In a triangle ABC, perpendiculars BD and CE are drawn to the sides AC and AB. Points D and E are joined, then the ratio of the area of ADE to the area of ABC is:

1) Cos²A 2) Sin²A 3) Cot²A 4) Tan²A 5) None of these

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its option 4) Tan^2 A

anupJuly 3, 2009 at 12:53 pm

Sorry, it is incorrect

RahulJuly 3, 2009 at 1:11 pm

hello bro..m in 3rd yr (engg)..

i hv read ur blog..it is very inspiring 4 me…can u plz suggest me sumways to improve in quant…

anupJuly 3, 2009 at 6:58 pm

Practice any and every material u can get hold of.

Follow forums. If you can come to testfunda.com, great!

Good Luck

RahulJuly 3, 2009 at 7:35 pm

First time here.🙂

Feels good to get hold of at least one problem.🙂

my take:

Option (1) Cos²A

I assumed the triangle to be equilateral which really simplified things.

And I got the ratio of areas as 1:4 which is Cos²60

Hence, ratio =Cos²A

Can we assume it so, where there are really no restrictions regarding the dimensions of the traingle?

AnishJuly 4, 2009 at 6:28 pm

Correct answer is Cos^2 A

@ Anish, you made a big assumption,

generally we can make such assumptions when there are no restrictions given..

only in rare cases the answers differ

RahulJuly 8, 2009 at 3:59 pm

The trignometric representation of area of a triangle is 1/2*length of side 1*length of side 2* sine of angle formed by these 2 sides …

We can use this for the smaller triangle and proceed to get the answer without any assumptions

HariJuly 20, 2009 at 3:34 pm

AD = AB cos A

AE = AC cos a

ar(AED) = (1/2) AE X AD cos A (cos A)^2

——- ——————- =

ar(ABC) (1/2) AB X AC cos A

ananymousSeptember 12, 2009 at 5:22 pm

AD = AB cos A

AE = AC cos a

ar(AED)/ar(ABC)

= [(1/2) AE X AD cos A ]/[(1/2) AB X AC cos A]

= (cos A)^2

ananymousSeptember 12, 2009 at 5:24 pm