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## geometry

This topic has 2 member replies
kandelaki Senior | Next Rank: 100 Posts
Joined
05 Feb 2007
Posted:
37 messages

#### geometry

Thu Mar 01, 2007 4:09 am
10. If a circle, regular hexagon and a regular octagon have the same area and if the perimeter of the circle is represented by "c", that of the hexagon by "h" and that of the octagon by "o", then which of the following is true?
c > o > h
c > h > o
h > c > o
o > h > c
h > o > c (correct answer)

source: http://free-quiz.4gmat.com/
i kmow the source is not quite reliable.
but still

can you explain to me how to solve it?

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kandelaki Senior | Next Rank: 100 Posts
Joined
05 Feb 2007
Posted:
37 messages
Thu Mar 01, 2007 5:43 am
great explanation! thanks a lot!

rjfNC Newbie | Next Rank: 10 Posts
Joined
01 Mar 2007
Posted:
4 messages
Thu Mar 01, 2007 5:35 am
Great question. The approach I would take, however, would not be technically math driven. Give the following reasoning a shot and see if is works for you.

There is a mathmatic principle regarding approximation of circles. As you increase the number of sides of a two dimensional object, you come "closer" to the image of a circle. Image a square then a pentagon, then a hexagon, all inscribed inside of a circle. With each increase in the number of sides you get closer to the sides of a circle. Take this out to an obect with seeminly infinte sides, you will, essentially, have a circle.

It has also been proven that if you take the ratio of the area of an object to that object perimeter the best you can do is a circle. The maximum amount of area you can get for a given perimeter is a circle. Also, visa-versa, for a given area the shortest perimeter you can have will be a circle. Give it a shot comparing a circle to a square.

Now for the question. All the shapes have the same area. That means that the perimeter will change in accordance with each object. Applying the principle above the circle will have to smallest perimeter. Then, based on the number of sides, the octagon and finally the hexagon.

You can try to solve this mathematically, but that is lengthy and you wouldn't have that kind of time in the test anyway. Another, commonly used method, would involve picking an area and solving for the side length. Either way, this is an obscure math principle and I doubt it would come up on a real exam.
Goodluck!

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