The National Student Research Center
E-Journal of Student Research: Math
Volume 5, Number 1, March, 1998
The National Student Research Center
is dedicated to promoting student research and the use of the
scientific method in all subject areas across the curriculum,
especially science and math.
For more information contact:
- John I. Swang, Ph.D.
- Founder/Director
- National Student Research Center
- 2024 Livingston Street
- Mandeville, Louisiana 70448
- U.S.A.
- E-Mail: nsrcmms@communique.net
- http://youth.net/nsrc/nsrc.html
TABLE OF CONTENTS
- Is The Formula For Finding the Volume
of a Rectangular Prism Always Right?
- Does The Formula For Finding The
Perimeter Of A Rectangle Really Work?
- Do The Sum Of The Angles In A Four
Sided Polygon Equal 360 Degrees?
- Does the Formula SA = 2(Pi x r2)
+ 2(Pi x rh) Really Work For All.Cylinders?
- Does The Formula For Finding The
Surface Area Of A Triangular Prism Really Work?
- Is The Theory Of Probability Correct?
- Is The Formula For Finding The Value
of Pi Really Accurate?
- Does Euler's Formula Work?
TITLE: Is The Formula For Finding the Volume of a Rectangular
Prism Always Right?
STUDENT RESEARCHER: Jack Bell and Adam Osborn
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
We would like to do mathematical research project to see if the
formula of length times width times height always equals the
volume of a rectangular prism, no matter how large or small the
prism. Our hypothesis states that the formula, V=LxWxH, always
equals the volume of a rectangular prism no matter how large or
small it is.
II. METHODOLOGY:
First, we chose our topic. Next, we wrote our statement of
purpose. Then we reviewed the literature about volume,
geometry, prism, cube, rectangular prism, mathematics, and
Archimedes. After that we developed our hypothesis and wrote
our methodology to test our hypothesis. Then we identified our
variables. We then gathered our materials. Next, we found five
objects of different sizes having the shape of a rectangular
prism. Then we used the formula, V=LxWxH, to find the volume of
the eight objects. After that we placed them, one at a time, in
a bucket completely full of water. The bucket was placed in a
metal tray. The water displaced by the rectangular object was
collected in the tray. Next, we measured how many milliliters
of water each object displaced. This represented the volume of
the object. One milliliter was equal to one cubic centimeter.
Then we recorded the results on our data collection sheet. Then
we compared the value for the two ways for computing volume.
Next we wrote our analysis of data. Last, we wrote our summary
and conclusion, and application.
III. ANALYSIS OF DATA:
Our data show that when we used the formula V=LxWxH to find the
volume of our first rectangular prism, it was 515 cubic
centimeters. When we put our first rectangular prism in the
water, it displaced 525 milliliters of water, which indicated
that the volume was 525 cm3. The difference of 10 cm3 between
the two volumes is due to measurement error.
Our data show that when we used the formula V=LxWxH to find the
volume of our second rectangular prism, it was 15 cm3 cubic
centimeters. When we put our second rectangular prism in the
water, it displaced 17 milliliters of water, which indicated
that the volume was 17 cm3. The difference of 2 cm3 between the
two volumes is due to measurement error.
Our data show that when we used the formula V=LxWxH to find the
volume of our third rectangular prism, it was 929 cubic
centimeters. When we put our third rectangular prism in the
water, it displaced 975 milliliters of water, which indicated
that the volume was 975 cm3. The difference of 46 cm3 between
the two volumes is due to measurement error.
Our data show that when we used the formula V=LxWxH to find the
volume of our fourth rectangular prism, it was 790 cubic
centimeters. When we put our fourth rectangular prism in the
water, it displaced 800 milliliters of water, which indicated
that the volume was 800 cm3. The difference of 10 cm3 between
the two volumes is due to measurement error.
Our data show that when we used the formula V=LxWxH to find the
volume of our fifth rectangular prism, it was 2072.67 cubic
centimeters. When we put our fifth rectangular prism in the
water, it displaced 2000 milliliters of water, which indicated
that the volume was 2000 cm3. The difference of 72.67 cm3
between the two volumes is due to measurement error.
Our data show that when we used the formula V=LxWxH to find the
volume of our sixth rectangular prism, it was 848.25 cubic
centimeters. When we put our sixth rectangular prism in the
water, it displaced 900 milliliters of water, which indicated
that the volume was 900 cm3. The difference of 51.75 cm3
between the two volumes is due to measurement error.
Our data show that when we used the formula V=LxWxH to find the
volume of our seventh rectangular prism, it was 2808 cubic
centimeters. When we put our seventh rectangular prism in the
water, it displaced 2700 milliliters of water, which indicated
that the volume was 2700 cm3. The difference of 108 cm3 between
the two volumes is due to measurement error.
Our data show that when we used the formula V=LxWxH to find the
volume of our eighth rectangular prism, it was 164.375 cubic
centimeters. When we put our fourth rectangular prism in the
water, it displaced 150 milliliters of water, which indicated
that the volume was 150 cm3. The difference of 14.375 cm3
between the two volumes is due to measurement error.
| Object | Tested Volume | Calculated Volume | Difference |
| object 1| 525 cm3 | 515 cm3 | 10 cm3 |
| object 2| 17 cm3 | 15 cm3 | 2 cm3 |
| object 3| 975 cm3 | 929 cm3 | 46 cm3 |
| object 4| 800 cm3 | 790 cm3 | 10 cm3 |
| object 5| 2000 cm3 | 2072 cm3 | 72 cm3 |
| object 6| 900 cm3 | 848 cm3 | 51 cm3 |
| object 7| 2700 cm3 | 2808 cm3 | 108 cm3 |
| object 8| 150 cm3 | 164 cm3 | 14 cm3 |
|Average Difference| | 39 cm3 |
IV. SUMMARY AND CONCLUSION:
The average difference between the volume of our rectangular
prisms when we used the formula, V=LxWxH, and when we used the
displacement of water was 39.35 cm3. The difference was due to
measurement error.
Our data show that the formula V=LxWxH does work. Therefore, we
accept our hypothesis which states that the formula, V=LxWxH,
always equals the volume of a rectangular prism no matter how
large or small it is.
This research needs to be repeated in such a way as to
significantly reduce the measurement error.
V. APPLICATION:
We could apply our findings to the world by telling people that
V=LxWxH gives you the correct volume for a rectangular prism.
This could help shipping companies to figure out how much of
their goods they could fit in the cargo holds of the trucks,
trains, and ships.
TITLE: Does The Formula For Finding The Perimeter Of A
Rectangle Really Work?
STUDENT RESEARCHER: Lalita Mondkar and Barrett Ainsworth
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
We would like to do a mathematical proof to see if the formula
for finding the perimeter of a rectangle, P=2(l+w), really
works. Our hypothesis states that the formula for finding the
perimeter of a rectangle, P=2(l+w), really works.
II. METHODOLOGY:
First, we chose our topic. Next, we wrote our statement of
purpose. Then we conducted a review of literature about
mathematics, perimeter, and rectangles. Next, we wrote our
hypothesis. We then developed our methodology to test our
hypothesis. Next, we made a data collection sheet. Then we
made six different sized rectangles using Lego's. Then we
measured the length and width and used the formula to find the
perimeter of a rectangle. Next, after using the formula, we
covered rectangle with graph paper. Then we counted the squares
on each side and added them together. We wrote our findings on
our data collection sheet. Then we compared the values for
perimeter found by both methods. Then we analyzed our data
using statistics, charts, and graphs. Next, we wrote our
summary and conclusion where we accepted or rejected our
hypothesis. Finally, we applied our findings to the world
outside the classroom.
III.ANALYSIS OF DATA:
Does The Formula For Finding The Perimeter Of A Rectangle Really
Work?
| | Formula P=2(l+w) | Sides Added |
| |Length | Width|Perimeter| Length| Width|Perimeter|Diff|
| 1 | 10.67 | 8.00 | 37.34 | 10.67 | 8.00 | 37.34 | 0 |
| 2 | 13.00 | 6.40 | 38.80 | 13.00 | 6.40 | 38.80 | 0 |
| 3 | 6.20 | 4.00 | 20.40 | 6.20 | 4.00 | 20.40 | 0 |
| 4 | 6.40 | 4.80 | 22.40 | 6.40 | 4.80 | 22.40 | 0 |
| 5 | 6.60 | 6.00 | 25.20 | 6.60 | 6.00 | 25.20 | 0 |
| 6 | 8.00 | 2.60 | 19.20 | 8.00 | 2.60 | 19.20 | 0 |
In rectangle one, when the length was 10.67 cm and the width was
8.00 cm the perimeter was 37.34 cm when using the formula. When
the sides were measured and added together the perimeter was
37.34 cm. In rectangle two, when the length was 13.00 cm and
the width was 6.40 cm, the perimeter was 38.80 cm when using the
formula. When the sides were measured and added together the
perimeter was 38.80 cm. In rectangle three, when the length
was 6.20 cm and the width was 4.00 cm the perimeter was 20.40 cm
when using the formula. When the sides were measured and added
together the perimeter was 20.40 cm. In rectangle four, when
the length was 6.40 cm and the width was 4.80 cm the perimeter
was 22.40 cm when using the formula. When the sides were
measured and added together the perimeter was 22.40 cm. In
rectangle five, when the length was 6.60 cm and the width was
6.00 cm the perimeter the perimeter was 25.20 cm when using the
formula. When the sides were measured and added together the
perimeter was 25.20 cm. In rectangle six, when the length was
8.00 cm and the width was 2.60 cm the perimeter was 19.20 cm
when using the formula. When the sides were measured and added
together the perimeter was 19.20 cm.
IV. SUMMARY AND CONCLUSION:
Our data indicate that the formula, P=2(l+w), will give a true
measurement of the perimeter of a rectangle. Therefore we
accept our hypothesis which states that the perimeter of a
rectangle is equal to P=2(l+w).
V. APPLICATION:
Our findings can be applied in the classroom by using the
formula on tests and experiments and also by informing teachers
that the formula, P=2(l+w), is correct.
TITLE: Do The Sum Of The Angles In A Four Sided Polygon Equal
360 Degrees?
STUDENT RESEARCHER: George Davis McPherson, Jr
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
I would like to do a mathematical research project to see if the
sum of all angles in a polygons equal 360 degrees. My
hypothesis states that the angles of a four sided polygon will
add up to 360 degrees.
II. METHODOLOGY:
First, I chose my topic. Then I wrote a statement of purpose.
I then conducted a review of literature about topics having to
do with angles, lines, and such. Next, I came up with a
hypothesis. Then I developed a methodology to test my
hypothesis. Then I gathered the materials necessary to conduct
my experiment. Then I drew a polygon on graph paper, measured
the angles, and added them together. Then I repeated this
process with 6 other polygons. I used a rectangle, square,
rhombus, parallelogram, octagon, pentagon, and trapezoid. Next,
I recorded the results on a data collection sheet. I then
analyzed my data. Next, I accepted or rejected my hypothesis,
wrote a brief summary and conclusion, and applied my findings to
the world outside of the classroom.
III. ANALYSIS OF DATA:
Total of all angles in each polygon.
|Shapes |Number Of Angles | Total Of Degrees |
|Rectangle | 4 | 360 |
|Square | 4 | 360 |
|Rhombus | 4 | 360 |
|Parallelogram | 4 | 360 |
|Octagon | 8 | 1084 |
|Pentagon | 5 | 544 |
|Trapezoid | 4 | 360 |
In my research, I found that the angles of each of the four
sided shapes added up to 360 degrees, but not for any of the
other shapes.
IV. SUMMARY AND CONCLUSION:
While conducting my experiment I found that the formula stating
that the sum of the angles of a quadrilateral is 360 degrees
proves to be true. This is not true for other polygons such as
the octagon with eight angles and the pentagon with five angles.
Therefore I accept my hypothesis which stated that the formula
would work only with four sided shapes and not with other
shapes.
V. APPLICATION:
If I have the measurements of three of the angles of a
quadrilateral, I could add them together and then subtract them
from 360 to find the measurement of the fourth angle.
TITLE: Does the Formula SA = 2(Pi x r2) + 2(Pi x rh) Really
Work For All Cylinders?
STUDENT RESEARCHERS: Whitney Stoppel and Jack Bell
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
We would like to do a mathematical proof to see if the formula
for finding the surface area of a cylinder SA = 2(Pi x r2) +
2(Pi x rh) is correct. Our hypothesis states that the formula
for finding the surface area of a cylinder, SA = 2(Pi x r2) +
2(Pi x rh) is correct.
II. METHODOLOGY:
First, we chose our topic. Next, we wrote our statement of
purpose. Then we reviewed the literature about cylinders,
mathematics, geometry, circles, Pi, radius, diameter, and
surface area. After that we developed our hypothesis.
Then we wrote a methodology to test our hypothesis. Then we
identified our variables and gathered our materials. Next, we
found five cylinders of different sizes. Then we used the
formula SA = 2(Pi x r2) + 2(Pi x rh) to find the surface area
of the ten cylinders. After that, we wrapped graph paper around
the cylinders and both of the ends. Then we counted all of the
square centimeters on the graph paper that covered the cylinder.
Then we recorded the results on our data collection sheet.
Next, we compared the values for the surface area which we
obtained with the formula to the value we obtained with the
graph paper. Next we wrote our analysis of data where we
compared the surface areas calculated from the two procedures
for each cylinder. Last, we wrote our summary, conclusion, and
application.
III. ANALYSIS OF DATA:
For cylinder one, the surface area was 356.5 square centimeters
when counted with the graph paper and 334.60 square centimeters
when calculated with the formula 2SA = 2(Pi x r2) + 2(Pi x rh).
The difference was 21.9 square centimeters. For cylinder two,
we counted 93.52 square centimeters and calculated 89.68 square
centimeters. The difference was 3.84 square centimeters. For
cylinder three, we counted 436.04 square centimeters and
calculated 422.80 square centimeters. The difference was 13.24
square centimeters. For cylinder four, we counted 41.36 square
centimeters and calculated 50.554 square centimeters. The
difference was 9.194 square centimeters. For cylinder five, we
counted 32.028 square centimeters and calculated 35.52 square
centimeters. The difference was 3.492 square centimeters.
Surface Area of Cylinders Using Two Methods
(cm2)
| Cyl | Formula | Graph Paper | Difference |
| 1 | 334.60 | 356.50 | 21.90 |
| 2 | 89.68 | 93.52 | 3.84 |
| 3 | 422.80 | 436.04 | 13.24 |
| 4 | 50.55 | 41.36 | 9.19 |
| 5 | 32.02 | 35.52 | 3.49 |
|Average Difference | 10.33 |
IV. SUMMARY AND CONCLUSION
We found that this formula does work, therefore we accept our
hypothesis which stated that the formula to find the surface
area of a cylinder, SA = 2(Pi x r2) + 2(Pi x rh), will always
work for any size of cylinder. The small difference between
the counted and calculated values for the cylinder's surface a
rea was due to the inaccuracy of our measurements using a ruler
and graph paper. The research should be done again using more
accurate instruments of measurement.
V. APPLICATION:
We are able to apply our findings by using this information when
we are in school. We now know that this formula really does
work. Another time this finding can be used is when a
construction worker wants to find out how much wood he needs for
a cylindrical building.
TITLE: Does The Formula For Finding The Surface Area Of A
Triangular Prism Really Work?
STUDENT RESEARCHERS: Chris Chugden and Adam Osborn
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
We would like to do a mathematical research project to see if
the formula for calculating the surface area of a triangular
prism, SA = (bh1+bh2+bh3) + 2(1/2bh), is always correct. Our
hypothesis states that the formula for calculating the surface
area of a triangular prism, (bh1 + bh2 + bh3) + 2(1/2bh), will
always work no matter what the size of the triangular prism is.
II. METHODOLOGY:
First, we chose a topic. Next, we wrote our statement of
purpose. Then we conducted a review of literature about
mathematics, geometry, triangular prisms, and prisms. Then we
developed a hypothesis.
Next, we developed a methodology to test our hypothesis. First,
we gathered our materials. Next, we outlined an unfolded
triangular prism on a sheet of graph paper. Then we counted the
square centimeters inside the outline of the triangular prism.
We also calculated the surface area using the formula. Then we
compared and calculated the difference between the two surface
areas. We repeated this procedure with 7 more prisms of
different sizes.
We recorded our data in a systematic way on a data collection
sheet and we analyzed our data using simple statistics, charts,
and graphs. Next, we wrote our summary and conclusion where we
accepted or rejected our hypothesis. Finally, we applied our
findings to the world outside the classroom.
III. ANALYSIS OF DATA:
Triangular prism number one had a surface area of 78 cm2 when
calculated with the formula. It's surface area with manual
measurement was also 78 cm2. Triangular prism number two had a
surface area of 63 cm2 when calculated with the formula. It's
surface area with manual measurement was also 63 cm2.
Triangular prism number three had a surface area of 37 cm2 when
calculated with the formula. It's surface area with manual
measurement was also 37 cm2. Triangular prism number four had a
surface area of 98 cm2 when calculated with the formula. It's
surface area with manual measurement was also 98 cm2.
Triangular prism number five had a surface area of 26 cm2 when
calculated with the formula. It's surface area with manual
measurement was also 26 cm2. Triangular prism number six had a
surface area of 19 cm2 when calculated with the formula. It's
surface area with manual measurement was also 19 cm2.
Triangular prism number seven had a surface area of 20 cm2 when
calculated with the formula. It's surface area with manual
measurement was also 20 cm2. Triangular prism number eight had
a surface area of 56 cm2 when calculated with the formula. It's
surface area with manual measurement was also 56 cm2. All the
differences in the calculations were zero.
The Manual And Calculated Surface Area Of The Triangular Prisms
| Triangular Prism | Area Formula | Area Manual | Diff |
|2(7x3)+7x4+2(.5x1x2) | 78 cm | 78 cm | 0 |
|2(3x4.5)+3x6+2(.5x6x3) | 63 cm | 63 cm | 0 |
|2(7x1.5)+7x2+2(.5x1x2) | 37 cm | 37 cm | 0 |
|2(9x3)+9x4+2(.5x2x4) | 98 cm | 98 cm | 0 |
|2(6x1)+6x2+2(.5x2x1) | 26 cm | 26 cm | 0 |
|3(6x1)+2(.5x1x1) | 19 cm | 19 cm | 0 |
|2(2x2)+2x4+2(.5x1x4) | 20 cm | 20 cm | 0 |
|2(2x6)+4x6+2(.5x2x4) | 56 cm | 56 cm | 0 |
| Average Difference | 0 |
SA = (bh1+bh2+bh3) + 2(1/2bh)
IV. SUMMARY AND CONCLUSION:
We found from our data that the formula for calculating the
surface area of a triangular prism works. There was no
difference in the surface area when calculated using the formula
and manual measurement. Therefore we accept our hypothesis
which stated that the formula for calculating the surface area
of a triangular prism, (bh1 + bh2 + bh3)+2(1/2bh), will always
work no matter what the size of the triangular prism is.
V. APPLICATION:
We can apply our findings to the world outside of the classroom
by using the formula when a triangular prism is in the
blueprints of a building. Architects can use the formula to
calculate how much space and material will be needed for certain
sections of the building, such as the roof.
TITLE: Is The Theory Of Probability Correct?
STUDENT RESEARCHER: George McPherson
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
I would like to do a mathematical proof to see if the theory of
probability is correct. My hypothesis states that, if I throw a
die three thousand times, it will land on each number five
hundred times.
II. METHODOLOGY:
First, I chose my topic. Then I wrote a statement of purpose.
I then conducted a review of literature about the probability
theory. Next, I came up with a hypothesis. Then I developed a
methodology to test my hypotheses. Then I gathered the
materials necessary to conduct my survey. Then I rolled one die
six hundred times. Next, I recorded the results on a data
collection sheet. I then analyzed my data. Next, I accepted or
rejected my hypothesis, wrote a brief summary and conclusion,
and applied my findings to the world outside of the classroom.
III. ANALYSIS OF DATA:
In my research, I found that in trial one, the die landed on
number one 102 times, on number two 100 times, on number three
100 times, on number four 100 times, on number five 100 times,
and on number six 98 times. In trial two, the die landed on
number one 104 times, on number two 101 times, on number three
100 times, on number four 104 times, on number five 93 times,
and on number six 98 times. In trial three, the die landed on
number one 100 times, on number two 99 times, on number three 99
times, on number four 102 times, on number five 100 times, and
on number six 100 times. In trial four, the die landed on
number one 100 times, on number two 99 times, on number three 98
times, on number four 100 times, on number five 103 times, and
on number six 100 times. In trial five, the die landed on
number one 94 times, on number two 101 times, on number three
103 times, on number four 98 times, on number five 104 times,
and on number six 100 times. I also found in my research, that
out of all five trials the total amount of times that it landed
on number one was 500 times, number two was 500 times, number
three was 500 times, number four was 504, number five was 500
times, and number six was 496 times.
Sides Of Die
| Trials | 1 | 2 | 3 | 4 | 5 | 6 | Totals|
| 1 | 102 | 100 | 100 | 100 | 100 | 98 | 600 |
| 2 | 104 | 101 | 100 | 104 | 93 | 98 | 600 |
| 3 | 100 | 99 | 99 | 102 | 100 | 100 | 600 |
| 4 | 100 | 99 | 98 | 100 | 103 | 100 | 600 |
| 5 | 94 | 101 | 103 | 98 | 104 | 100 | 600 |
| Totals | 500 | 500 | 500 | 504 | 500 | 496 | 3000 |
IV. SUMMARY AND CONCLUSION:
In conclusion, I found that I reject my hypothesis which stated
that the die would land on each number five hundred times. If I
repeated this research more times the likelihood is that the
results would more closely approximate the values predicted by
probability theory.
V. APPLICATION:
I can relate this information to the world outside of the
classroom because the probability theory can help us determine
the outcome of games of chance especially gambling.
TITLE: Is The Formula For Finding The Value of Pi Really
Accurate?
STUDENT RESEARCHER: Christine O'Rourke And Brant Linde
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
We would like to do a mathematical proof research project to see
if the formula for finding the value of Pi (3.14), Pi= C/D, is
accurate for circles of all sizes. Our hypothesis states that
the formula, Pi= C/D, is accurate for circles of all sizes.
II. METHODOLOGY:
First, we came up with a topic. Then we wrote our statement of
purpose. Next, we wrote our review of literature about Pi,
Mathematics, and Circles. After that we developed a hypothesis.
Then we came up with a methodology to test our hypothesis.
Next, we conducted our research. First, we found ten common
circle shaped objects of all sizes such as a Frisbee, bottle,
soup can, Sprite can, Slinky, poker chip, jumbo marker, circular
clock, bowl, and cup. Then we put a little mark on the edge of
the object and put it length wise on our piece of white paper.
After that, we rolled the object across the paper until the mark
touched the paper again. Then we measured to see how long the
mark was with a ruler to find the circumference of that circular
object. Next, we divided the circumference by the diameter to
see if the formula Pi= C/D is always equal to 3.14.
Then we recorded our data on a data collection sheet. After
that we analyzed our data. Then we wrote our summary and
conclusion, accepted or rejected our hypothesis, and applied our
findings to the world outside the classroom.
III. ANALYSIS OF DATA:
We found that the Frisbee had a circumference of 93.12 cm and a
diameter of 29.7 cm. When we divided the circumference by the
diameter, we got a value for Pi of 3.13 cm. The 118 mL bottle
of rubber glue had a circumference of 17 cm and a diameter of 5
cm. When we divided the circumference by the diameter, we got a
value for Pi of 3.40 cm. The 15 oz can of soup had a
circumference of 23.6 cm and a diameter of 7.5 cm. When we
divided the circumference by the diameter, we got a value for Pi
of 3.13 cm. The next object that we used was the Sprite can.
It had a circumference of 22 cm and a diameter of 7 cm. When we
divided the circumference by the diameter, we got a value for Pi
of 3.14. Next, we used the Slinky. It had a circumference of
81.3 cm and a diameter of 26 cm. When we divided the
circumference by the diameter, we got a value for Pi of 3.13 cm.
The next object that we used was the poker chip. We found that
it had a circumference of 12 cm and a diameter of 3.8 cm. When
we divided the circumference by the diameter, we got a value for
Pi of 3.15 cm. Then we used a jumbo marker with a circumference
of 4.2 cm and a diameter of 1.3 cm. When we divided the
circumference by the diameter, we got a value for Pi of 3.23 cm.
Next came the circular clock with a circumference of 67 cm and a
diameter of 19.5 cm. When we divided the circumference by the
diameter, we got a value for Pi of 3.43 cm. Then we did the
bowl and we found that it had a circumference of 50.1 cm and a
diameter of 16.7 cm. When we divided the circumference by the
diameter, we got a value for Pi of 3 cm. The last one that we
did was the cup. We found that the circumference was 25.7 cm
and it had a diameter of 7.4 cm. When we divided the
circumference by the diameter, we got a value for Pi of 3.4 cm.
The value of Pi which we calculated for each circle was not
exactly 3.14 cm. The average difference between the calculated
value of Pi and the known value of Pi was .12 cm. The
difference between the two values of Pi was due to the fact that
our measurements were not exact. We did not have the proper
tools to do an exact research project.
|Difference|
| pik & pic| Pik | Pic | C / D |
Frisbee | .01 cm | 3.14 | 3.13 |93.12 | 29.7|
118mL bottle | .26 cm | 3.14 | 3.4 | 17 | 5 |
of rubber glue|
15oz soup | .01 cm | 3.14 | 3.13 |23.6 | 7.5 |
can
Sprite can | 0 cm | 3.14 | 3.14 | 22 | 7 |
Slinky | .01 cm | 3.14 | 3.13 |8l.3 | 26 |
Poker chip | .01 cm | 3.14 | 3.15 | 12 | 3.8 |
Jumbo marker | .11 cm | 3.14 | 3.23 | 4.2 | 1.3 |
Circular clock| .29 cm | 3.14 | 3.43 | 67 |19.5 |
Bottom of a | .14 cm | 3.14 | 3.0 |50.1 |16.7 |
bowl
Cup | .26 cm | 3.14 | 3.4 |25.7 | 7.4 |
Average | .12 cm | | | | |
Difference
Pik= the known value of Pi
Pic= the calculation of Pi
C= the circumference of the circle
D= The diameter of the circle
IV. SUMMARY AND CONCLUSION:
We have found in this experiment that the formula, Pi= C/D, is
accurate. Therefore we except our hypothesis which stated that
the formula, Pi= C/D, is accurate for circles of all sizes.
This research needs to be repeated using more precise
instruments of measurement to reduce measurement error.
V. APPLICATION:
We can apply this to the world by telling math text book
companies that they don't have to change their books because the
formula is correct. We can also use this information to find
the circumference of a circle if we know it's diameter, or you
can find the diameter of a circle if you know it's
circumference.
TITLE: Does Euler's Formula Work?
STUDENT RESEARCHER: Matt Kubicek and Alex Manuel
SCHOOL: Mandeville Middle School
Mandeville, Louisiana
GRADE: 6
TEACHER: John I. Swang, Ph.D.
I. STATEMENT OF PURPOSE AND HYPOTHESIS:
We would like to do a mathematical proof to see if Euler's
Formula works. We would like to see if the number of vertices
plus the number of faces minus 2 always equals the number of
edges of all polyhedrons. Our hypothesis states that Euler's
Formula, (V+F-2=E), works.
II. METHODOLOGY:
First, we chose our topic. Next, we wrote our statement of
purpose. Then we gathered our information and wrote our review
of literature on Euler's Formula, Euler, geometry, faces,
vertices, edges, formulas, and polyhedrons. Next, we developed
our hypothesis.
After that, we created our methodology to test our hypothesis.
Next, we listed all the materials we needed to be able to
perform our experiment which were a data collection sheet, eight
different shaped polyhedrons, a pencil, and a sheet of paper.
Then we took eight different sized and different shaped
polyhedrons and counted the number of vertices and faces on each
of them. We tested Euler's Formula on each polyhedron by adding
the number of vertices to the number of faces and subtracting 2
to see if the overall total equaled the number of edges. We
then counted the number of edges to see if the formula was
correct. Then we recorded the information on our data
collection sheet.
After that, we analyzed our data using charts and graphs. Next,
we wrote our summary and conclusion. Finally, we applied our
findings to the world outside our classroom.
III. ANALYSIS OF DATA:
We counted 8 edges in polyhedron one which had 5 faces and 5
vertices. Using Euler's Formula, V+F-2=E, we calculated that
polyhedron one had 8 edges. We counted 16 edges in polyhedron
two which had 9 faces and 9 vertices. Using Euler's Formula,
V+F-2=E, we calculated that polyhedron two had 16 edges. We
counted 12 edges in polyhedron three which had 6 faces and 8
vertices. Using Euler's Formula, V+F-2=E, we calculated that
polyhedron three had 12 edges. We counted 24 edges in
polyhedron four which had 10 faces and 16 vertices. Using
Euler's Formula, V+F-2=E, we calculated that polyhedron four had
24 edges. We counted 9 edges in polyhedron five which had 5
faces and 6 vertices. Using Euler's Formula, V+F-2=E, we
calculated that polyhedron five had 9 edges. We counted 4 edges
in polyhedron six which had 3 faces and 3 vertices. Using
Euler's Formula, V+F-2=E, we calculated that polyhedron six had
4 edges. We counted 7 edges in polyhedron seven which had 3
faces and 6 vertices. Using Euler's Formula, V+F-2=E, we
calculated that polyhedron seven had 7 edges. We counted 11
edges in polyhedron eight which had 6 faces and 7 vertices.
Using Euler's Formula, V+F-2=E, we calculated that polyhedron
eight had 11 edges. We counted 34 edges in polyhedron nine
which had 20 faces and 16 vertices. Using Euler's Formula, V+F-
2=E, we calculated that polyhedron nine had 34 edges. We
counted 18 edges in polyhedron ten which had 8 faces and 12
vertices. Using Euler's Formula, V+F-2=E, we calculated that
polyhedron ten had 18 edges.
|Polyhedrons | EA | EF = V + f - 2 |
| 1 | 8 | 8 | 5 | 5 | 2 |
| 2 | 16 | 16 | 9 | 9 | 2 |
| 3 | 12 | 12 | 8 | 6 | 2 |
| 4 | 24 | 24 | 16 | 10 | 2 |
| 5 | 9 | 9 | 6 | 5 | 2 |
| 6 | 4 | 4 | 3 | 3 | 2 |
| 7 | 7 | 7 | 6 | 3 | 2 |
| 8 | 11 | 11 | 7 | 6 | 2 |
| 9 | 34 | 34 | 16 | 20 | 2 |
| 10 | 18 | 18 | 12 | 8 | 2 |
EA = actual edges
EF = edges calculated from formula
IV. SUMMARY AND CONCLUSION:
Based upon the tests on the ten different polyhedrons, we
conclude that Euler's Formula accurately predicts the number of
edges on all polyhedrons. We accept our hypothesis which stated
that Euler's Formula, V+F-2=E, would really works.
V. APPLICATION:
We can apply our findings by using Euler's Formula in math to
solve mathematics problems that deal with three dimensional
figures.
© 1998 John I. Swang, Ph.D.