I’m proud of my students. I’m proud of what we do in room 15. My classroom. My home away from home for the last 11 years. These bopping teenagers, sullen one minute bubbly the next, hormonal but invincible. They don’t all love math like I do. (Not everyone loves college football, but we get along.)

What is happening in room 15 is the loud and proud math culture that we have set in place. We build it from day one — then we continue to do, say, and write stuff to sustain and strengthen the culture because we know our behaviors are our best evidence that this culture exists.

Here’s one piece of that evidence:

Her frustration resulted in her loving the problem. Her last sentence is an enormous celebration of how much we honor the process of problem solving. Her classmates had their own reflections — short snippets of how they engaged in the problem.

They worked hard on the problem because it was driving them nuts. It’s not unusual to hear kids blurt out, “This problem is making me crazy!” Or, “I won’t be able to think about anything else until I get this!” Now, they’ve owned it. This isn’t about a letter grade any more; and it certainly isn’t about me.

This was the problem they’d worked on.

**The Missing Area**: A 10 by 16 rectangle is attached to a triangle as shown below. If the purple section is 24 square units, then what is the area of the yellow section of the rectangle?

[John Golden @mathhombre, GeoGebra extraordinaire, created an animated gif of this problem.]

[Mike Lawler @mikeandallie solved this problem using similar triangles.]

## 9 Comments

I saw a triangle with base 16 and height 10, hence area 80

This triangle is the purple + the upper white

So upper white triangle area is 80 – 24 = 56

Its base is 16, its height is the vertical side of the purple triangle

So 56 = 16 x vertical side of the purple triangle x 1/2

Then vertical side of the purple triangle has length 7

We now have a trapezoid with base 16 and sides 10 and 10 – 7 = 3

and this is the yellow bit

So its area is 16 x (10 + 3)/2 = 8 x 13 = 104

No algebra to speak of !

Thanks for sharing Howard, but I don’t see the “upside down” obtuse triangle with a base of 16 and a height of 10. Could you help me here?

thanks!

Howard,

I was trying to do it with algebra but kept going in circles. I looked at the first line of your reply without reading the rest of your explanation. When I saw that the large upside down obtuse triangle had area 80, I subtracted 24 from 80 to get 56, which was the area of the white part of the rectangle. Subtracting 56 from the area of the rectangle, 160, I got the answer you go without going through the extra step that you did. Thanks for the hint. When I started down that path, I was amazed at how easy it was and kicking myself for not seeing it. I wonder if the whole figure were rotated 180°, your approach would be easier to see. That’s a good strategy for students (and me) that I will use from now on. Rotate the figure to see if anything jumps out at you.

Fawn,

As always, thank you for sharing your students’ work and for showing that they can be challenged to a point beyond what most middle school teachers would believe. The culture that you create in your classroom allows your students to understand that there are multiple approaches to problems…and they wouldn’t be “problems” unless they made us a bit uncomfortable.

you know what’s fun about this sort of problem is that the approach a student takes will be dictated by what the most recent thing they’ve worked on might be. If they’ve just worked on finding the area of obtuse triangles they might approach it as Howard did.

I myself solved a system of equations probably because I am most confident in algebra. But I can imagine that if you introduced this problem around the time students were studying trigonometry, they’d all try to use trig. Same with similar triangles.

This of course means it can be perplexing at any age! The fact that your students love this challenge means you are indeed winning!

Hi Fawn,

I’m enjoying your website and am sure many other teachers are as well – great resources.

Your problem caught my eye, so I made a GeoGebra worksheet for the problem.

You can use it online, or you can download it if you wish to modify any of the drawing components. I’d be happy to help if needed. This problem has lots of room for great

problem solving discussion and variations. Thanks for sharing.

Bill Lombard

http://tube.geogebra.org/material/show/id/266561

What a terrific question. You should be really proud of the children for showing such perserverance. I love questions that, at first sight, look quite straightforward and then you get the “hang on a minute” issue. A wonderful resource. I shall try it out.

I loved this problem too! Gave it to students this week. One student found the simplest solution. Label the corners of the rectangle ABCD (with A as bottom left). Let E be the vertex of the triangle that is not a shared vertex with the rectangle.

The triangle BCE has a base 16, and height 10, so it’s area is 80. This triangle is composed of the triangle with area 24, and a triangle within ABCD. The triangle within ABCD then must have area 80 – 24: 56. Now we know that the area of ABCD is 10(16) = 160. So the area of the shaded region is 160 – 56 = 104.

I’m so surprised when students find the simple road where I would have shown them something much more difficult. Thanks for sharing this problem!

Hi Fawn

I have been reading your blog for a while now, but almost weekly come back to this post, not because of the problem (although I do really like the problem) it is more about how you talk about your class, the culture you have developed and the learning that is going on. They passion you have for this type of work is evident in every post but I think is particularly pronounced in this one.

It is a post that helps to keep me focussed on what I am trying to achieve each day, it helps to keep me accountable for every task I give to my class and every conversation I have with my students.

Thanks

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