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
For All Your Quant Queries
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
its option 4) Tan^2 A
anup
July 3, 2009 at 12:53 pm
Sorry, it is incorrect
Rahul
July 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…
anup
July 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
Rahul
July 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?
Anish
July 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
Rahul
July 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
Hari
July 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
ananymous
September 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
ananymous
September 12, 2009 at 5:24 pm