Maths Homework
# 2 son was working through his year 7 maths homework.
He said: "Dad, how do I do this one?"
He was being asked to divide various object up into fifths by drawing lines on the provided objects:
1) A rectangel. No problem, measure the long side (50), divide by five (10) and draw in the lines every 10 units.
2) A circle. A bit harder, divide 360 degrees by five (72) and draw the lines out from the centre at 72 degreee increments.
3) An irregular right angle triangle, sides are 3, 4 and 5. Ummmmmmm.
4) A regular hexagon. Errrrrrrrrrrr......
*starts to wonder how he would have in fact gone in "Are you smarter that a 5th grader?"*
Ok, I'm going to phone a friend - anyone give me a clue how to approach 3 and 4?
posted by wysiwyg @ 11:03 AM
5 comments
5 Comments:
ummmm....gonna have to ask for a clarification ...i'm assuming the 5 pieces have to be equal in some way. the solutions you describe have the five pieces 'congruent' (the same shape and equal area). If you just need equal area pieces then divide the length 5 side into 5 length-one intervals and draw the lines from the opposite vertex. That gives 5 triangles with the same height and base ,so they have the same areas though not the same shape.
i'll keep thinking , it's probably possible to do with 5 congruent pieces...
The easiest way I can think of to do the hexagon would be to first divide each side into 5 equal segments. Then draw lines from the center of the hexagon to every 6th mark. Like the triangle, the parts won't be congruent, but they'll be equal.
nice solution, bis!
*is totally and in all ways enamored of bis's and insom's math prowess*
My mind just doesn't think in these kinds of terms.
~neophyte
Thanks for that guys - I knew I could rely on you! The original question was "Divide the following shapes into fifths", so I think your assumption about the answer being able to be different shapes but the same area was correct.
I had in fact figured out the triangle one, but started to doubt myself because I had forgotten the basic "area = base * height" piece of geometry.
And the hexagon solution is very clever Bis! Having drawn up the answer by hand using your technique I suddenly realize that I could have also solved in in the same manner as the circle one by drawing lines at 72 degree angles, or alternatively drawn a regular pentagon outside the hexagon and joined the corners into the middle of the hexagon.
But of course its always easier to see those things once you have the answer....
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