My 5th grader has a math problem I cannot seem to figure out. I'm thinking there is a formula I don't know. The question is: Pamela used 6 triangles to make a hexagon. How many triangles would she need to make a similar hexagon with sides that are 3 times as long? The choices are 9, 18, 36, or 72. How can we figure it out without drawing a picture to scale and filling it in with triangles? Thanks!!!!!
Geometry
- Thread starter ksullyy
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JeffM
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ksullyy said:
Frequently a sketch is sufficient to put you on the right path. There is usually no need for drawing to scale.
First off, does the question say something about the triangles used to make the larger hexagon being the same as the triangles made to make the smaller hexagon? If it does not, the correct answer is 6 despite the choices given. If Pamela can make a hexagon out of six triangles in the first instance, she can in the second instance also by using bigger triangles. If the question does specify that it is congruent triangles that are to be used, then tkhunny has given you a clue that is mathmatically rigorous.
But a sketch may be clearer to your 5th grader. Try sketching a small hexagon using six small EQUILATERAL triangles. (The equilateral is not necessary; it just makes it a lot easier to sketch.) The sketch does not need to be very exact. OK. Now imagine that you were to use six EQUILATERAL triangles with sides 3 times longer to build the bigger hexagon. Look at just one of those bigger triangles. Can you make that bigger triangle from the smaller ones? How many does it take? So if it takes that many small triangles to build one big triangle, how many small triangles does it take to build the six big triangles that make up the big hexagon.This approach is not mathmatically rigorous, but it may help your child understand the mathmatics involved. If all this seems too involved, just pay attention to tkhunny, who is giving you great information about all kinds of scaling problems.
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I had not imagined that we might change the size of the triangles. Excellent counter interpretation.
Note to OP: When you wander around in the land of "There is only one possible way to answer the question", like I just did, you WILL learn LESS. A useful exploration considers more than the first impression.
Note to OP: When you wander around in the land of "There is only one possible way to answer the question", like I just did, you WILL learn LESS. A useful exploration considers more than the first impression.
ksullyy said:
While the manner in which the problem is stated is not clear, I choose to interpret that the triangles are intended to be equilateral.
This allows you to manipulate cardboard equilateral triangles (sides of 1 inch length) to form any chosen multiple of the basic regular hexagon.
If you do this, you will soon realize that the number of triangles required to configure a hexagon whose sides are 3 times the initial hexagon's side length is 54. And yes, there is a formula in the form of N = 6n^2, n being the number of unit sides of the hexagon. This would lead to numbers of unit triangles as follows.
n.....1.....2.....3.....4.....5...
N.....6....24...54....96...150...
If, per chance, you would be interested in a non-regular hexagon, the number of unit triangles would derive from N = 2n(n + 2). this is based upon side 1 = 1, side 2 = 2. side 3 = 2, side 4 = 1, etc., n being the length of the side adjacent to the unit triangle.
..........1
......2......2
......2......2
.........1
Granted, while this follows from the equilateral assumption, other assumptions will lead to similar conclusions.
Thanks to all who took the time to answer the question. Two responses were 54 which was not listed as a choice. The private answer I got with the picture which showed why 54 should be the answer makes sense. Perhaps the teacher needs to re-look at the question. I appreciate all of your time!
Lucky guess.