35 ft. Mrs. Copeland wanted a 4X2 ft. door with half a window on the top. The width of

the door was 2 ft. the height was 4 ft. The area of the rectangular part of the door was 8 ft2. I

found the dimensions for the half window by using Sine and Cosine. Sin(18) = y/1 was the

equation I used to find the side length of the window. I solved this and got that y = 0.31. The

answer I got (0.31) is only ½ of the side length so I multiplied it by 2. (0.31*2 = 0.62) the side

length of the window was 0.62 ft. in order to find the apothem of the window I used Cosine.

Cos(180 = x/1 was the formula I used to find the apothem for the window. I solved this equation

and got that x = 0.95. The apothem of the window was 0.95 ft. I found of the window by

14’

20’

20’

3.68’

7.35’

.84’2’

.84’

4’

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multiplying the apothem by the sides and then divided by two, and multiplied by five. The

formula I used was A = (B*H/2)*5 I divided by two because a decagon is made up of triangles

and multiplied by five because there are ten sides in a decagon and ½ of that is five (10 *1/2) =5.

A = (0. 62*0.95/2)*5. I solved this to get that A = 1.47 ft. the area of the ½ window piece was

1.47 ft2. To find the area of the door I just added the two areas together 1.47+8= 9.47 the area of

the door was 9.47 ft2.

Figure 9. Window

I found the dimensions for the window by using Sine and Cosine. Sin(18) = y/1 was the

equation I used to find the side length of the window. I solved this and got that y = 0.31. The

answer I got (0.31) is only ½ of the side length so I multiplied it by 2. (0.31*2 = 0.62) the side

length of the window was 0.62 ft. in order to find the apothem of the window I used Cosine.

Cos(180 = x/1 was the formula I used to find the apothem for the window. I solved this equation

and got that x = 0.95. The apothem of the window was 0.95 ft. I found of the window by

multiplying the apothem by the sides and then divided by two, and multiplied by 10. The formula

I used was A = (B*H/2)*10 I divided by two because a decagon is made up of triangles and

multiplied by ten because there are ten sides in a decagon. A = (0. 62*0.95/2)*10. I solved this to

get that A = 2.94 ft. the area of the window piece was 2.94 ft2. I multiplied this by two because

there were two windows 2.95*2= 5.88. The area of the windows was 5 .88 ft2. I found the lateral

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surface area by first finding the area with the doors and windows and then subtracting the door

and windows off. The formula I used was A = B*H*10. I found the surface area of the whole

prism by doing (7.35*3.68*10) = 270.32. The area of the whole prism with the doors and

windows was 270.32 ft2. I then added 5.88 and 9.47 together to get the area of the windows and

doors 5.88+9.47 = 15.35. The area of the windows and door was 15.35 ft2. I then subtracted

15.35 from 270.32 to get the lateral surface area of the prism excluding the doors and windows

270.32 – 15.35 = 254.97. The lateral surface area of the prism was 254.97 ft2.

Part Five: Volume of the Inner Base Prism

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Figure 10. Volume of Inner prism

The figure above shows the base of the polygonal prism. The base of the prism is

polygon 3. I found the volume of the inner prism by multiplying the area of polygon 3 by the

height of the prism. The area of polygon 3 was 70.48 ft2 (Figure 4).The formula I used was to

find the volume was V = Area of Base*Height. The equation I used to find the volume of the

inner prism was V = 70.48*7.35 I solved this equation and got that V = 518.22 the volume of the

inner prism was 518.22 ft3.

Part Six: Pyramid Top of The Outer Pyramid

14’

14’

20’

20’

3.03’

7.35’

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Figure 11. Outer Pyramid

The figure above shows the outer pyramid. the height of the outer pyramid was 11.03 ft. I

found the height of the outer pyramid by multiplying one side of polygon 2 by 3 (3.68*3)

=11.03. The formula I used to find the slant height was the Pythagorean Theorem (a2+b2 = c2)

11.032+5.662 = c2 I solved this and found that c= 12.40. The slant height of the outer pyramid was

12.40 ft. The formula I used to find the angle measure between the prism base and the pyramid

face was (Tan(Ø) = Opposite/Adjacent) Tan(Ø)=11.03/5.66 I solved this to get that Ø= 62.85°.

The angle between the prism base and pyramid face is 62.85°.

Part Seven: One Lateral Face of Outer Pyramid

11.03’

12.40’

3.68’ 5.66’

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Figure 12. Lateral Face of Outer Pyramid

The figure above shows one lateral face of the outer pyramid. The height of the lateral

triangular face was 12.40 ft. The length of the base of the lateral triangular face was 3.68 ft. The

formula I used to find the angle measures was Tan(Ø)= Opposite/Adjacent Tan(Ø) = 12.40/1.84

I solved this and got that Ø= 81.57°. The angle measures of the two corners are both 81.57°

because the triangular face is isosceles. To find the missing angle I used the formula Angle = 180

– angle 1 –angle 2 Angle = 180-81.57-81.57 I solved this and got 16.86 for the angle. The angle

measures are 81.57°, 81.57°, and 16.86°. The formula I used to find the area of one lateral face

was (A= B*H/2) A=3.68*12.40/2 I solved this and got A = 22.79. The area of one lateral face

was 22.79 ft2. The formula I used to find the lateral surface area was (LSA = A*10) LSA =

22.79*10 I solved this and found that the Lateral Surface Area was 227.85 ft2.

Part Eight: Inner Pyramid

3.68’

12.40’

12.53’12.53’

81.57°81.57°

8.43°8.43°16.86°

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Figure 13. Inner Pyramid

The figure above shows the inner pyramid. The height of the inner pyramid was 9.08 ft.

The formula I used to find the volume of the inner pyramid was V =1/3(Polygon 3*Height) V=

1/3( 70.48*9.08) I solved this and got V = 213.31. The volume of the inner pyramid was

213.31 ft3.

Part Nine: My Tower

9.08’10.20’

3.03’ 4.66’

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Figure 14. My Tower

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The figure above shows a transparent model of my tower. The formula I used to find the

surface area of my tower was (A= Area of Prism + Area of Pyramid) A= 254.97 + 227.85 I

solved this and found that A = 482.82. The total surface area of my tower was 482.82 ft2. The

formula I used to find the total volume of my tower was (V = Volume of Prism + Volume of

Pyramid) V = 518.22 + 213.31 I solved this and got V = 731.53. The total volume of my tower

was 731.53 ft3.

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Part Ten: Conclusion

One problem I encountered with the Tower Project was miscalculation. I subtracted a

foot from the hypotenuse when I should have subtracted it from the altitude. Scot is very happy

with his new doghouse. Mrs. Copeland is happy because Scot is now busy with the aquarium and

staring at Beyoncé. The total volume of my tower was 731.53 ft3. The total surface area of my

tower was 482.82 ft2. The cost of the concrete to make the footing was $1,610. The cost of the

Plexiglas was $2,200. In my opinion the Tower Project is an excellent way to show how

geometry is used in everyday life. The Tower Project allows students to take the stand of an

architect and use math to solve real world problems in possible scenarios.