In the figures above, if the area of the triangle on the

by **Guest** Sun Aug 03, 2008 11:09 am

I know I have to use the concept of congruent triangles, but I think I am missing something.

The answer is C. I thought it was E. Can someone please explain how to get to C.

by **Kamlesh** Sun Aug 03, 2008 2:26 pm

wen two traingles are similar the area is proportional to the square of their sides

A1/ A2 = x /2x

s^2 / S^2 = x/2X

solve this u get C

by **Guest** Sun Aug 03, 2008 2:59 pm

Thank you Kamlesh. It makes sense.

by **RonPurewal** Tue Aug 12, 2008 4:07 am

Kamlesh Wrote:wen two traingles are similar the area is proportional to the square of their sides

A1/ A2 = x /2x

s^2 / S^2 = x/2X

solve this u get C

this is true for ANY similar figures, whether or not they are triangles.

in other words:
for similar figures, (area ratio) = (scale factor)^2, where "scale factor" is the ratio of any linear dimension (circumference, diameter, length, width, etc.) in the figures.

so, for instance, if 2 similar figures have corresponding lengths in the ratio 3 : 5, then their areas must be in the ratio 9 : 25, regardless of what they're shaped like.