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This is part of a homework question

I'm trying to figure out the perimeter of the black-filled object below (sorry, it's crudely drawn in paint, but I think you get the idea). To find the perimeter, I'm assuming that I should first find the perimeter of the rectangle (12*2+3*2 in this case) then subtract each little triangle (which is suppose to be all the same size by the way). The question mentions only one fact other than the size of each side- that it's trisected. I'm pretty sure the answer to this question lies in here...

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Be careful: the process you're describing calculates area, not perimeter. –  Isaac Oct 11 '10 at 21:23
    
@Issac- I see the error here, fixed. Should use * instead of ^. EDIT: I've tried to edit but since @Moron was nice enough to put up the image, I can't edit it without removing the image since I don't have enough rep. –  highy Oct 11 '10 at 21:24
    
@highy: I've edited the ^ into * for you. Given that, note that "subtract each little triangle" may not be quite right to get the perimeter of the black region. I'd suggest that you try to label each edge of the black region with its length and see how far that gets you. –  Isaac Oct 11 '10 at 22:55
    
@Issac- Thanks, I have already arrived at each answer. I guess what I should of set is subtract the hypotenuse of each small triangle. –  highy Oct 12 '10 at 1:47
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1 Answer

up vote 3 down vote accepted

Well, if each side is trisected, that means it's split into three identical parts. The base of each triangle is one of the parts of the 12 side, and the height of each triangle is one of the parts of the 3 side. You can figure it out from there.

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Thank you. Am I correct in saying that then the 12 metre side would be split as 4,4,4 and the other side 1,1,1? –  highy Oct 11 '10 at 21:17
    
Yeah, that's correct. –  Eugene Bulkin Oct 11 '10 at 21:20
    
That solves my question, thanks! What threw me off is the image I was given isn't drawn to scale. And I forgot that bisect means dividing something into two equal parts... –  highy Oct 11 '10 at 21:21
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