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Problem Of The Week 19

with 5 comments

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


Written by Implex

September 20, 2008 at 3:16 am

5 Responses

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  1. i m gettin 16:1


    September 20, 2008 at 6:02 am

  2. yupp right , but do post your method !


    September 20, 2008 at 10:24 am

  3. From the given conditions we can find that-


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

    Hence BC=8root3

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


    September 20, 2008 at 5:31 pm

  4. 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


    September 23, 2008 at 7:26 pm

  5. 16:1 it is ..



    September 5, 2010 at 8:12 pm

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