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area of circular region

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ifthyder Master | Next Rank: 500 Posts Default Avatar
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area of circular region

Post Wed Oct 15, 2008 8:44 am
please explain each bit of details?
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rohangupta83 Legendary Member Default Avatar
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Post Thu Oct 16, 2008 2:01 am
(D)

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Stuart Kovinsky GMAT Instructor
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Post Wed Oct 15, 2008 9:50 am
parallel_chase wrote:
Since 30,40 are the points on the circle, the distance from the origin to the point on the circle will be the radius.

r^2 = 30^2 + 40^2

r = sqrt 2500

r = 50

area = pi r^2 = 50^2 pi = 2500 pi.

Hence D.
Good solution using the Pythagorean formula.

We can shave some time off if we quickly recognize a 3x:4x:5x triangle in the mix. 3/4/5 is the most commonly tested Pythagorean triplet and almost always shows up in some form on test day.

As soon as you see that you have a 30/40/x right angle triangle, 50 should come to mind as the radius. Then calculate the area as above.

Other important triplets to know:

5x:12x:13x
x:x:xroot2 (right isosceles triangle)
x:xroot3:2x (30/60/90 triangle)

and less tested, but still comes up:

7x:24x:25x



Last edited by Stuart Kovinsky on Thu Oct 16, 2008 12:59 pm; edited 1 time in total

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parallel_chase Legendary Member Default Avatar
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Post Wed Oct 15, 2008 8:54 am
Since 30,40 are the points on the circle, the distance from the origin to the point on the circle will be the radius.

r^2 = 30^2 + 40^2

r = sqrt 2500

r = 50

area = pi r^2 = 50^2 pi = 2500 pi.

Hence D.

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No rest for the Wicked....

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