Note: Diagram has been modified from original.
For Figures 1 and 2, the following is given:
AD + AC = BD + BC
Perimeter of triangle = 36
AC = 15
Show that the length of AD = 3.
In other words, demonstrate that the length of AD is independent of sides AB and BC.
Instead of imposing or suggesting my way of using this question to build an investigation, how would you do it?
If you're new to this blog, I have published dozens of examples of investigations which are intended to develop process, conceptual understanding, generalization and a different view of what mathematics is for our students. An investigation allows students to explore particular cases before attempting to generalize and abstract. Some might call this scaffolding. I see it as creating an experimental environment in the classroom, encouraging our students to become mathematical researchers! I know every argument against this approach but, remember, I'm suggesting that this type of activity only be used perhaps once a month...
The question above can be given as is to some groups of students but may not be appropriate in its present form for many others. The question can be reworded or changed completely.
What would you do?
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8 comments:
If the triangle is equilateral, AC can't be 15 if the perimeter is 36.
Figures I would make a dumb error like that. I added the equilateral case last and I never checked the value. I definitely need you to proofread before I post! Thanks, Mathmom! Just ignore the equilateral case...
I have a confession to make. When I read this problem on Monday, it sounded hard. I kept it on a tab in Firefox until now. This evening I finally pulled out a piece of paper and drew the triangles and wrote down the conditions.
I was embarrassed at how easy it suddenly became. It almost feels like that hotel problem with the alleged missing $5. Something about the description made me think there'd be a deep, mysterious geometric connection.
I have no idea how to present it, but it was interesting to see how I avoided what I thought would be hard. ;^)
Why so defensive? Once a month is too little! I want to thank you for these investigations, they have really added to my teaching (more than once a month!)
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Sue,
I agree-- the problem appears harder than it is. Actually it was an SAT question which inspired this post. A right triangle was given with legs 9 and 12, no perimeter. Most students would determine the hypotenuse and go from there. Some might use a straight algebraic approach:
x + 15 = (9-x) + 12
Others might use"guess-test-revise". My instinct is that most would not use the perimeter approach --- this is why I changed the question.
Which version do you think is more challenging?
BTW, I appreciate your comments. If our students are told or believe that a question is difficult, they may look for something more than there is in the problem or some will just give up!
I found this question to be interesting on many levels:
(1) It reminded me of those "walk-around" problems you may have seen.
(2) The formula (semiperimeter) - c is fascinating in its own right. We can encourage our students to keep probing, make further observations and conjectures, then verify it.
In the end, pretty much any problem can lead to a discussion of how to enrich our instruction, the raison d'etre for this blog.
Ingo---
Thank you for those supportive comments. I'm not publishing as often anymore and I have been remiss in promoting the math carnivals and other great blogs. I feel guilty about this.
>Which version do you think is more challenging?
I think your version is more challenging, because it's less concrete.
(If I saw this along with lots of other problems, I'd just dive in, and (for me, anyway) the path was obvious as soon as I began to follow it. The challenge was in deciding to try it.)
We can show easily AD=3 for fig-1 by using pythagoras theorem and perimeter ..
But it is very difficult to show the same for fig-2 ...
How can i show AD=3 for fig-2 ????
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