Geometry PoW Packet
All Around the World
October 1, 2007
Welcome!
•
http://mathforum.org/geopow/
We hope you’ve been keeping up with the packets we’ve been writing and that they’ve been useful to
you. We’d love it if you’d post to the geopow-teachers discussion board and share how you’ve been
using the packets. What’s the most useful part to you?
This packet contains the text of the problem, the “answer check”, our solution and scoring rubric, a
note about common mistakes we are expecting to see, and ideas for implementing the problem in the
classroom. It also includes sample student solutions that we received the last time we used this
problem, which was in 2003. These will give you an idea of the range of things you might see if you
use this problem with your students.
The Problem
page 2
All Around the World is a slightly modified version of Problem 2955 from the Library. In this version,
students are asked to find the answer for both the Earth and either Jupiter or a basketball. In the
original, the Jupiter/basketball part was the Extra. In the original, many students looked up “the radius
of the Earth” and then solved the problem from there (though naturally they used all sorts of different
numbers, something which we failed to point out in our commentary the first time around). That’s fine,
and many of them arrived at the correct answer. It shows that they have an understanding of how the
problem works. But given that most geometry students have taken at least Algebra 1, our hope is that
you’ll encourage students to solve the problem for the general case, without using a specific radius for
the Earth (or the basketball or Jupiter).
If students aren’t sure where to start, suggest that they write an equation for the circumference of the
Earth (which is also the length of the wire), then write another equation that represents the length of the
wire after 100 meters has been added. Also, ask them to think about what measurement we need in
order to answer the actual question of “How high above the surface of the Earth will the wire be?”
For each problem, we will pick one category from the scoring rubric (see below) on which we’ll focus.
For All Around the World, we’re choosing “Strategy”, which generally means whether or not the
student has picked a strategy that is based on sound mathematics and doesn’t at all rely on luck. A full
range of strategy, from Novice to Expert, can be seen in the sample student solutions included in this
document.
Answer Check
page 3
Our Solution
page 3
Scoring Rubric
page 5
After students submit their solution, they can choose to “check” their answer by looking at the answer
that we provide. Along with the answer itself (which never explains how to actually get the answer,
and in this case only gives one of two possible answers), we provide hints and tips for those whose
answer doesn’t agree with ours, as well as for those whose answer does. You might use these as
prompts in the classroom to help students who are stuck and also to encourage those who are correct
to improve their explanation.
We’ve not made “our solution” available to anyone but mentors in the past, feeling that the student
solutions are much better examples of how kids actually solve the problem! But sometimes having the
solution ahead of time can be helpful, and we often include tips for mentors about how to support
students in different areas of their work or in thinking about different parts of the problem. When
appropriate, we also include multiple ways to solve the problem (which isn’t the case with this
problem).
The problem-specific rubric is something we write for every problem for use by those who are
assessing student work. It spells out what we expect from students in three areas of problem solving
and three of communication. The goal is to look at each category separately when evaluating the
student work. This way the assessment process provides more focused information regarding the
areas of strength and weakness in the student work. Completeness, especially, can affect other
scores. For example, incomplete or unclear communication can lead to lower scores in strategy
because it’s harder to understand what the strategy is if it’s not explained well.
Keep in mind also that a “Novice” score is not indicative of no work or a zero. It simply indicates that
the student is at a beginning level in that category.
Sample
Solutions
page 6
Common
Mistakes
right here!
The sample student work included in this packet represents a broad range of both writing and problem
solving skills. Our focus is on Strategy, and the solutions run the gamut from Novice to Expert. Note
that the original problem these students were answering was slightly different, in that the current
Question 2 was originally the Extra question. The result of this is that many students didn’t go beyond
finding the answer for the Earth situation. The scores for Strategy, however, are applied as if they are
solving the current problem so that you can get an idea of how the rubric would be applied.
Some students will use a specific radius for the Earth and then make some rounding or arithmetic
errors because of the large numbers. Others will use a specific radius, but not be organized enough to
reach the end successfully. Still others won’t understand what they need to find—whether the change
it in the radius or the diameter or the circumference, or understand where to add the 100 meters.
Students who use variables and attempt to solve for the general case may also lose track of what
they’re doing. In both instances, organizing things clearly will help them make progress.
A few kids will find the change in the diameter instead of the radius.
Good luck!
We’re excited about providing these new resources to you. We hope to get feedback and ideas from
you on the geopow-teachers discussion group starting October 1, 2007.
—
Problem
Annie
All Around the World
1.
Imagine that the Earth is a perfect sphere, and that a metal wire is
snugly wrapped around its equator. Now imagine that we cut this
wire in one spot and splice in an additional 100 meters of wire. We
take up the slack by using posts to raise the wire an equal distance
all the way around.
How high above the surface of the earth will the wire be?
2.
Answer Check
Say the wire was first wrapped tightly around either a basketball or
the planet Jupiter (your choice). Now how far above the surface of
the object is the wire when 100 meters is added?
The wire will be about 15.9 meters above the surface of the Earth.
If your answer doesn’t match our answer,
•
•
•
recall that the formula for the circumference of a circle is 2 * π * r.
if your answer is about 31.8 meters, think about whether you found the change in the radius
of the circle formed by the wire or the change in diameter.
don’t round too much too soon.
If any of those ideas help you, you might revise your answer, and then leave a comment that tells
us what you did. If you're still stuck, leave a comment that tells us where you think you need help.
If your answer does match ours,
•
•
•
•
if you used a specific radius for the Earth, can you find a way to solve it without that? (Yes, it's
possible!)
is your explanation clear and complete?
did you make and correct any mistakes along the way? If so, how did you find them?
are there any hints that you would give another student?
Revise your work if you have any ideas to add. Otherwise leave us a comment that tells us how
you think you did - you might answer one or more of the questions above.
Copyright © 2007 by The Math Forum @ Drexel
2
Our Solution
The key concepts of this problem are really the manipulation of variables and solving a problem for
which you might initially think there isn’t enough information. But you can also get away with being
good at finding the circumference of a circle given the radius, or visa versa.
The wire will end up an unbelieveable 15.915 meters above the surface of the Earth!
It always amazes me when I do this problem that it’s that high. And of course, it’s that high for the
basketball and for Jupiter as well, since the answer is independent of the initial circumference.
Method 1: Using Specific Radii
Question 1: I’ll use the radius of the Earth, calculate the resulting circumference, add 100 meters to
that circumference, find the resulting radius, and subtract the original radius from the new one.
The radius of the Earth at the equator is 6378 km, which is 6,378,000 meters. The circumference is
found by 2 * pi * r.
original circumference = 2 * π * 6,378,000 meters
= 40,074,155.9 meters
new circumference = original + 100 meters
40,074,255.9 meters
new radius = new circumference/2π
= 6,378,015.9 meters
change in radius = new radius - original radius
= 15.9 meters
The wire will be 15.9 meters above the surface of the Earth, all the way around.
Question 2: I found the radius of both Jupiter and a basketball. Jupiter’s is 71,492 km, or 71,492,000
meters. A basketball has a radius of 0.1213 meters. Let’s do the same math for both of these (though
the question only asks you to do one of them).
Jupiter
original circumference = 2 * π * 71,492,000
= 449,197,484 meters
new circumference = original + 100
= 449,197,584 meters
new radius = new circumference/2π
= 71,492,015.9 meters
change in radius = new radius - original radius
= 15.9 meters
It’s the same answer! No way!
So let’s move on and try the basketball and see what happens.
Basketball
original circumference = 2 * pi * 0.1213
= 0.762 meters
new circumference = original + 100
= 100.762 meters
new radius = new circumference/2pi
= 16.04 meters
change in radius = new radius - original radius
= 15.9 meters
It’s the same answer again! No way!
It looks like the starting radius doesn’t matter. That’s pretty cool.
Method 2: The General Case
The problem doesn’t give a specific radius, so I am going to use r and hope that the unknowns cancel
out in the end.
Copyright © 2007 by The Math Forum @ Drexel
3
We want to find the new radius minus the original radius, as that will give us the height of the wire
above the ground.
We know that circumference, C, is 2 * π * r, so r = C/(2π).
100 meters is being added to the circumference, so that becomes C + 100. The equation for the new
radius, R, will be R = (C + 100)/2π.
Subtracting the original radius from the new radius, we get
C + 100
C
R - r = --------- - ---2π
2π
C + 100 - C
= ------------2π
100
= ----- = 15.9 meters
2π
Another way to look at the same thing is to focus on the diameter instead of the radius. We can
represent the initial (unknown) circumference of the Earth as C, and then the diameter becomes C/π. If
we add 100 meters to the circumference, our new circumference is C + 100, and our new diameter is
new diameter = (C + 100)/π
= C/π + 100/π
Since our old diameter was C/π, we’re adding 100/π to the diameter. We divide that by 2 to see how
much we’ve added on each side and we end up with 50/π, or about 15.915 meters.
Since we did not use a specific radius to start the problem, the radius of the object doesn’t matter. It
can be the Earth, Jupiter, a basketball, or anything. We don’t have to calculate anything for the second
question, since we know the answer will still be 15.9 meters.
Copyright © 2007 by The Math Forum @ Drexel
4
Apprentice
has made several mistakes or
misstatements (those who use a
specific radius may make rounding
errors with the large numbers)
makes no mistakes of consequence and uses
largely correct vocabulary and notation
solves the general case, but doesn’t explicitly
exhibit understanding of why it’s general
may use a specific radius for the Earth and
Jupiter/basketball
has a strategy that relies on sound reasoning, not
luck
attempts to answer both questions
shows understanding that the 100 m are added to
the circumference and that the height is the
difference in radius
Practitioner
© 2007 by The Math Forum @ Drexel
does nothing reflective
Reflection The items in the columns
to the right are
considered reflective, and
could be in the solution or
their comment:
does one reflective thing
does two reflective things
http://mathforum.org/geopow/
does three or more reflective things or
great job with two
revising their answer and improving
anything counts as reflection
comments that they’re surprised at the
answer!
explains where they're stuck
reflects on the reasonableness of their
answer
summarizes the process they used
comments on and explains the ease or
difficulty of the problem
answer is very readable and appealing
formats things exceptionally clearly
connects the problem to prior knowledge or
experience
makes an effort to check their formatting, spelling,
and typing (a few errors are okay)
explains the steps that they do explain in such a
way that another student would understand
(needn’t be complete to be clear)
includes additional helpful information,
doesn’t just add more for the sake of
adding more
[generally not possible]
answers both questions using the
general case and exhibits understanding
of the fact that it’s the general case by
saying something such as, “It doesn’t
matter how big the object is, because we
didn’t use a specific radius anywhere.”
as there is no Extra for this problem,
students can’t be an Expert in this
category
Expert
checks their answer (not the same as
viewing our "answer check")
lots of spelling errors/typos
length warrants separation into more
paragraphs
Clarity explanation is very
another student might have trouble
difficult to read and follow following the explanation
Completeness has written almost
shows work with no explanation, or
shows and explains the steps that they take and
nothing that tells you how gives an explanation and shows no work why they are reasonable steps
they found their answer
Communication
Accuracy has made many errors
Strategy has no ideas that will lead understands that there are two
them toward a successful situations (before 100 m added and
solution
after) and that something is changing,
but isn’t sure how to put it all together
Interpretation does neither of the things does one of the things listed under
listed under Practitioner
Practitioner
Problem Solving
Novice
For each category, choose the level that best describes the student's work
Geometry Problem of the Week Scoring Rubric for All Around the World
Student Solutions
Mark
age 14
Strategy
Novice
Dara
age 13
Strategy
Novice
Katie
age 15
Strategy
Novice
I think the wire will be raised about 1mm.
Knowing that the earth has a very great equator and that 100 m is only
.1km I came up with the solution that the wire would be raised about 1
mm (at most).
The wire around the earth would be approximately 0.008 cm off of the
ground.
First, I found the diameter of the earth. Then, I converted it into meters.
After that, I added 100 meters to the total. Since the 100 meters would
we dispersed evenly around the earth, I divided 100 by the diameter+100
to find how out how high off of the ground it was.
The wire would be about .0003798833 meters above the Earth’s surface.
First, I found that the diameter of Earth is 83,834 meters. I then added 100
meters to get the length of the wire, which is 83,934. I wanted to find the
circumference of the world, so I multiplied 3.14 and 83,834 to get
263,238.76 meters. Then I figured that you really didn’t need to know the
length of the wire. You already know that it is 100 meters longer than the
Earth’s circumference. So I divided 100 by the Earth’s circumference and
got about .0003798833 meters.
Copyright © 2007 by The Math Forum @ Drexel
Mark doesn’t have any ideas
about how to solve the
problem, though he is using
some reasonable thinking—
it doesn’t seem like it would
be very far! I would ask Mark
to write an expression for
the circumference of a
circle, and then an
expression for the
circumference of a circle to
which has been added 100
meters.
It doesn’t look like she’s
considering the
circumference in any way.
And I can’t quite figure out
why she is dividing 100 by
the “new diameter”. I would
ask her to consider the fact
that the wire is going around
the circumference of the
Earth and not just through
the diameter, and point out
that the 100 meter is being
added to the circumference,
not to the diameter.
While Katie is at least
considering the
circumference, she’s adding
the 100 meters to the
diameter, so she doesn’t
entirely understand the
problem and doesn’t have a
sound idea about how to
solve it. I would start by
pointing out to her that the
100 meters is being added
to the circumference, not
the diameter, and ask her
how that will change her
work.
6
Katelyn
age 13
Strategy
Novice
John
age 16
It is an incomplete problem.
You will need to know how long the peice of wire you started out with
was, and how big around the earth is. All we know is that you cut the
rope and spliced in an additional 100 meters. We need to know more
information.
the wire will go 49 meter high above the surface of the earth.
the Circumfrance of the sphere is 2(PI)r and there are 100 m was added.
So I put 2(PI)(r+50) - 2(PI)r = 2(PI)r(50 -1) that is how get the number 49
Strategy
Apprentice
Andy
age 13
Strategy
Apprentice
My final answer was the wire would be suspended 1.026525819 meters
above the Earth’s surface. Also the extra was the same procedure would
be followed only with the radius of the object.
First I plugged in a radius for the Earth. Next I used the circumference
equation C=2*pi*radius and dubbed the answer C1. Then I repeated the
equation only this time I added in 100 meters of wire and dubbed the
answer C2. I then divided C2 by C1 to get the answer of: 1.026525819
meters. My first circumference was 3769.91184. The second was
3869.91184.
Copyright © 2007 by The Math Forum @ Drexel
While her explanation
sounds reasonable, it
doesn’t sound like she dove
in and tried to see if, in fact,
she could solve the problem
without more information. I
would simply suggest that
she try it, and that she start
by writing two expressions—
one that represents the
circumference of the earth,
based on a radius of r1, and
one that represented the
circumference of the earth
plus 100, using a radius of
r2.
John has the right idea in
that you can represent two
quantities with variables,
without knowing the actual
quantities, and then find a
value in the end. He just
hasn’t added the 100
meters to the right thing. I
would ask him to tell me
more about the expression
2(PI)(r+50) and explain what
the different parts represent.
He may be mixing up
circumference and diameter.
Andy is in the same place as
John, in that he knows to do
something with
circumference, then add
100 to that, and then do
something with the two
results. He just hasn’t made
the right decision with the
results. I would ask him to
tell me what value will tell us
how far the wire is above the
earth. The radius? diameter?
circumference? Helping him
organize his thinking will get
him going in the right
direction.
7
Vineeth
age 15
The wire is 16.175 meters from the surface fo the earth.
Diameter of the earth= 12,756,300 meters
The Circumference or the Perimeter of the earth is 40075.1 kilometers.
Strategy
Practitioner
1kilometer=1000meters
The Circumference or the Perimeter of the earth in meters is
40075.1*1000= 40075100 meters
In other words you could also say that the 2*pi*radius = 40075100 meters.
The problem told you that you are supposed to add another 100 meters to
the circle wire.
The circumference of earth +100meters = the Circumference of the wire
around the Earth.
40075100 meters + 100 meters = 40075200 meters.
So that we know the circumference of the circle we can find the radius or
the diameter by the circumference formula, which is 2*pi*r =
circumference.
It’s true he doesn’t have the
right answer, but the
strategy he chose is not
based on luck. He’s wrong
because of
rounding/accuracy issues.
While we know you don’t
need to know the radius or
circumference or other
dimensions, if you have one,
you can get the right
answer. You just won’t know
that it isn’t based on any
particular radius. I would ask
Vineeth to look at the
circumference he got for the
Earth and see if he can find
a more exact value and then
go from there again.
Circumference of the wire=40075200 meters.
2*pi*r = 40075200
Radius= 40075200/2*pi
Radius= 6378166.175 meters
Diameter= 2r= 6378166.175*2= 12756332.35 meters
To find the how much longer the is from the surface you just have to
subtract the diameter of earth from the diameter of the wire and then
divide that by two because you only want to find one side not the
difference of two sides.
12756332.35 meters-12,756,300 meters= 32.35 meters then you divide
32.35/2=16.175meters
Therefore this proves that the distance from the surface of the earth to the
wire is 16.175 meters.
Rowdy
age 13
The wire is 52.216188288 feet above the earth.
radius of earth plus 100 meters divided by 2(pi) - radius of earth/2pi then
conversions to feet
Strategy
Practitioner
Copyright © 2007 by The Math Forum @ Drexel
This is interesting. What he’s
written makes it sound like
he’s adding the 100 meters
to the radius and then doing
some stuff. But his answer is
correct, right down to the
last decimal place (why he’s
converting to feet, I dunno).
I’m thinking he has written
“radius” when he meant to
write “circumference”. I
would ask him to show the
numbers/calculations that
he used so that another
student could learn from
reading his work, and hope
that the process of doing so
helps him move his
explanation closer to what
he actually did.
8
Nadia
age 15
Strategy
Practitioner
Kush and
Yizheng
age 14
Since the earths radius is 6370 km, then the wire will be 1210.62 km
above the surface of the earth.
Its just basically formulas and solving, but i did have to find out the radius
of the earth which was 6370 km, then i multiplied that by 2 to get the
diameter which was 12,740. Then i multiplied it by pi which is 3.14 and i
got 40003.60. Now since the wireloop is 100 meters i added it and got
40004.60. Then i divided it by pi and subtracted the original radius and
got my awnser.
The height above the surface is approximately 16 meters.
I took the diameter of the Earth, 8000 miles,I converted it to meters. Then
I multiplied that by Pi. I took that result added 100 to it and then I divided
it by Pi. I divided the last two digits of the resulting number, about 32, by
2. That would give me 16 meters
Strategy
Practitioner
Julian
age 16
50/pi metres, same answer for a sphere of any radius!
2piR=l, l+100=2pi(R+r), r=100/2pi qed
Strategy
Practitioner
Copyright © 2007 by The Math Forum @ Drexel
Her strategy is sound. But
she needs work with her
organization so that she’ll
see that she’s subtracting a
radius from a diameter
instead of from a radius. I
would ask her to work
through her solution and
label each quantity with a
unit and also its purpose—is
it a radius? circumference?
diameter? something else?
It’s a sound way to solve the
problem. I would ask them
to include more information
to explain why they did the
things they did and
emphasize that that sort of
information is important to
include. For example, I might
ask, “Why did you multiply
the diameter by pi?”
He might easily be an Expert
if he includes more
information, since it seems
that he understands this will
be true for any sphere. His
solution needs quite a bit
more detail added, though. I
would ask him to add more
information to his
explanation, pretending that
he is explaining his work to a
student who isn’t sure how
to solve the problem.
9
Pedro
age 15
Strategy
Expert
15.915 metros en todos los casos, la tierra, balón, júpiter, etc.
Sorry, but I speak spanish
Dadas la condiciones del problema podemos usar la fórmula de la
circunferencia:
C=2pir donde C= la circunferencia del objeto, pi=3.1415... y r es el radio.
Si modificamos la circunferencia añadiendo un segmento de la siguiente
manera C+d donde d el la distancia añadida tenemos que r resulta
modificado así: r+h donde h es el valor que estamos buscando.
Sustituyendo queda
C+d=2pi(r+h)
Por lo que h = (C+d)/ 2pi-r pero r=C/2pi
Así que h= (C+d)/ 2pi- C/2pi= d/2pi
Nuestra fórmula final es h= d/2pi
Para el ejemplo d=100metros, por lo que h=100metros/2pi= 15.915
metros.
Como se ve en la fórmula el resultado no depende de la circunferencia del
objeto sino sólo del segmento añadido. Así que el resultado es igual para
todos los casos, la tierra, un balón de baloncesto, júpiter, etc.
Keone
age 14
Strategy
Expert
Since The wire will be 50/pi, or about 15.92, meters above the earth.
For the extra, the wire will still be 50/pi = 15.92 meters above the ground,
regardless of whether the initial object is a basketball or Jupiter.
Let us call the radius of the earth r. The circumference C of the earth is
given by the forumla C = 2*pi*r.
When we splice in the 100 meters and raise the wire equidistant around
the earth, the radius of the new shape is equal to the radius of the earth
plus whatever height we raised, which we'll call x. Thus, the new radius
equals r + x. Also, by splicing in 100 meters, we are simply adding 100 m
to the circumference. Thus, the new circumference is C + 100. Using the
formula for circumference, we have:
It’s fun to realize that even if
you don’t know all the
Spanish words, you can still
understand what he’s done
mathematically. He says at
the end that the result
doesn’t depend on the
circumference, but only on
the amount added, so he’s
understand what the general
case is and what it means to
have solved the problem for
the general case. It might be
an interesting solution to
share with students who
understood the problem and
have them see how many of
the words they can figure
out just based on the
context. (I wouldn't make
any suggestions to Pedro,
since I don't speak Spanish,
and while online translators
will give you the gist of the
text, they won't give you
anything you could actually
critique from a
communication standpoint.)
His solution is really good all
around. It’s very clear and
complete, and he
understands what it means
to solve for the general case
without relying on specific
numbers. I would probably
suggest that he make the
whole thing shorter by not
repeating so much work in
the Extra. I’m not convinced
that it adds much.
(C+100) = 2*pi*(r+x)
Our original formula for the circumference of the earth was
C = 2*pi*r.
Subtracting the second equation from the first one gives:
(C+100) - C = 2*pi*(r+x) - 2*pi*r
100 = 2*pi*r + 2*pi*x - 2*pi*r
100 = 2*pi*x
x = 100/(2*pi) = 50/pi
So the wire will be 50/pi, or about 15.92, meters above the earth.
EXTRA:
In the main question, the actual circumference and radius of the Earth
never mattered, since they canceled out. Thus, regardless of the size of
Copyright © 2007 by The Math Forum @ Drexel
10
the sphere in question, the wire will always be 50/pi = 15.92 meters above
the ground.
We can show this by calling the circumference of the sphere in question
C, and the radius r. The equation relating the two is always C = 2*pi*r.
If we add 100 m to C, we get C+100. This makes the radius bigger by x
meters, so the radius is r+x. The equation relating these two is C+100 =
2*pi*(r+x). As before, subtracting one from the other cancels the C and
the r, and yields x = 100/2pi = 50/pi = 15.92 m
Copyright © 2007 by The Math Forum @ Drexel
11