both the triangles are similiarjapan1453 wrote:Could you solve this problem. Thanks
OA: A
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
gmat prep coordinate system
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piyush_nitt
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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
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
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
