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We don't have to do number 9 because we did the proof in class. -Jennie
 
We don't have to do number 9 because we did the proof in class. -Jennie
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Does anyone have a hint on number 5? I'm not sure if it's something obvious and I'm overlooking it or if there is a little trick.  Any help would be greatly appreciated!! ~Janelle

Revision as of 12:56, 22 September 2009

  • For problem number one, I see that the sum of the areas of the four small triangles is equal to the area of quadrilateral MNPQ. Is this the result we're looking for? It seems a little too obvious but I can't figure anything else out.

I also stated for number one that the area of the little paralllogram is half that of the big parallelogram and that the area of all of the little triangles i equal to the area of the little parallelogram


I am really bad with sketchpad and having trouble with number 2. Can anyone help me get started? -chris [sure. at what point are you getting hung up on? - sue]

  • thanks anyways i got it. For any one need help with it, just calculate the lengths and think of thm 28 (multiply)

any hints for # 10 - sue

for # 3, if anyone has gotten it, is it a bunch of summing of areas?

it is!

for #3, what areas might we be summing? I remember with the example, the intersection was inside the triangle, so we could use the area of the triangle... any hints? ~Lauren

with p being outside, there are now 2 little triangles outside the bigger eq triangle. sum all of the triangles in one direction than all of them in another direction - if that makes sense. what I did was write down the areas for all of the triangles ad see which ones give me what I needed - Sue

I'm down to 9 and 10 now. no clue on 9 - Sue

We don't have to do number 9 because we did the proof in class. -Jennie

Does anyone have a hint on number 5? I'm not sure if it's something obvious and I'm overlooking it or if there is a little trick. Any help would be greatly appreciated!! ~Janelle

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Correspondence Chess Grandmaster and Purdue Alumni

Prof. Dan Fleetwood