Saturday, September 20, 2008

Problem Of The Week 19

New Problem: A fresh one, I created it this morning, trying to find the most adequate data!

In a triangle ABC, altitude AD=6 is drawn to cut BC at D. From D, altitude DE=3 is drawn to cut AC at E. If it is know that AB =12. Find the ratio of the area of ABC to area of DEC?

A) 4:1 B) 16:1 C) 25:1 D) 5:1 E) Cannot be determined
Posted by Unknown at 8:46 AM
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5 comments:

  1. crushkillerSeptember 20, 2008 at 11:32 AM

    i m gettin 16:1

    ReplyDelete
  2. outtimedSeptember 20, 2008 at 3:54 PM

    yupp right , but do post your method !

    ReplyDelete
  3. AmitSeptember 20, 2008 at 11:01 PM

    From the given conditions we can find that-

    BD=6root3
    AE=3root3

    Taking EC as x we get DC=root(9+x^2)
    Applying pythagorus on tri(ADC) we get x=root3

    Hence BC=8root3
    EC=root3

    Hence ar(ABC)/ar(DEC) = 6*8root3/3*root3 = 16:1

    ReplyDelete
  4. riyaSeptember 24, 2008 at 12:56 AM

    We can say that triangle CEB is similar to triangle CAB. Then using the property of similar triangle that if two triangles are similar then the ratio of their squares is equal to the ratio of the squares of their sides . Which in this case is 12^2/3^2 = 16:1

    ReplyDelete
  5. SujaSeptember 6, 2010 at 1:42 AM

    16:1 it is ..

    Suja

    ReplyDelete

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