Given,
A triangle ABC in which DE is parallel to BC. DC and BE intersect at O such that
area of triangle ADE = 3 sq. units and area of triangle DOE = 1 sq. unit.
Find the area of triangle ABC....
Thinking over the problem since 2 days but m not able to work it out.
Some one help....
30See, we know by mid point theorem,
DE||BC and DE=1/2 BC
Now since, ΔADE and ΔABC are similar,
(DE)2/(BC)2 = arΔADE/arΔABC
Since, DE=1/2 BC,
arΔADE/arΔABC = 1/4
Hence, arΔABC=12 sq units. [1]
36Absolutely wrong.
I haven't posted D and E to be the mid point of sides AB and AC in my question.
If it was mentioned than why mid point theorem?...
Simply by Similarity property...
DE/BC = AE/AC
=> DE/BC = 1/2
=> Ar(tri. ADE)/Ar(tri.ABC) = DE2/BC2
=> 3/Ar(tri.ABC) = 1/4
=>Ar(tri.ABC) = 12 sq. units..........
this is a 2010 kvpy question........
the answer is correct but approach is incorrect
36Got the solution to this question....
Absolutely not a general solution....
Got it by the resonance solution.....
11Is it the first step?
ar ADE/ar ABC=ar DOE/ar BOC=DE2/BC2
so,ar ABC=3 ar BOC