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Problem of the Day 3.07.09

with 9 comments

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


Written by Implex

July 2, 2009 at 6:00 pm

9 Responses

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  1. its option 4) Tan^2 A


    July 3, 2009 at 12:53 pm

  2. Sorry, it is incorrect


    July 3, 2009 at 1:11 pm

  3. 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…


    July 3, 2009 at 6:58 pm

  4. Practice any and every material u can get hold of.
    Follow forums. If you can come to testfunda.com, great!

    Good Luck


    July 3, 2009 at 7:35 pm

  5. 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?


    July 4, 2009 at 6:28 pm

  6. 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


    July 8, 2009 at 3:59 pm

  7. 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


    July 20, 2009 at 3:34 pm

  8. 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


    September 12, 2009 at 5:22 pm

  9. AD = AB cos A
    AE = AC cos a

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


    September 12, 2009 at 5:24 pm

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