Thursday, July 2, 2009

Problem of the Day 3.07.09

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

9 comments:

  1. its option 4) Tan^2 A

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  2. Sorry, it is incorrect

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  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...

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

    Good Luck

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  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?

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  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

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  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

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  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

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  9. 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

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