similar triangles...

Asked by geethika | 19th Jan, 2010, 07:49: AM

Expert Answer:

Let ABC and DEF be the two similar triangles.

AB/ DE = k     ⇒  AB = kDE      ................(1)

BC/ EF = k     ⇒  BC = kEF     ..................(2)

AC/ DF = k     ⇒  AC = kDF     ..................(3)

Adding (1), (2) and (3), we get

AB + BC + AC = k (DE + EF + DF)

25   = k (15)

⇒ k = 5/3

Now substituting in equation (1), we get

9 = (5/3) DE

⇒ DE = 27/5 = 5.4

Answered by  | 19th Jan, 2010, 09:09: AM

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