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math help thread


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Split it into four triangles, four rectangles, and a square. Work out the area's of each independently, and add them together.


draw a square around the entire Octagon. Then use 52.4176 and subtract the area of the four triangles you create on each corner (They're right triangles where H=3 Edit: and B=2.12).

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For the first problem set, I think #3 is possible given that the interior lines intersect exactly halfway (I forgot the math terminology...). From there you could find the length of the hypotenuse and then multiply that by 4.

But that's a long complicated process so I'm not doing it to see what the actual answer is.

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this one is really easy once you realize that you don't need to know much geometry. The only reason they'd give you the numbers they do is if the triangles are similar, so its just a few proportions





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For 26.) the two triangles are similar, so you can setup a ratio of the big triangle's perimeter to AC and set that equal to x (the small triangle's perimeter) to DF, so that:

72/22 = x/14

solve for x and you get 45.82

For 27.) I think the ratio 6/5 carries through for the entire triangle, since as far as I can tell they're similar. So I believe you can just multiply 6/5 by the smaller triangle's area to get the larger triangle's area, which results in 90 cm^2. It's been a long time since I've done geometry though, so I may be wrong.

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