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Tower

Jillian Maceroni & Yana Pasinos9C GATMr. Acre6-1-15

Maceroni - Pasinos 1Introduction:

One day in May, there was a challenge that was put out there, by a millionaire,

which had to be faced. A tower had to be built by hand and all the dimensions had to be

calculated by one’s mind. Challenge accepted.

A butterfly garden tower was requested to be built to shelter all the various types

of butterflies. It was also asked be built with an aquarium underneath the floor of this

tower so that as the millionaire walks through this garden she could be able to look

down and feel like she is walking on water.

She assigned the builders a 10-sided tower and was asked to build it upon a plot

size of 27x27. She wanted the tower to be built at the most maximized size, still

remaining within the boundaries of the “working area”, and still remaining perpendicular

to the available plot. The main requirements that also had to be met while creating the

tower was an aquarium built into the flooring created with strong foundations, and the

walls one foot thick. There are several more sizing requirements and other standards

that were told to be met.

The tower project is a very intimidating project and we can already anticipate that

there will be a lot of time and effort put into this challenge in order to complete this and

meet all the requirements. With that being said we are up to the challenge and hope

that you are pleased with the outcome of the tower that is being built for you.

Maceroni - Pasinos 2

Part Two:

Figure 1. Base of the Tower in the Plot

Figure 1, above, shows the base of the tower in the plot. The base is made out of

the four polygons that will be shown in the upcoming figures. The polygons are used to

show the footings and the outer and inner walls.

Maximize the plot=plot size-6

Maximized plot=27-6 x 27-6

Maximized plot=21 x 21

Figure 2. How to Maximize the Plot Size

Figure 2 shows how the plot size was maximized. To get the working space,

27 x 27 is subtracted by 3 from each side, which makes it 6, because the tower cannot

be built within three feet of the boundaries of the plot. 21 x 21 feet would be the

maximized plot size.

Central angle=360/n

Central angle=360/10

Central angle=36°

Figure 3. How to Find the Central Angle

Maceroni - Pasinos 3Figure 3 above shows how to find the central angle measure of the polygon. To

find the central angle, 360 is divided by the amount of sides (n) of the polygon. In this

case, 360 is divided by 10 and the central angle is found to be 36°. The central angle is

used to find the bases and the heights for the four polygons.

Figure 4.Outermost Polygon Measurement

Figure 4 shows the outermost polygon or polygon 1. The length of one side and

the height of one triangle in polygon 1 are used to find the area of polygon 1. The base

of the triangle(x in figure 4) is used to find half the side length of the complete polygon.

The height of one triangle (h in figure 4) in this polygon is drawn from the center of the

polygon to one side of the polygon.

Height of Polygon 1 =(hypotenuse of one triangle*cosine of angle)

H of P1= (10.5*(cos(18°))

H of P1≈ 9.99 ft.

Figure 5. Height of Polygon 1

Figure 5 shows how to find the height of polygon 1. First half of the central angle

was found to be 18° and the hypotenuse of the right triangle was previously found to be

10.5. knowing these two measurements the trig function cosine can be used to find the

Maceroni - Pasinos 4adjacent side, or the height of the polygon. When the equation 10.5 cos 18° was solved

the was height found to be approximately 9.99 feet.

Base of Polygon 1 =2(hypotenuse of one triangle*sine of angle)

B of P1= 2(10.5*(sin(18°))

B of P1≈6.49 ft.

Figure 6. One Side Length of Polygon 1

Figure 6 shows how to find one side length of polygon 1, also known as the

outermost polygon. The base of one triangle was needed to be found in order to

complete this. First the central angle of 36°, which is shown in figure 3, was cut in half

and was found to be 18°. The vertex of the polygon or the hypotenuse of the right

triangle is equal to 10.5 ft because it is half of the side of the “working area.” With the

knowledge of those two measurements the trig function sine can be used to find the

opposite side or the base of the triangle. The equation 10.5 sin 18° was multiplied by

two because the triangle is half of the base of the polygon. When the previous equation

is solved the base of the polygon is found to be approximately 6.49 feet.

Area of Polygon 1 =(number of sides*½)(base*height)

A of P1= 10*½(2(10.5*(sin(18°)))(10.5*(cos(18°))

A of P1≈324.02 ft2

Figure 7.How to Find the Area of Polygon 1

Figure 7 shows how to find the area of polygon 1. The base and height of one

triangle was previously found. This means the area formula for a triangle ½ base *

height can be used. Furthermore, that equation was multiplied by ten because the

polygon is made up of ten triangles. When that equation is solved the area is found to

be approximately 324.02 feet2.


Figure 8. Second Outermost Polygon Measurements

Figure shows the second outermost polygon, or other wise known as polygon 2.

The length of one side and the height of one triangle in polygon 2 are used to find the

area of polygon 2. The base of the triangle(x in figure 8) is used to find half of one side

length. The height of one triangle (h in figure 8) in this polygon is drawn from the center

of the to the middle one side of the polygon.

Height of Polygon 2 =(hypotenuse of one triangle*cosine of angle-1)

H of P2=(10.5*cos(18°)-1)

H of P2≈8.99 ft.

Figure 9. How to find the height of polygon 2

Figure 9 shows the height of polygon 2. Polygon 2 has a height of one foot less

than polygon 1. Knowing this, the the equation for the height of polygon 1, 10.5 cos 18°,

was used and 1 was subtracted from it. When the equation for the height, (10.5 cos

18°)-1, was solved it is found to be approximately 8.99 feet.

Base of Polygon 2 =2(tangent of angle*hypotenuse of one triangle*cosine of angle-1)

B of P2=2(tan(18°))*(10.5*cos(18°)-1)

B of P2≈ 5.84 ft.

Maceroni - Pasinos 6Figure 10. One Side Length of Polygon 2

Figure 10 shows how to find one side length of polygon 2, also known as the

second outermost polygon. The base of one triangle was needed to be found in order to

complete this. First the central angle of 36°, which is shown in figure 3, was cut in half

and was found to be 18°. Since the side that was trying to be found was opposite the

angle and the height, or the side adjacent, was previously found to be (10.5 cos 18°)-1

the equation tan 18° ((10.5 cos 18°)-1) can be used to find the base of one right triangle.

That equation then has to multiplied by 2, 2(tan 18°) (10.5 cos 18°)-1, to find the entire

length of the base for polygon 2. When the previous equation is solved the length of the

base is found to be approximately 5.84 feet.


A of P2= 10*½(2(tan(18°)))*(10.5*cos(18°)-1)

A of P2≈262.37 feet2








Figure 12. Third Outermost Polygon Measurements

Figure 12 shows the third outermost polygon, or otherwise known as polygon 3.

The length of one side and the height of one triangle in polygon 3 are used to find the

area of polygon 3. The base of the triangle (x in figure 12) is used to find one side

length. The height of one triangle (h in figure 12) in this polygon is drawn from the

center of the polygon to one side of the polygon.


H of P3=(10.5*cos(18°)-2)

H of P3≈7.99 ft.

Figure 13. How to find the height of polygon 3

Figure 13 shows the height of polygon 3. Polygon 3 has a height of two feet less

than polygon 1. Knowing this, the the equation for the height of polygon 1, 10.5 cos 18°,

was used and 2 was subtracted from it. When the equation for the height, (10.5 cos




B of P3=2(tan(18°))*(10.5*cos(18°)-2)

B of P3≈5.19 ft.

Figure 14. How to find the base of polygon 3

Figure 14 shows how to find one side length of polygon 3. The base of one

triangle was needed to be found in order to complete this. First the central angle of 36°,

which is shown in figure 3, was cut in half and was found to be 18°. Since the side that

was trying to be found was opposite the angle and the height, or the side adjacent, was

previously found to be (10.5 cos 18°)-2 the equation tan 18° ((10.5 cos 18°)-2) can be

used to find the base of one right triangle. That equation then has to multiplied by 2,

2(tan 18°) (10.5 cos 18°)-2, to find the entire length of the base for polygon 3. When the

previous equation is solved the length of the base is found to be approximately 5.19

feet.


A of P3= 10*½(2(tan(18°)))*(10.5*cos(18°)-2)(10.5*cos(18°)-2)

A of P3≈207.23 units2

Figure 15. How to Find the Area of Polygon 3







Figure 16. Fourth Outermost Polygon Measurements

Figure 16 shows the fourth outermost polygon. The length of one side and the

height of one triangle in polygon 4 are used to find the area of polygon 4. The base of

the triangle(x in figure 16) is used to find one side length. The height of one triangle (h in

figure 16) in this polygon is drawn from the center of the polygon to one side of the

polygon.


H of P4=(10.5*cos(18°)-3)

H of P4≈6.99 ft.

Figure 17. How to find the height for polygon 4

Figure 17 shows the height of polygon 4. Polygon 4 has a height of three feet

less than polygon 1. Knowing this, the the equation for the height of polygon 1, 10.5 cos

18°, was used and 3 was subtracted from it. When the equation for the height, (10.5 cos




B of P4=2(tan(18°))*(10.5*cos(18°)-3)

B of P4≈4.54 ft.

Figure 18. How to find the base of polygon 4.

Figure 18 shows how to find one side length of polygon 4. The base of one

triangle was needed to be found in order to complete this. First the central angle of 36°,

which is shown in figure 3, was cut in half and was found to be 18°. Since the side that

was trying to be found was opposite the angle and the height, or the side adjacent, was

previously found to be (10.5 cos 18°)-3 the equation tan 18° ((10.5 cos 18°)-3) can be

used to find the base of one right triangle. That equation then has to multiplied by 2,

2(tan 18°) (10.5 cos 18°)-3, to find the entire length of the base for polygon 4. When the

previous equation is solved the length of the base is found to be approximately 4.54

feet.


A of P4= 10*½(2(tan(18°)))*(10.5*cos(18°)-3)(10.5*cos(18°)-3)

A of P4≈158.58 units2







Part Three:


Figure 20. Dimensions of Footing

Figure 20 shows the dimensions used to construct the footing. It had to be 3 feet

deep and extend from polygon 1 to polygon 4 making it 3 feet wide. The length of the

inner base, or polygon 4. was previously found to have a base of 4.54 feet and the

length of the outer base, or polygon 1, was previously found to have a base of 6.49 feet.

Volume of Footing= (A of P1*height of prism)-(A of P4*height of prism)

V of footing= (324.02*3.5)-(158.58*3.5)

V of footing= 1134.06-555.03

V of footing≈579.03 ft3

Figure 21. Volume of the Concrete Needed for Footing

Figure 21 shows how to find the volume of the footing. The footing extends from

polygon one to polygon four. The depth of the footing which is labeled in figure 20, was

3.5 feet deep, so that was used as the height of the prism. The volume of the footing

was found by finding the product of the area of polygon one and the height of the

prism(A of P1*height of prism), then finding the product of the area of polygon four and

the height of the prism(A of P4*height of prism). Once those two were found, they were

subtracted from each other((A of P1*height of prism)-(A of P4*height of prism)) and lead

Maceroni - Pasinos 12us to finding the volume of the footing which told us the volume of the concrete needed

for the footing. The volume of the footing was found to be approximately 579.03 feet3.

Figure 22. Dimensions of Flooring

Figure 22 shows the dimensions used to construct the floor. The floor had to

extend to polygon 4 making the length of the base the same as polygon 4, 4.54 ft. The

was also required to be 4 inches thick.

Volume of Floor= A of P4*Hprism

V of floor=158.58*1/3

V of floor≈52.86 ft3

Figure 23.Volume of the Plexiglass needed for Floor

Figure 23 shows the volume of the floor. The floor extended to polygon 4 and

was 4 inches thick, as shown in figure 22. 4 inches then had to be converted into feet

making the height ⅓ of a foot. Knowing that information, the volume formula for prisms

(area of base)(height of prism) can be used. The area of polygon 4 which was

previously found in figure 14 was 158.86 and when multiplied by ⅓ the volume of the

floor was found to be approximately 52.86 feet3.


Figure 24. Dimensions of Aquarium

Figure 24 shows the dimensions needed to construct the aquarium. The

aquarium had to be located inside the footing of polygon 4, making it 3.5 feet deep and

a side length of 4.54 feet.

Volume of Water= (75%*Hprism)*A of P4

V of water=(0.75*3.5)158.58

V of water≈416.27 ft3

Figure 25. Volume of Water Needed for Aquarium

Figure 25 shows the volume of the water needed for the aquarium. The aquarium

needed to be 75% filled. To find the water needed, 75% of the height of the aquarium

which was 3.5 feet (labeled in figure 24) was calculated and then multiplied by the area

of polygon 4 which was previously found to be 158.58 in figure 14. When the equation,

(0.75*3.5) 158.58 was solved the volume of water needed for the aquarium was found

to be approximately 416.27 feet3.

Cost of Concrete= (Vfooting*115)/27


Cost of Concrete= (579.03*115)/27

Cost of Concrete¿$2530

Figure 26. Cost Analysis of Concrete needed

Figure 26 shows how to find the cost of the concrete. The concrete is sold for

$115 per cubic yard. To find the cost of the concrete, the volume of the footing which

was previously found in figure 21 to be 579.03, was multiplied by the cost of the

concrete, 115. That was then rounded up because part of a bag of concrete and divided

by 27 to convert it into feet. When the math was solved the cost of the concrete was

$2530.

Cost of Plexiglas=(A of P4/32)*1100

Cost of Plexiglas=(158.58/32)*1100

Cost of Plexiglas¿4.96*1100

Cost of Plexiglas¿$5500

Figure 27. Cost Analysis of Plexiglas needed

Figure 27 shows how much the plexiglass would cost. For each 48” x 96” x 4”

sheet of plexiglass the cost was $1100. Since the plexiglass being sold and the height

of the plexiglass for floor had the same thickness, the width can be disregarded. The

plexiglass that was sold needed to converted from inches into feet. That was done

dividing 48 and 96 by 12 making it 4’ x 8’. When 4 and 8 are multiplied together it is a

total of 32 feet. Now knowing this information the area of polygon 4 which was 158.58

can be divided by 32 and multiplied by the cost which was 1100.The cost was rounded

up to 5500 because part of a sheet cannot be purchased.

Part Four:


Figure 28. Dimensions of the Door

Figure 28 shows the dimensions needed to construct the door. The base and

height of the door were required to be 3 feet and 5 feet respectively. The radius of the

polygon was half the length of the base of the door, 1.5 feet, because the vertices

touch. The side length of the polygon of the polygon was 0.93 feet. How that number

was calculated will be explained later.

Area of the door= (base*height)+(5½base*height)

Adoor=(5*3)+(5½*0.93*1.43)

Adoor≈18.31 ft2

Figure 29. Area of the Door

Figure 29 shows how to find the area of the door. The length of the door needed

to be 3 feet and the height needed to be 5 feet. Half of a decagon needed to top the

door as well. First the area of the top of the door needed to be found. To do this the

base and height of the polygon needed to be found. The radius of the polygon was 1.5

feet because it was half of the length of the door. Half of the central angle which, was

found to be 18, was also used. Knowing that information, the equation 1.5 cos 18 can

be used to find the height of the polygon on top of the door. As shown in figure 28, the

Maceroni - Pasinos 16height of the polygon on top of the door was found to be about 1.43 feet. Also knowing

this information, the equation 1.5 sin 18 can be used to find the length of the base of the

triangle. When 1.5 sin 18 is solved the length was found to be 0.46. When multiplied by

2 to find the length of the base for polygon it was found to be 0.93 feet (side length of

polygon shown in figure 28). Once the base was found, the equation ½ base*height can

be used and multiplied by 5 creating the 5 sided polygon used. That was then added to

the base of the door 5*3 to find the area of the door to be 18.30 feet2.

Figure 30. Dimensions of Window

Figure 30 shows the dimensions used to construct the windows. The side length

and height of the polygon were found to be 0.93 feet and 1.43 feet respectively. How

those numbers were calculated will be explained later in figure 31.

Area of window= ½*base*height

Awindow=10(½*0.93*1.43)

Awindow= 6.61*2


Awindow≈13.23 ft2

Figure 31. Area of the Window

Figure 31 shows how to find the area of the window. To get the area of the

window, you would simply use the equation ½*base*height. The equation

½*base*height is used to find the area of triangles. Since the window had to be a 10

sided polygon, it was broken up into 10 triangles.The radius of the polygon was 1.5 feet

because it was half of the length of the door. Half of the central angle which was found

to be 18 was also used. Knowing that information, the equation 1.5 cos 18 can be used

to find the height of the window. When that equation was solved the height was found to

be 1.43. Also knowing this information, the equation 1.5 sin 18 can be used to find the

length of the base of the triangle. When that equation is solved the length was found to

be 0.46. When multiplied by 2 to find the length of the base for polygon it was found to

be 0.93 feet. After finding the height and the base of the window, it was plugged into

the equation ½*base*height. Since there were 10 triangles, the overall equation had had

to by 10. Once that was solved the area one one window was found to be 6.61. It was

required to have at least 2 windows so the area of one window, 6.61 was multiplied by

2. The area of both windows was found to be approximately 13.23 feet2.


Figure 32. Outer Base Polygon and One Outer Lateral Wall Dimensions

Figure 32 shows the outer polygon used for the base with one outer lateral wall

and their dimensions. The length of the base for the outer polygon, or polygon 2, was

5.84 feet. That was then doubled to find the height of the walls to be 11.68 feet.

Lateral Surface Area= 10(base*height)-(Adoor+Awindow)

LSA= 10(5.84*11.68)-(18.31+13.23)

LSA= 682.00-31.53

LSA≈ 650.47 ft2

Figure 33. Lateral Surface Area of the Outer Prism

Figure 33 shows how to find the lateral surface area of the outer prism. To find

this, the equation 10(base*height)-(Adoor+Awindow) was used. To find the base, we took

the base of polygon two. The height of the prism was found by doubling the length of

the base, 5.84, which was found to be 11.68 (shown in figure 32). When the product of

the base and height was found, it was then multiplied by ten because there were 10

sides to the polygon. The area of the door was found in figure 29 and the area of the

window was found in figure 31. Since the area of the door and the window was

Maceroni - Pasinos 19supposed to be removed when calculating the lateral surface area, the area of the door

and window were added together and then subtracted from the equation

10(base*height). The lateral surface area of the outer prism was found to be

approximately 650.47 feet2.

Part Five:

Figure 34. Dimensions of Inner Prism

Figure 34 shows the dimensions used to construct the inner walls. The inner

walls extended to polygon 3 making the length of the base 5.19 feet. The height was the

same used in figure 31 at 11.68 feet.


Figure 35. Dimensions of Lateral Wall of Inner Prism

Figure 35 shows the dimensions used to construct the lateral wall of the inner

prism. The inner walls were constructed on polygon 3 making the length of the base

5.19 feet. The height was the same as the outer walls used in figure 31.

Volume= Abase*Hprism

V=207.23*11.68

V≈2420.20 ft3

Figure 36. Volume of the Inner Prism

Figure 36 shows the volume of the inner prism. The area of the inner base, or

polygon 2 was previously found to be 207.23. To find the height of the prism, the length

of the base 5.84 was doubled to 11.68. That was then multiplied by the area of the

base, 207.23, to find the volume of the inner prism to be 2420.20 feet3.

Part Six:


Angle Measure:cosΘ=adjacent/hypotenuse

cosΘ=8.99/19.69

cosΘ=cos-1(8.99/19.69)

Θ≈62.84°

Figure 37. Angle Measure Between the Prism Base and the Pyramid Face Found at the Foot of the Slant Height

Figure 37 shows the angle between the base and the roof. To find this, the trig

function cosine was used. The height of polygon 2, 8.99, was used as the adjacent side

and the slant height of the outer pyramid, 19.69, was used as the opposite side. when

the equation cos-1(8.99/19.69) was solved the angle was approximately 62.84°.

Figure 38. Dimensions of Outer Pyramid

Figure 38 shows the dimensions used to construct the outer pyramid of the roof.

The height was 17.52 feet and the slant height was 19.69 feet. How these

measurements were calculated will be explained later.

Slant Height of Outer Pyramid= H of P2/cosine of angle


Slant Height of Outer Pyramid= 8.99/ cos(62.84)

Slant Height of Outer Pyramid≈19.69 ft

Figure 39. Slant Height of Outer Pyramid

Figure 39 shows the slant height of the outer pyramid. To find the slant height the

trig function cosine was used. The angle between the roof and the base which was

previously found to be 62.84 was used along with the height of polygon 2, or the

adjacent side, previously found to be 8.99. When the equation 8.99/cos(62.84) was

solved the slant height of the outer pyramid was found to be approximately 19.69 feet.

Height =B of P2*3

Height= 5.84*3

Height= 17.52 ft.

Figure 40. Height of Outer Pyramid

Figure 40 shows the height of the outer pyramid. The base of polygon 2 which

was 5.84 was multiplied by 3 to find height to be 17.52.

Part Seven:


Figure 41. Dimensions of One Outer Lateral Face of the Pyramid

Figure 41 shows the dimensions used to construct one lateral face of the outer

pyramid. The outer pyramid extends to polygon 3 making the length of the base 5.84

feet. The height of the triangle was found to 19.69 feet which will be explained how that

was calculated later. It also shows that angle measures of the face. Angle A was

approximately 16.87° and angle B and angle C were found to be approximately 81.56°.

An explanation of how those angles were calculated will also be latter in the paper.

ΘAngle A= tan-1(½*base/ height)

ΘAngle A= tan-1(2.92/19.69)

ΘAngle A≈16.87°

Figure 42. One Angle Measure of the Lateral Face

Figure 42 shows how to find the measure of angle A. To find this angle the trig

function tangent was used. Half of the base, 2.92, was used for the opposite side and

the height of the triangle, 19.69, was used for the adjacent side. When the equation tan -1

(2.92/19.69) was solved the measure of angle A was found to be approximately 16.87°.


ΘAngle B= tan-1( height/ ½*base)

ΘAngle B= tan-1(19.69/2.92)

ΘAngle B≈81.56°

Figure 43. Second Angle Measure of the Lateral Face

Figure 43 shows how to find the angle measure for angle B. to find this angle the

trig function tangent was used. The height of the triangle, 19.69, was used for the

opposite side and half of the base, 2.92, was used for the adjacent side. When the

equation tan-1 (19.69/2.92) was solved the measure of angle B was found to be

approximately 81.56°

ΘAngle C= tan-1( height/ ½*base)

ΘAngle C= tan-1(19.69/ 2.92)

ΘAngle C≈81.56°

Figure 44. Third Angle Measure of the Lateral Face

Figure 44 shows how to find the angle measure for angle C. Since the lateral

face being used is isosceles the equation and angle are the same. When that height

and half of the base are plugged into inverse tangent the equation is tan-1(19.69/ 2.92).

When the equation is solved the measure of angle C is found to be approximately

81.56°.

Area of one Lateral Face=½(b*h)

A=½(5.84*19.69)

A≈57.49 ft2

Maceroni - Pasinos 25Figure 45. Area of One Lateral Face of the Outer Pyramid

Figure 45 shows how to find the area of one lateral face of the outer pyramid. To

find this, the equation ½(b*h) was used because that is how you find the area of a

triangle. The base of polygon two was used as the base. The slant height of the outer

pyramid, which was found in figure 39, was used as the height to find one lateral face of

the outer pyramid. Once the product of the base and height was found, it was then

multiplied by ½ . The area of one lateral face of the outer pyramid was found to be

approximately 57.49 feet2.

Lateral Surface Area= 10*½(base*height)

LSA=10*½(5.84*19.69)

LSA≈574.87 ft2

Figure 46. Lateral Surface Area of the Outer Pyramid

Figure 46 shows how to find the lateral surface area of the outer pyramid. To find

this, the equation 10*½(base*height) was used. Since we had a 10 sided polygon, We

had 10 triangles in order to make up the pyramid for this tower. The area of one lateral

face of the outer pyramid, which was found in figure 45, was multiplied by 10 because

there was a total of 10 triangles. The lateral surface area of the outer pyramid was

found to be approximately 574.87 feet2.

Part Eight:


Figure 47. Dimensions of Inner Pyramid

Figure 47 shows the dimensions of the inner pyramid of the roof. The inner

pyramid extended to polygon 3 making the length of the base 5.19 feet. The height was

then required to be 3 time the length of the base. When the base was tripled the height

was found to be 15.57 ft. This was shown in figure 48 below.

Height of the Inner Pyramid= 3* B of P3

Height of the Inner Pyramid=3*5.19

Height of the Inner Pyramid≈15.57 ft.

Figure 48. Height of Inner Pyramid

Figure 48 shows how to find the height of the inner pyramid. To find the height of

the pyramid the base of polygon 3 which was 5.19, was multiplied by 3 to find the height

to be approximately 15.57 feet.

Volume= ⅓(Abase*Hpyramid)


V= ⅓(207.23*15.57)

V≈1075.44 ft3

Figure 49. Volume of the Inner Pyramid

Figure 49 shows the volume for the inner pyramid. To find the volume of the

pyramid, the formula ⅓ (area of base)(height of pyramid) can be used. The area of the

base of polygon 3 was previously found to be 207.23 and the height was previously

found to be 15.57. When the equation ⅓ (207.23*15.57) was solved the volume of the

inner pyramid was found to be approximately 1075.44 feet3.

Part Nine:

Figure 50. Completed Tower

Figure 50 shows how the tower will look when the walls and roof are built.

LSA of Outer Tower= LSA of outer prism + LSA of outer pyramid


LSA of Outer Tower= 650.47+574.87

LSA of Outer Tower≈1225.33 ft.2

Figure 51. Lateral Surface Area of Outer Tower

Figure 51 shows how to find the lateral surface area of the outer tower. To find

the lateral surface area of the outer tower, the lateral surface area of the outer prism

and the lateral surface area of the outer pyramid were added together. The solution of

the lateral surface area of the outer prism was found in figure 33 and the lateral surface

area of the outer pyramid was found in figure 46. The lateral surface area of the outer

tower was found to be approximately 1225.33 feet2.

Volume of Inner Tower= Volume of inner prism + Volume of inner pyramid

Volume of Inner Tower= 2420.2 + 1075.44

Volume of Inner Tower≈3495.64 ft.3

Figure 52. Volume of Inner Tower

Figure 52 shows how to find the volume of the inner tower. To find the volume of

the inner tower, the volume of the inner prism and the volume of the inner pyramid

were added together. The solution of the volume of the inner prism was found in figure

36 and the volume of the inner pyramid was found in figure 49. The volume of the inner

tower was found to be approximately 3495.64 feet3.

Maceroni - Pasinos 29Conclusion:

Acknowledging the fact that there was such a large amount of space in the

interior of the tower, which is approximately 3495.64 feet3, there will be a lot of room

available to redesign a beautiful interior and add in some decorations of your choosing.

There was also such a large amount of space outside of the tower that is entitled to be

decorated. There is approximately an area of 1225.33 feet2 to be decorated. The scale

model was designed with a theme, but the millionaire out there is entitled to switch

things and redesign the tower to their liking. This tower is probably one of the best

butterfly garden towers that have been built and designed. In order to build this tower, it

took thousands of dollars to put up. The money should be reimbursed and the builders

should also be paid a significant amount of profit.

While calculation the math used to build this tower for the measurements and the

costs, there were some issues that were encountered. One of the issues was plugging

in the incorrect numbers, for example the door and the window. For the height, we

plugged in the measurement for the radius, but we were really supposed to find cosine

of the angle. Another issue is that since the math was continuous and you had to reflect

off of other calculations that were already done, we had to change the other calculations

that reflected off of that certain issue. This lead to the answer being off by a couple

decimals or even whole numbers. There were also problems with building the tower

itself. While drawing the outline of the plot the wrong dimensions were used at first. The

whole plot had to been trashed and restarted. The last problem that was encountered

while building the tower occurred while trying to build the cross section. The glue that

Maceroni - Pasinos 30was being used was not holding the dowels in place. Multiple types of glue were used

until the resolution of using pipe cleaners was reached.

So there you have it. The butterfly tower has been successfully created. We

hope that the millionaire enjoys the tower and tells all of her friends about us. We love

to do custom work!

Maceroni - Pasinos 31Works Cited

"Butterflies - Daytime Dreamy Winged Creatures." Butterflies. Web. 31 May 2015.<http://www.factzoo.com/insects/butterflies-daytime-dream-winged-creatures.html>.

"Butterflies Wallpaper: Butterfly Wallpaper." Butterfly Wallpaper. Web. 31 May 2015. <http://www.fanpop.com/clubs/butterflies/images/7451076/title/butterfly-wallpaper-wallpaper>.

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