My Other Math Sites
Lessons From the Classroom
"Working Together" Problem
Heya, back-to-back post about a problem from Five Triangles mathematics.
When I tweeted how much I love this problem, a few people did not feel the same at all. Here are my reasons for appreciating this problem:
It's a notched up "working together" problem that I have not seen before.
It has percentages and fractions.
I can use rectangles to solve this. (I was asked on Twitter how I would solve this using rectangles, hence this post.)
I had to work on this problem. This is a big reason for me. We should assume that if we're teaching a particular math subject — Geometry, Statistics, or Calculus — that we're able to easily do all the exercises in the textbook. A set of exercises allows us to practice a particular skill. But a problem should require us to think. I hope I've encouraged problem-solving enough with my students that they value a problem more when they have to struggle with it, when they don't know immediately how to start it, when they get stuck and become frustrated, when they seek others for help, when they can leave the problem and come back to it another day.
While I'm at it, I also love the site Five Triangles in general for a couple of reasons:
The Geometry problems are simply stated and interesting. They make me pause and think, very few have been automatic gimmes.
The solutions are not posted. I really appreciate this because if they were, we might be tempted (mainly due to lack of time) to check the answers too early before we allow ourselves a chance to work through the problem and perhaps struggle with it. "Anticipating" is the first of 5 Practices that gives us insight on how students might solve the problem.
I did, however, retype the question above so it's easier to read and track information. I also numbered the paragraphs for quicker reference.
How we worked through this problem. Colors and all.
Draw a rectangle to represent the task. It has an area of 80 square units because that's the LCD of the three fractions in the problem.
Because this grid represents the task, we use it to fill in the amount of work done. Paragraph [3] is the first concrete piece of information that allows us to do this.
We continue to fill in the work done as described in paragraph [4].
Paragraph [5] is the first piece of information that allows us to figure out C's rate. Knowing that C can do 16 boxes in 8 hours means C can do the task — 80 boxes — in 40 hours.
With C's rate, we can now take on paragraph [2]. We know from the last step that C's hourly rate working alone is 2 boxes per hour or 10 boxes in 5 hours. But when working with A, C's rate is 40% faster, therefore instead of getting just 10 boxes done, C can get 14 boxes done in 5 hours when working with A.
From picture above in green, we know A and C did 24 boxes in 5 hours, and since C was responsible for 14 of those, the remaining 10 boxes were done by A.
Then A's hourly rate when working with C is 2 boxes per hour. Because this hourly rate represents a 20% increase than if A were to work alone, the math we need to do is 2 boxes divided by 1.2 to get 5/3 boxes. Solving for x in the proportion below gives us the answer that A completes the task in 48 hours.
Lastly we use paragraph [1] to figure out rate for B. We know A's alone rate is 5/3 boxes per hour, but when working with B, A's rate is 40% faster. Thus we multiply 5/3 by 1.4 to get 7/3. If A can do 7/3 in 1 hour, then A can do 35/3 in 5 hours when working with B.
The yellow boxes show that A and B can do 25 boxes in 5 hours, so subtracting 35/3 from 25, we see that B did 40/3.
To get B's alone rate, we divide 40/3 by 1.2 (because B is 20% faster when working with A) to get 100/9. Solving the proportion below gives us the answer of B completing the task in 36 hours.
Technology and Construction Papers are Cool
Big-time struggles for my 8th graders on this problem from Five Triangles mathematics.
The one student who got it also struggled, but he was good about our rule of "never tell an answer." What I gathered from seeing their boards and listening to them explain:
They wanted to find the area of the parallelogram first.
So they needed to find the height.
To find the height, they reasoned that triangle ABC was isosceles, making the perpendicular bisector AY also be the height of the parallelogram. Then they could use the Pythagorean theorem to find AY. (I can't get AB and DC to be exactly 6.00 cm.)
So I showed them the parallelogram below just to be less helpful and to remind them about assuming something is isosceles because it looks it.
Without a clear way to find the height, frustration mounted.
But major props to them for persevering as you can tell from their boards that they tried to dissect the parallelogram into even more pieces in hoping they'd find what the shaded piece would be equal to.
The next day... I let them struggle some more. They weren't seeing the key pieces (at least to me they were key), so I told them to look at the "relationships" among the pieces. They were pretty sure the diagonals of a parallelogram cut it into 4 triangles of equal area. Then they were stuck again. I then asked them to construct this parallelogram using Geometer's Sketchpad (GSP) and find the answer to the question posed. Soon one student asked, "What should the angles be?" I replied, "It doesn't say in the problem, so I don't know. But it is a parallelogram, so make sure you have the same side lengths." Without much trouble, they found the answer. They also realized that even though they may not have created congruent parallelograms among themselves, they all came up with the answer of 0.17 or 1/6.
Me: So why did I ask you to use GSP to do this?
Student: Because we couldn't do it by hand?
M: Maybe. Or maybe you didn't have enough time. But why do we use GSP in general? What's the difference between doing a construction by hand and doing it using GSP?
S: So we can move things around.
M: Right! It's dynamic! Then this is what I want you all to do. Move things around. Drag the side lengths. Change the side lengths. But while you're doing all this messing around with the parallelogram, I want you to pay attention to the numbers, the measurements.
Not long before many of them chimed in excitedly: The ratio doesn't change! I came by each kid's screen to check their work. (These are my reconstructions of some that I saw.)
To make the most of their construction (and a good problem), I then asked them to find the ratios among all the pieces that they saw. I wish you could hear them. A bunch of them kept making the same exclamations: So cool! Mind blown. While only one student found the solution by hand, the rest of them felt pretty good about their construction of parallelograms, and they discovered important relationships among the pieces. Technology came to the rescue. Here's how I used colored construction papers to help students see the relationships better.
Advantages:
Instead of having to write down or say "triangle ABC," "triangle BEC," or "quadrilateral FDGE," I can just refer to each one by color.
By not having to follow the letters of ABC or GCE, kids can focus on the visuals and see the relationships more quickly.
I can show congruence by placing one piece on top of another.
Something like this:
The diagonals cut the parallelogram into two pairs of congruent triangles. Therefore, each blue is 1/4 of whole. (I reflected each pair so they can see the congruence.)
Below left: green is 1/2 of whole.
Below right: red + yellow = 1/2 of green, or 1/4 of whole, because they share the same height and half the base.
Below right: red + pink is 1/4 of whole (same size as a blue). Since left and right images below are equal in area, and red is red, therefore yellow equals pink.
Blue + yellow (combined) is a similar triangle with pink due to AA postulate. And because their side lengths are 6 to 3, or 2 to 1, then their areas are 4 to 1. Let's call the area of pink as 1, and we learned from above that yellow equals pink, so yellow is also 1. This leaves blue to be 3.
Putting all the pieces together, we have a total area of 12, so the shaded part (red) is 2/12 or 1/6.
Kate
Add to what Robert said, your kids have enough GSP skills to use it to investigate a problem. Kudos! That's a tough trick to pull off. I did it a different way! I found constructing a supporting parallel line dissected the quadrilateral in a helpful way. http://www.geogebratube.org/student/m83241
February 10, 2014 12:10 PMfawnnguyen wrote:
Hi Kate. Yay, beautiful work!! We did this activity just last week and one of the kids asked at the start, "How do you find the area of a non-parallelogram or trapezoid [FDEG] anyway?" I replied, "Well, can you divide that shape into other shapes that you can find the areas for?" So, I saw the groups draw in that parallel line that you have. They were able to find the area of pink triangle as 1/8 of whole but struggled with finding area of blue triangle during their paper/pencil work.Thank you, Kate.