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gmat prep coordinate system

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Source: — Problem Solving |
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Re: gmat prep coordinate system

by piyush_nitt » Sun Feb 01, 2009 7:18 pm
japan1453 wrote:Could you solve this problem. Thanks
OA: A
both the triangles are similiar

from therom

ratio of the areas of similiar triangle = ratio of the square of corresponding sides

therefore

ratio = (PR/LN)^2
= (4/12)^2
= 1/9
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by gmat.2008 » Sun Feb 01, 2009 9:03 pm
one more solution to your prob: (w/o using any theorem)

A=0.5XbXh


base= X2-X1
height= Y2-Y1

just substitute the values of X2,X1,Y2,Y1 for both the triangles.

area of bigger triangle = 0.5 X 12 X 12
= 72

area of smaller= 0.5 X 4 X 4
= 8

ratio=
8/72=1/9
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