- 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
Janelle - for # 5, think areas of litle triangles and big triangles all adding up. and rearrange the equation to show +'s on both sides. makes more sense to me that way. - Sue
Thanks Sue. As soon as I wrote that I figured it out:-) I just needed to draw one more line so I could make all the connections with the triangles. ~Janelle
help # 4. I thought I had it, now I don't! Sue
It seems to me that for number five there about 5 different cases we have to consider, based on whether or not the perpendiculars from P intersect the vertices of triangle. Or does that not have much bearing on the solution? - Tim
I'm not sure if that matters or not. My guess is that it doesn't. The reason I say this is because whatever point you choose as the intersection could always be proven to be or not to be the vertex. However, like I said before, I'm not entirely sure. ~Janelle
I did make the perpendiculars intersect with the vertices. Only way I could get it to work out. Not sure if this is the cheater way though - Sue
Any hints for 10? 4? or 5?
There is a hint for 10 when you click on the link below:
maybe I'm slow, but I don't follow the second hint. Does it have to do with the sides adding up?
so for number ten if you draw lines through A and D perpendicular to BE you can use similar triangles and parallelograms--Jrhaynie 13:53, 23 September 2009 (UTC)
