Geometry PoW Packet
Divided Rectangle
September 8, 2008
Welcome!
•
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This packet contains a copy of the problem, the “answer check,” our solutions, teaching suggestions,
a problem-specific scoring rubric, and some samples of the student work we received in September
1998, when Divided Rectangle first appeared. It is LibraryPoW #315.
We invite you to visit the PoW discussion groups to explore these topics with colleagues. From the
Teacher Office use the link to “PoW Members” or use this URL to go to geopow-teachers directly:
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The Problem
In Divided Rectangle, students determine the area of a region of a rectangle that has been split into
four smaller rectangles. They use knowledge of area of rectangles, but also about variables and
general cases – the dimensions of the small rectangle in question don’t need to be specified, and in
fact shouldn’t be.
The text of the problem is included below. A print-friendly version is available from the “Print this
Problem” link on the current GeoPoW problem page.
Divided Rectangle
A rectangle is divided into four rectangles with areas of 45, 25, 15, and x square units. Find x.
Answer Check
The value of x is 27 square units.
If your answer does not match our answer,
• did you remember that the area of a rectangle can be found by multiplying its length by its
width?
• notice that we don’t know the area of the surrounding rectangle.
• did you try picking dimensions for, say, the rectangle with an area of 45 square units and seeing
what happened to the rest of the rectangles?
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,
• did you use 9 units and 5 units as the dimensions of the rectangle with an area of 45 square
units? What happens if you use different dimensions? What happens if you use a variable as
one of the dimensions?
• did you explain your work as well as you can?
• what hints would you 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.
Our Solutions
The key concept in this problem is the area of a rectangle.
Method 1: Algebraic Reasoning
Assign the variables to the four interior segments as shown (so that ab = x, ac =
45, dc = 25, and bd = 15). We want to find the product of a and b, so let’s find
expressions for them.
From the equations above, we can say that a = 45/c and that b = 15/d. Looking at
the product:
45
15
a * b = ---- * ---c
d
45 * 15
a * b = --------c * d
45 * 15
a * b = --------25
a * b = 27
So the value of x is 27, no matter what a and b are individually.
Method 2: Guess and Check
Assign the variables p, q, r, and s to the four interior segments as shown. Let’s pick 5 units for p, since
it’s a common factor of 45 square units and 25 square
units. Then q must be 5 units and s must be 9 units (since
the area of a rectangle is found by multiplying length *
width, we can find the width by dividing the known area by
the length). If q is 5 units, then r must be 3 units. So rs = 9
units * 3 units = 27 square units. That’s one possible value
for x.
What if we don’t use factors of the given areas? Let’s say that p is 10 units. Then q is 2.5 units and s is
4.5 units. If q is 2.5 units, then r is 15/2.5, or 6 units. Then rs = 6 units * 4.5 units = 27 square units. So
it’s still 27 square units!
To find out if it’s always 27 square units, let’s just use p instead of a number. Then s is 45/s and q is
25/s. If q is 25/p, then r is 15/25/p, which is (3p)/5. Then rs is (3p)/5 * 45/p, which is 27. So it’s always
27 square units.
Method 3: Proportional Reasoning
The rectangles with areas of 25 square units and 15 square units share an edge, so the ratio of their
lengths must equal the ratio of their areas (see below for an example). Their lengths are in a ratio of
25:15, or 5:3. Similarly, the rectangles with areas of 45 square units and x square units share an edge.
So the ratio of their lengths must equal the ratio of their areas. But their lengths are the same as the
lengths of the 25 and 15 rectangles. This means they must be in a ratio of 5:3. So to find x, we find a
number that's in a 5:3 ratio with 45 (weird wording).
5
45
--- = ---3
x
x = 27
The same thing could be done finding the ratio of 45 and 25, then the ratio of x and 15.
*Example: Below are two rectangles. Both have a height of 4 units. The area of the small rectangle is 5
* 4 = 20 square units, and the area of the big one is 7 * 4 = 28 square units. The ratio of their areas is
20:28, or 5:7. This is the same as the ratio of their lengths.
Copyright © 2008 by The Math Forum @ Drexel
2
Teaching
Suggestions
Many students will find one possible solution, specifying dimensions for the four small rectangles, and
assume they’re done. While it’s true that they may have found the correct answer, it’s important for
them to be able to show that while they relied on specific dimensions to find a solution, choosing any
dimensions at all will produce the same value for x. If students chose integer dimensions for the
known rectangles, ask them what happens if they uses non-integers. Once they’ve explored that and
hopefully gotten the same answer every time, you might ask if they can think of a way to show that the
actual dimensions don’t matter.
Look for opportunities in the classroom where different students find “different” specific solutions, in
terms of the dimensions they choose. That’s a great opportunity to ask them if they think the other
solutions are right, and whether they can think of a way to show that the answer will always be 27
square units.
You can find more online resources to go along with this problem at
http://mathforum.org/geopow/puzzles/supportpage.ehtml?puzzle=401
All the resources to go with this and all other Current Geometry PoWs from this year are linked to from
http://mathforum.org/pow/support/
Sample
Student
Solutions
Focus on
Strategy
In the solutions below, we’ve focused on students’ strategies. Generally speaking, it’s important that
students pick a sound strategy that doesn’t rely on luck. In this case, a sound strategy will show that
the answer must always be 27 square units. Students who pick specific dimensions and show that the
resulting solution is 27 square units are considered Apprentices – they have some ideas about how to
solve the problem, but they haven’t yet found a way to solve it completely. With this problem, that
might simply be that they haven’t realized that they should be looking for a general solution.
The student work below illustrates the range of possible strategies (though it doesn’t cover all the
possible ways students might attempt to solve the problem). Each group of solutions is preceded by
information about that category of the scoring rubric and the ways in which the solutions in that set are
similar. For each individual solution, I’ve offered some comments about the solution, plus what I think
a good next question might be for that student.
Copyright © 2008 by The Math Forum @ Drexel
3
Novices
Ben
age 12
Students who are considered a Novice in Strategy haven’t exhibited any ideas that might eventually
lead to a successful solution. Usually these students have the incorrect answer (as is the case with
these examples). It’s sometimes possible to work with their ideas, even if it’s by asking questions that
might help them see that they need to take a different tack.
The answer for X is 15.
45+25+15= 85
100-85= 15
Strategy
Novice
Andrew
age 9
Strategy
I think the answer is 20.
I knew that x was bigger than 15 so the answer had to be more than 15. It
also was smaller than 25. I thought it looked like it was exactly in the
middle of 15 and 25.
Novice
Aaron
age 15
Strategy
Novice
The area of rectangle x is 35 square units.
First, I noticed that the dark yellow rectangle's area was ten square units
less than the yellow rectangle's area. Then I noticed that the yellow
rectangle's area is twenty square units less than the pale yellow rectangle.
I then realized that the lower rectangles' area is ten square units less than
the upper rectangles' area, and the left rectangles' area is twenty square
units greater than the right rectangles' area. Finally, I subtracted 10 from
the area of the pale rectangle's area, and got 35 units squared for x.
Copyright © 2008 by The Math Forum @ Drexel
It looks like Ben might be
thinking that the given areas
are actually percents. At the
very least, he thinks that the
area of the large rectangle is
known. This might be an
indication that he’s
misinterpreted the problem
(which would mean that his
strategy is actually sound,
as he’s correctly solving a
different problem). But I’m
inclined to think that he just
wasn’t sure what to do with
the given information. I
might simply ask him how
he knew that the area of the
whole rectangle is 100
square units.
Andrew seems to be going
by how the picture looks –
that and that fact that all the
given areas are multiples of
5. I’d suggest that he can’t
just go by how the picture
looks, but needs to come up
with some “math” reasons
for his choices. I might ask
him to find some dimensions
for the small rectangles that
would work with his areas,
hopefully leading him to see
that 20 isn’t a possible
answer.
Aaron has done a great job
looking for relationships
between the numbers. This
is a handy skill that
unfortunately doesn’t help
so much in this situation.
He doesn’t seem to be
using the idea of area at all,
so I might do as I did with
Andrew and ask him to find
some dimensions for the
small rectangles that would
work with his areas.
4
Apprentice,
Group 1
Jaewoo
age 14
These solutions are examples of students finding a single possible solution. It’s true that they’ve found
the only answer, but they don’t show that they know that, or even that that might be important.
They’re on their way to a sound strategy, but they’re not there quite yet.
X= 9 times 3 (27)
OK! X's segment have 9 on the width, and 3 on the length
square which has a 45, have 9 on the width and 5 on the length
Strategy
Apprentice
square which has a 25, have 5 on the width and 5 on the length
square which has a 15, have 5 on the width and 3 on the length
So if you guys fill this numbers on that picture, you guys will get an
ANSWER.............
Jaewoo’s solution is very
typical of the solutions that
we received – assume
dimensions of 9 units and 5
units for the rectangle with
an area of 45 square units,
and go from there, while not
saying much about why any
of the steps were taken.
I’d ask him if there are other
possible solutions, either for
the unknown rectangle or for
the dimensions of the known
rectangles.
Jared
age 12
Strategy
Apprentice
The area of X is 27 square units.
First we know that because all of the shapes are rectangles and not
squares that the figure to the upper right cannot be 5l(length) by 5w
(width). Thus proving that the one to the bottom right cannot be 3l by 5w
and that the one to the upper left cannot be 5l by 9w. That tells us that
some of our numbers will be decimals. By process of elimination we find
that the upper right figure is 10l by 2.5w. That tells us that the one to the
lower right is 6l by 2.5w and that the one to the upper left is 10l by 4.5w.
That means the dimensions of X are 2.5l by 4.5w and an area of 27 sq.
units.
Copyright © 2008 by The Math Forum @ Drexel
Jared has assumed that
since the figures are
rectangles, they can’t be
squares, hence the
rectangle with an area of 25
square units can’t have
dimensions of 5 units and 5
units. So he’s picked
another pair of numbers
(each number is followed by
an el or a double-u) to use
that multiply to 25. It’s not
clear what he means when
he talks about the process
of elimination, so I would
probably ask him about that.
5
Apprentice,
Group 2
Matthew
These students have discovered more than one way to show that 27 is the unknown area, but they’ve
not yet shown that the area must always be 27 square units.
x equels 27.
________9__________5_____
|
|
|
|
|
|
5|
45
|
25
|5
|
|
|
|______________|________|
|
|
|
3|
x=27
|
15
|3
|
|
|
|______________|________|
9
5
age 15
Strategy
Apprentice
The way I figured this problem out was:
It’s interesting that he
checked his numbers with
decimals – it’s not clear how
that’s a check, unless he
thinks that the answer really
will be the same no matter
what the dimensions are of
the individual rectangles. I
might ask him whether he
was surprised to get the
same answer using two
different sets of numbers.
I started with 25. I knew that if each side had to be whole numbers, that
the dimensions of the 25 rectangle must be 5 and 5. If you put the 25
rectangle together with the 15 rectangle, then the short sides are 5 each,
since the opposite sides are congruent. Then in the 15 rectangle, the
other side is 3, since 15 divided by 5 is 3. Then in the 15 + x ractangle,
the shorter sides are 3. In the 25 + 45 rectangle, the shorter sides are
both 5. In the 45 rectangle alone, the other side is 9, because 45 divided
by 5 is 9. So, in the 45 + x rectangle, both the shorter sides are 9. Now,
in the x rectangle alone, the dimensions are 3 and 9. 3 times 9 is 27, so x
equels 27.
______0.9__________0.5___
|
|
|
|
|
|
50|
45
|
25
|50
|
|
|
|______________|________|
|
|
|
30|
x=27
|
15
|30
|
|
|
|______________|________|
0.9
0.5
Now, to check my answer, I substituted the whole numbers and used
some decimals. 0.5 times 50 is still 25, o.5 times 30 is still 15, 50 times
0.9 is still 45, so 30 times 0.9 still equels 27.
John
age 13
Strategy
Apprentice
There are at least 4 solutions to this problem. The area of X is always 27
and it consists of sides .6 and 45, 3 and 9, 6 and 4.5, 12 and 2.25
First, examine rectangle 45,its whole number factors are 1 X 45, 3 X 15,
and 5 X 9.
Second, examime rectangles 25 and 15 the same way.
Factors:
45
1
3
5
9
15
Rectangle
25
15
1
1
5
3
25
5
His belief that there “may be
more solutions” is really
interesting! And he’s sticking
to rational factors, for some
reason. He’s really close to a
more general solution, and I
might encourage him to
think about ways he might
show that—maybe by using
a variable for a dimension of
one of the known rectangles
and seeing what happens..
Next, determine the common factors of the rectangles 45 and 25,they are
1 and 5 using 1 as the height of 1 side, the lengths of the top rectangles
Copyright © 2008 by The Math Forum @ Drexel
6
are 45 and 25. Dividing Rect 15 by 25 results in the second side of height
.6. Rectangle X has sides of 45 and .6 giving an area of 27.
Using 5 as the height of the 2 top rectangles would have lengths of 9 and
5, this results in rect 15 having a length of 5 and a heigth of 3. This results
in Rect X having a length of 9 and a heigth of 3 giving an area of 27. The
third solution of 6 and 4.5 was given in my previous answer.Using a height
of 20 gives lengths of 2.25 and 1.25. Dividing 15 by length 1.25 gives a
heigth of 12 which gives Rect X a length of 2.25 and a height of 12 There
may be more solutions if Rectangles 45 and 25 have additional rational
common factors.
Apprentice,
Group 3
Sorin
age 17
These are strategies that are sound, and do show that 27 square units must be the answer all the time,
but the students have given no indication that they know that. Consequently, while the strategies could
be mathematically sound, it's entirely possible that the student stumbled on the right answer by luck.
It’s sometimes hard to tell since, for example, randomly guessing proportions and knowing how to set
up correct proportions can look identical. This is a case of where the Completeness of a student’s
explanation affects their Strategy score.
x=27
25/15=45/x ==> x=15*45/25=27
Strategy
Apprentice
Regan &
Chelsea
age 11
The answer to this week's problem is 27.
Divide the two top numbers, to find the ratio. We then multiplied the ratio
times the bottom number. the answer was 27.
Sorin has used what’s
possibly the most efficient
way to solve the problem. I
would ask him to explain
how he knows that his
original proportion works.
Same deal here – it works,
but do they know that? I’d
ask them to tell me more
about why they did the
things they did.
Strategy
Apprentice
Chung
age 15
The answer is 27.
x+45:15+25=45:25
x+45:40=45:25
Strategy
Apprentice
And another one! I’d ask
Chung why these are valid
steps.
25(x+45)=40*45
25x+1125=1800
25x=675
x=27
Copyright © 2008 by The Math Forum @ Drexel
7
Practitioner
The idea that you can find the area of that chunk without finding its dimensions is a level of
sophistication that not all students in geometry have reached. These three kids obviously get it. Brian
is the only one who explicitly says, “We want to know the product of [the length] and [the width].”
Finding the product of two quantities without finding the actual quantities is a handy little strategy.
Each of these is an example of a sound strategy and is a Practitioner. An Expert would use more than
one strategy, or talk about different strategies or the choices they made.
Tiffanie
age 13
Strategy
Practitioner
The number for x is 27.
Let the side that
the side that
the side that
the side that
Then
ab = x
ad = 15
bc = 45
cd = 25.
is
is
is
is
common
common
common
common
to
to
to
to
x and 15 be a
x and 45 be b
45 and 25 be c
25 and 15 be d.
Since abcd = 15*45 = 675 and cd = 25, ab(25)= 675 or
x = ab = 675/25 = 27.
Morgan
age 14
Strategy
Practitioner
Brian
age 19
The value for x is 27.
Label the segments as follows. Top a and c, vertical segments b and d.
a(b) = 45, c(b) = 25, c(d) = 15,and a(d) = x. I will use substitution to solve.
b= 45/a and c = 15/d therefore (45/a)(15/d)= 25. Mulitplying both sides
by (ad) you get 675 = 25ad . Dividing by 25 gives the answer of 27.
The area of the rectangle in the lower left hand corner is equal to 27.
L1 x W1 = 45
L1 x W2 = 25
Strategy
Practitioner
L2 x W2 = 15
We want to know the product of L2 and W1.
L2 = 15 / W2
W2 = 25 / L1
Therefore L2 = 15/(25/L1) = 15xL1/25 = 3/5(L1) By substituting into the
first equation L1 x W1 = 45 L2 x W1 = 3/5(L1) x W1 = 3/5 (45) = 27.
Another way to do this problem is to check the proportions of the widths
which would be a 9 to 5 ratio. 9/5 x 15 = 27.
Tiffanie has clearly shown
the work that she did. She
doesn’t talk explicitly about a
general solution, but has
shown that it doesn’t matter
what a and b (or c and d)
actually are. I might ask her
if she has any hints she
might give another student,
or if she could say a little
more about why she chose
to use a variable.
Morgan’s solution is a lot like
Tiffanie’s, only not quite so
easy to read because of the
paragraph format. I might
ask her the same questions I
would ask Tiffanie.
This is interesting! Another
unique method that does,
indeed work. I especially like
that he explicitly said that we
want to know that product.
It also brings up the 3/5
relationship between the
“top” length and the
“bottom” length. [Further
exploration shows a 9/5
relationship between the
“left” width and the “right”
width – the 3 and 9 in those
fractions are related to the
27, so do some exploring!]
I might ask him to say more
about his closing idea about
the ratio of the width.
Copyright © 2008 by The Math Forum @ Drexel
8
Scoring Rubric
On the last page is the problem-specific rubric, to help in assessing student solutions. We consider
each category separately when evaluating the students’ work, thereby providing more focused
information regarding the strengths and weaknesses in the work. A generic student-friendly rubric
can be downloaded from the Scoring Guide link on any problem page. We encourage you to share it
with your students to help them understand our criteria for good problem solving and communication.
We hope these packets are useful in helping you make the most of Geometry PoWs. Please let me
know if you have ideas for making them more useful.
~ Annie
Copyright © 2008 by The Math Forum @ Drexel
9
Practitioner
solves for the general case, not just by assigning
specific dimensions to one of the rectangles and
working from there
makes a number of mistakes that makes no mistakes of consequence
matter
deals with units appropriately when they use units
thinks that the dimensions of the
rectangles must be whole
numbers
finds the value of x using specific picks a sound strategy—success achieved through
dimensions for the rectangles
skill, not luck
attempts to find the area of the rectangle labeled x
uses the given numbers as
understands that the given numbers are areas (as
percentages – assumes that the opposed to percents or something else)
area of the large rectangle is 100
understands that the figures are rectangles (which
means they could be squares)
Apprentice
© 2006 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 the comment
they leave after viewing our
answer:
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)
does one reflective thing
does two reflective things
explains where they're stuck
reflects on the reasonableness of
summarizes the process they used
their answer
checks their answer (not the
same as viewing our "answer
check")
lots of spelling errors/typos
long and written in one paragraph
Clarity explanation is very difficult to another student might have
read and follow
trouble following the explanation
Completeness has written nothing that tells doesn’t include enough
explains almost all of the steps taken to solve the
you how they found their
information for another student to problem, which might include
answer
follow their explanation
• how any dimensions were chosen
• how the area of a rectangle is calculated
makes statements of fact without
supporting explanations
Communication
Accuracy has made many errors
Strategy has no ideas that will lead
them toward a successful
solution
Interpretation doesn’t understand what’s
being asked in the problem
Problem Solving
Novice
For each category, choose the level that best describes the student's work
Expert
http://mathforum.org/geopow/
does three or more reflective things or
does a great job with two of them
revising their answer and improving
anything counts as reflection
comments on and explains the ease or
difficulty of the problem
answer is very readable and appealing
formats things exceptionally clearly
the additions are helpful, not just “I’ll
say more to get more credit”
explains all the steps taken and adds
in useful extensions and further
explanation of some of the ideas
involved
[generally not possible – can’t be more
accurate than Practitioner]
solves the problem in two different
ways
since there is no Extra, students won’t
score Expert in this category
Geometry Problem of the Week Scoring Rubric for Divided Rectangle