## 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**

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i m gettin 16:1

crushkillerSeptember 20, 2008 at 6:02 am

yupp right , but do post your method !

outtimedSeptember 20, 2008 at 10:24 am

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

AmitSeptember 20, 2008 at 5:31 pm

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

riyaSeptember 23, 2008 at 7:26 pm

16:1 it is ..

Suja

SujaSeptember 5, 2010 at 8:12 pm