PoW-TER Problem Packet
Building on Water (Author: Ashley Miller)
1. The Problem: Building on Water [Problem #1394]
The region surrounding Tampa, Florida has many retention ponds. These
ponds are designed to catch rainwater that runs off from paved areas, or to
divert the water from reclaimed land. At a local shopping center, a business
was allowed to construct a building over such a pond. Concrete pillars
(pylons) were used to keep the building above the water line.
For ease of calculation, we will assume that the pond is in the shape of a perfect circle and that
the shape of the building is a rectangle. To assist in communicating details, we will label the
corners of the building ABCD. Drawing a good picture will help you to solve this problem.
The ratio of side AB to side BC is 4 to 3, and the perimeter of the building is 280 units. The edge
of the circular pond is 4 units beyond each corner of the building on the diagonal when viewed in
an aerial photo.
What is the area of the pond that is not covered by the building?
Hint: It will help if you first calculate the length of each side of the building and the radius of the
circular pond. You will be expected to use appropriate rounding at the end of the problem.
Bonus: The building owners plan to put water plants in the area of the pond that will receive
sunlight at noon. The plants cost $1.25 each and require a surface area of 30 square units to be
healthy when mature. What is the total cost of the plants needed to cover the available area?
Note: This problem, Building on Water, is one of many from the Math Forum @ Drexel's
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2. About the Problem
This problem is a great enrichment problem after students have been introduced to Pythagorean
theorem and the area of two-dimensional shapes. Presenting this problem in groups will
generate brainstorming with the noticing and wondering method. While working in groups,
encourage students to move to drawing a picture of the situation. At that point I may end the
group effort to see if individuals could then come up with a solution. I would possibly extend this
problem to develop the communication and representation part of problem solving. Students
could write their answer as a proposal for a company as if they were the plant company.
I love how this problem shows a real application of how the Pythagorean theorem can be used in
conjunction with finding area of multiple shapes. I also love this problem because it is one where
Building on Water 1
drawing a picture seems almost a necessity which is a skill that students are sometimes reluctant
to use. The bonus also reminds students why this would be a necessary situation. We would have
a discussion about other situations where one would need to know an exact measurement before
making a purchase such as purchasing paint or purchasing fabric.
Students need to be familiar with ratios and equations in addition to the two main mathematics
goals: Pythagorean theorem and area. Some students could use guess and check to find the
dimensions of the building, but I think they will find frustration if they try to only guess and
check for the diameter or area of the circle.
After successful completion of this problem, students will be ready to start looking at areas of
composite shapes and shapes that are enclosed within other shapes (shaded area).
3. Math Goals (6-8)
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Ratio
Multi-Step Equations
Pythagorean Theorem
Area of Circles
Area of Rectangles
Surface Area
Multiplication ($)
Geometry Vocabulary
Scale Drawings (possibly)
4. NCTM Mathematics Standards (6-8)
Problem Solving NCTM Standards (6-8)
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build new mathematical knowledge through problem solving;
solve problems that arise in mathematics and in other contexts;
apply and adapt a variety of appropriate strategies to solve problems;
monitor and reflect on the process of mathematical problem solving.
Communication NCTM Standards (6-8)
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organize and consolidate their mathematical thinking through communication;
communicate their mathematical thinking coherently and clearly to peers, teachers, and
others;
analyze and evaluate the mathematical thinking and strategies of others;
use the language of mathematics to express mathematical ideas precisely.
Representation NCTM Standards (6-8)
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create and use representations to organize, record, and communicate mathematical ideas;
select, apply, and translate among mathematical representations to solve problems;
use representations to model and interpret physical, social, and mathematical phenomena.
Building on Water 2
5. Common Misconceptions
1. What was the question again?
The answer is 9156 units squared.
The first thing I did was figure out the dimensions and since the dimensions had to be 3 to 4. The
length was 60 and the height was 80. Then I had to find the length from one corner to the other
so then I would know the circles diameter after adding on the extra 8. I made a line from one
corner to the other and I used the formula for finding the side lengths of the triangle. 80 squared+
60 squared=1000. Then 100*100=side C. 100+8=108 and the diameter of the circle is 108. Then
the radius is 54 and 54 squared is 2916. Then I multiplyed it by 3.14 and I got 9156.
This student found the area of the circle and stopped. Often students will find an answer and
circle it and move on without taking the time to look back at the question. The work the student
did so far is correct but he would not receive any credit on a standardized test because he didn’t
finish the problem. This is a significant misconception because students need to complete the
cycle of problem solving by going back and rereading the question, looking over their answer,
and even trying other methods to find their answer in another way. I would pair this student up
with another and have them look at the question again, have them draw a picture if one was not
drawn the first time and ask them, “Was the question answered?”
2. Calculator is not a replacement for knowledge
There will be 23,951 units not covered by the building.
First, I needed to calculate the perimeter of the building. Since the building is a rectangle, I drew
a diagram with the rectangle and the circular pond around it. Then I used guess and check. I
started making a list of the possible side lengths. I started with 4 and 3, then 8 and 6, then 12 and
9, and so on. I also just tacked on a zero to each of the numbers to see if I needed to make the
numbers greater or lesser. If 2 sides of a rectangle are 40 and 30, the perimeter of the rectangle is
140. I noticed that this is exactly half of 280 units, the building's perimeter. So now I knew that 2
sides of the rectangle were 80 units, and the other 2 sides were both 60 units. Next, I had to find
the radius of the circle. Since the pond was a perfect circle, and it was 4 units from each corner to
the edge of the pond, I knew the building was exactly in the center. The Pythagorean theorem
states that a2+b2 (the two legs)=c2(the hypotenuse). The two legs form a right angle, so 80
squared(6400) + 60 squared (3600) equals 10,000 (the hypotenuse squared). That means the
hypotenuse is 100. If I add 100 plus 8(the distance to the edges of the pond) my diameter is 108.
Now I had to find the area of the pond. I know the formula for circle areas- Πr2.The problem
instructed me to round, so I approximated pi to about 3.14. My equation was..... 3.14x542. When
I solved the equation, my answer was 28,750.5936. I rounded that to 18,750.6. Then I needed the
area of the building. The formula for that is simply length times width. So, 80x60 should give me
my area. The answer is 4800. Then I subtracted the two to find the area of the pond not being
built on- 28,750.6-4800=23,950.6, which I rounded to 23,951 units. So in conclusion, there will
be 23,951 units not being built on.
Building on Water 3
This student worked so hard on this question and his explanation was in great detail but a simple
calculator and/or formula error early on gave him the incorrect answer. It is unclear whether he
did not know how to use the calculator properly or if he did not understand the circle formula. In
either case the student squared the radius and pi and therefore calculated a much larger answer.
A picture drawn to scale would help him see that his answer was not feasible. The building takes
up a larger percentage of the water than his answer would have shown.
I would give the student graph paper and ask him to draw the building to scale and the circle
around it to see if the very large numbers gave him a reason to pause. Then we would look at the
formula and see if he could see the error of his calculations.
Student Solutions
• Why the particular solution type is significant for the current problem and/or with regards to a
particular mathematical agenda or process.
• How it relates to the misconceptions discussed above
• Why it might be useful for the solution to be presented in class (how might it be useful for
students who used different strategies to solve the problem – either more or less
sophisticated or efficient strategies)
Stuck At Step One
I’m not quite sure of my answer yet. The first thing I did was to highlight the key phrases in the
problem. I read in the problem that drawing a good picture will help me, so I borrowed a
compass from my teacher and drew a perfect circle. I also read at the bottom of the page that it
would help me if I first calculated the length of each side of the building. So I looked at the ratio
for side AB to side BC, but I couldn’t understand it. My teacher hasn’t taught ratios yet, and my
fifth grade teacher only touched on them. I was wondering if you could help me understand
them.
This student needs guidance about where to begin. He is stuck on the word ratio and is
convinced he has never learned them but with a little vocabulary clarity he will see he probably
knows more about ratios than he realized. I would start by asking a few questions to gage what
the student knows about fractions and ratios to see if he can make the connection to the
relationship of 3 to 4. Since this student seems to be younger (referring to 5th grade) I would
maybe focus on his drawing and see what generalities you could draw from it. I would also
suggest the student try guess and check for the sides of the building to come up with sides that
would make sense for the parameters given.
This problem is important to share with students so they see they are not the only one that has
ever been stuck and that being stuck is ok.
Building on Water 4
Read Between the Lines
The area of the pond not covered by building is 4356.25 units.
First I figured out the length of the sides of the rectangle. I used algebra to get 60*80. Then I
found the area of the rectangle by length times with. then I found out what the diagnol was of the
square by (w^2+L^2)^0.5. It was 108. then I found the area of that by multipying it by 3.14.
Finnaly I subtracted the areas to get my ansewer.
This student gave a correct answer but did not give a lot of explanation of how he solved the
problem. There are a lot of places where he assumed the reader would know what he is talking
about. This is an important lesson in communication. Students need to see how hard it is to
follow an answer that is incomplete. I would use this student solution when students are resistant
to showing their work or to explaining their answer.
Ratio and Proportion Approach
The area of the pond not covered by the building is 4360.8842 square units. The cost to plant the
145 plants which will nt be under the building is $181.25.
The ratio of AB to BC is 4 to 3. The total units of half the perimeter of the building are 7. 4/7 =
x/140 The total periemter is 280 therefore half the perimeter is 140. To find the length of one of
the longest sides do 4/7 = x/140. x = 80. The longest sides of the rectangle are 80 units ech and
the shortest are 60 units each. The area of the rectangular building is 60 x 80 = 4800 square units.
To find the area of the circular pond do (60)(60) + (80)(80) = 10000 Find the square root of
10000 = 100 This is the hypotenuse of half the rectangle. Now add 8 to that and you will have
the diameter of the circle. The radius is then 54. Then the area of the circle is (54)(54)(pi) =
approx. 9160.8842 square units Now do the subtraction 9160.8842 - 4800 = approx 4360.8842
square units So 4360.8842 is the approx area not covered by the building
Bonus
How many plants can be planted in that area given that each plant needs a 30 units 4360 / 30 =
145.3333...
Therefore one may plant 145 plants at $1.25 each = $181.25
This student decided to focus on the ratio instead of relating a ratio to an equation and the whole
perimeter. This is a great solution to share with students who think there is only one way to
solve a problem.
Building on Water 5
Good Algorithms, Rounding?
The area of the pond that is not covered by the building is 4356.24 square units. For the bonus,
the total cost of the plants needed to cover the available area is $181.25
I did this by first finding the length of each of the sides of the rectangle. I found the length of
each side by doing: 3x + 4x +3x + 4x= 280
When I solved it, i got x=20
Then i replaced x with 20, and got 3*20=60 and 4*20=80. Therefore, the sides are 60 units and
80 units.
Then, i found the diagonal length (z) of the rectangle to find the diameter of the circle.
I found an equation that gave me the length of all the sides of the triangle formed when the
rectangle is sliced diagonally:
z2 = x2 + y2
I alreay knew that x was 80 and y was 60, so when i simplified the problem i got : z2=10000 so z
equals 100.
Since there were 4 units from each corner of the rectangle to the edge of the pond, i added 8 to
100 which gave me 108. Because the radius is half the diameter, i cut 108 in half which gave me
54. Then, to find the area of the pond, i used the formula:
r2 π= area
542 π= area
2916*3.14=area
9156.24=area
THEN, i found the area of the rectangle: 80*60=4800
To find the area not covered by the building, i did: 9156.24 - 4800=4356.24 square units
So the area not covered by the building was 4356.24 square units.
For the bonus, i did, 4356.24/30 then times the cost, 1.25 i got 145*1.25=$181.25
This is the solution I would use to start the discussion about rounding and whether or not it
matters when and if we round. This student rounded to the hundredths place from the beginning
and may have had a reason for doing so but if so, there needs to be an explanation for doing this.
Drawn to Scale
The area of the pond that is not covered by the building is ~4,800 units squared.
First, I knew that I needed to know the area of the building because once I figured out the area of
the pond; I would need to know the area of the building to subtract it from the area of the pond.
So, I used some simple math. On grid paper with one centimeter representing ten units, I
measured eight centimeters by six centimeters, because if you multiply both of those by ten and
then add the base and the height and multiply that by two, it would equal 280 units, which is
what the perimeter is according to the problem. From there, I found the area. I multiplied base
times height (60 *80) and got 4,800 units squared, so I knew the area of thebuilding. Then, I took
a ruler and measured 0.4 centimeters beyond each corner of the building on my grid paper
Building on Water 6
because the problem said that the edge of the pond was 4 units beyond each corner of the
building, so if I were using one centimeter for 10 units, then 0.4 centimeters would be equal to 4
units. Then, I found the radius of the circle by connecting opposite edges of the pond diagonally,
and I marked the point where they met as the radius. Next, I measured the radius and it measured
to be about 5.5 centimeters, or 55 units. I squared 55 because the area of a circle equals the
radius squared times pi. So, after squaring 55(radius), I got 3,025 and multiplied that by pi to get
the area of the pond, and got 9,498. To get my final answer, I subtracted 4,800 units(area of
building) from 9,498 units(area of pond) because that would five me the area of the pond without
the building on it. Finally, I got 4,698, so I knew that the area of the pond without the building
was ~4,698 units squared.
This student solution is a great example for a student who wants to give up because he claims to
not know the algorithms. This student found a more creative way around this problem by
drawing it out.
Picture Worth 1000 Words-But Words Are Good,Too
There is 4,360 units^2 of uncovered pond area. It would cost $181.25 to cover exactly 145
plants.
My solution includes some diagrams and pictures. Therefore, I attached a document including all
those diagrams. To locate which diagram goes where I places the word "FIGURE(& its number)"
when a diagram is meant to be shown. The number of the figure here will match the number of
the figure on the attached document.
---------------------------------------------------------------- Building on WaterIn order for me to find out what they are asking for, the area of the pond that is not covered by
the building, I will first need to find out the area of both the building and of the pond. Then with
that information I will subtract the area of the building from the area of the pond, leading my to
my final answer.
Finding out the area of the building:
Useful information from given situation:
- The perimeter of the building is 280 units
- Ratio of side AB to side BC is 4 to 3
4+4+3+3 = 12 units
280 / 12 = 20 units
This means that the ratios were reduced by (dividing it by) 20.
Therefore:
Side AB: 4*20 =80 units
Side BC: 3*20= 60 units
FIGURE #1
In order to check myself I added 80+80+60+60= 280 units to make sure the perimeter is actually
equal to 280 units.
To figure out the area I multiplied 80*60=4,800 units^2.
Building on Water 7
Finding out the area of the pond (a circle):
Useful information know known:
- 4 units from the edge of the building to the outside of the pond
- The area of the building is 4,800 units^2
Diameter:
FIGURE #2
147984 10993
Now I am going to figure out the diameter of the pond by using the Pythagorean theorem, which
is
FIGURE #3
The sum of the areas of the two squares on the legs (a and b) equals to the area of the square on
the hypotenuse (c). Also, the known formula is:
FIGURE #4
In this case, A= 60 units and B=80.
Therefore I did:
FIGURE #5
FIGURE #6
FIGURE #7
C = 100 units
However, the diameter is not only 100 because there is that extra space (4 units)
between the pond and the building. So the diameter really is: 100+4+4=108 units
Now that I have the diameter of the pond I can figure out its area using the formula
R^2*PI:
108/2= (radius) 54 units
54^2= 2,916 units
2916*PI= ~9,160 units
AREA OF THE CIRCLE = ~9,160 units^2
Now that I have both the area of the building and the area of the circle I can figure out the area of
the uncovered pond by:
FIGURE #8
ANSWER: 4,360 units^2 of uncovered pond area
Cost of plants needed to cover the available area:
4,360 units^2 / 30 needed space per plant= 145⅓ plants but you can’t buy ⅓ of a plant so they
could only get 145 plants
145 plants * $1.25 = $181.25
Building on Water 8
This is an excellent solution to demonstrate how using diagrams really makes the solution much
clearer. In addition, the student explained every step of the process. I would use this solution as
an exemplar of what I hope my students will work toward.
Building on Water 9
Conversation with References
At a bird's-eye veiw, 4,361 units2 of the retention pond will be showing. And to put in water
plants, it would cost about $181.70
First, we must find the lengths of the sides of the building, I fould half of the perimeter, which is
280units (since the other two sides would be the same lengths) 280/2=140units. So the lengths of
the two pairs have to be something X*3 and X*4 (because of our 3 to 4 Ratio) and the products
have to add up to 140, where we can add the other side to make the full rectangle with a 280-unit
perimeter. I dicided to guess a bit and start with 20 as our 'X' (it happened to be a wonderful
guess, btw) So I tried it. 20*3=60, and 20*4=80. 60+80=140. BINGO!! the rectangle is 60 by 80
units
--Sorry, but here I thought I'd clarify that the ratio 3 to 4 means every three units I add to one
pair of sides, I have to add four to the other pair, I'm not sure if it is right though, I never really
learned ratios... Now that we know the side-lengths of the rectangle, we can find the area, or b*h.
60*80=4800 units squared. I don't think I need to explain much on this. (We need to find the area
of the rectangle because once we know the area of the pond, we can subtract it from the rectangle
to get only the veiwed area of the circle, like in problem #1538, Partial Area.)
We're half-way there. Now we need to find the area of the circlepond, to do that we need the
diameter of the circle. It says that the circle reaches 4 units beyond the diagonal of the building.
But to figure that out we need to find the diagonal of the rectangle. I chopped the circle in half on
the diagonal and found that it was, to my reief, a right-triangle. I used the Pythagorean Theorum
to find the length of the 3rd side, or diagonal. Luckily I was semi-familiar with it from Partial
Area. (If you don't know it, I'd visit Ask Dr. Math's article on the Theorum, it is how I found how
to use it) I found that the length of the diagonal or third side was 100 units, a nice round number.
BUT the circle's diameter reaches 4 units beyond 116852 353 each side, so 100+4+4= 108. Now
for the area of the circle. We need the Radius, so 108/2=54,*itself to get r2=2,916. Area = r2*pi,
so 2,916*3.14159= about 9,161 units2. That's it for the circle, now for the simpler part.
We want to know the area that is not covered by the rectangle so the circle's area minus the
building's will give us that. 9161u2- 4800u2= 4,361 units2. That is our first answer.
Bonus: we need to find how many water plants needed, so because they cover 30u2, 4361/30=
about 145, that is how many we will need to cover the area that recieves sunshine at noon. (Noon
is when the sun is straight above you, so it would be the bird's-eye.) Each plant costs $1.25, so
145*1.25= about 181.70 dollars. That is the cost needed to cover it
These students also did a great job of explaining their thoughts. They demonstrated their
metacognitive skills with each step. They even included references they had to use when their
mathematical abilities were not advanced enough. These students also showed perseverance by
not giving up when they didn’t know how to do something but instead they used other resources
to help them overcome those obstacles.
Building on Water 10
6. Supporting Classroom Discussions
Two of the themes that seemed to present in most, if not all, of the student solutions are the dilemma of rounding and how to communicate the answer effectively. The solutions above are organized in order of communication development. I put the ones that seemed to struggle with how to communicate their answer first and the ones that were mostly successful at sharing their complete mathematical process I placed last. I would use this order of presentation in the classroom to show students a progression of how problem solving develops over time. Most students start out problem solving having no idea where to start (student 1) or if they do answer the question they do not know how to write clear enough to make others understand the process they use. I would use these to encourage students to move toward the ones that were clearer. The ones that were more developed weren’t better just because they were longer but because they included algorithms and in some cases diagrams. Students should be reminded that communicating their answer should be clear enough to teach someone else how to solve the question. One of the solutions that I thought was particularly interesting was Student 5. This student actually used graph paper to draw the building and water to scale. This is an interesting way to get around a possible lack of mathematical understanding. It is possible that this student had not been exposed to the Pythagorean theorem yet so he found a way to overcome that limitation. This solution shows flexibility and perseverance where many students would just say “I don’t get it.” One of the other goals I hope would come out from these selected solutions is the significance in proper rounding. So many students round haphazardly without any regard to significant digits or to what makes sense with the context of the problem. Some students used 3.14 for pi that is just an approximation and in this case can make a difference in the final answer. Students often see the importance of rounding but in this case, when you show them it makes a difference in the money amount, it might resonate the importance of rounding at the latest possible time
7. Closing Comments
I think this problem is a great springboard for students who are learning algebraic concepts but
who haven’t quite grasped the need for algebra. This problem gives the algebra a context that
most students can relate to in regards to territory and money. While the mathematics is fairly
complex for a 6th grade student, an advanced student may be able to get close to an answer using
diagrams and scale drawings.
Building on Water 11