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St 1 - Gives us area of rect but there can be n number of rectangles so we cant fnd any side of triangle - NS
St2 - Again it gives us the area of small triangles. With this we can find one side of rectangle and sides of small triangles but we can have diff sizes of such triangles.
Combining we can find sides of square , then sides of small triangles. With these we can find the side of triangle- so Suff.
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Area of equilateral triangle =sqrt(3)*AB^2 / 4

So AB=?

Statement 1:

DG*GF=8*sqrt(3)

Insufficient

Statement 2:

1/2*AG*DG=2*sqrt(3)

We know ADG is a 30-60-90 triangle

So AG*AG*sqrt(3)=4*sqrt(3)

AG=2

DG=2*sqrt(3)

Insufficient

Statement 1&2:

2*sqrt(3)*GF=8*sqrt(3)

GF=4

We know that

BF=AG=2

Therefore AB= 2+2+4=8

Area= sqrt(3)*8^2 /4 = 16*sqrt(3)

Sufficient

C


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Statement1: insufficient because we can have multiple values of sides of rectangle DEFG.
Statement2: Insufficient again.
Combining:
Draw a perpendicular CP that intersects DE at O and AB at P.
We can also see that area of triangle ADG is 1/4 the area of DEFG.
Triangles AGD and PGD can be proved congruent.
DE parallel to AB.
DE=1/2AB (AG=GP=PF=FB, DE. DE=GP+PF, AB=AG+GP+PF+FB)
by mid point theorum,points D and E are mid points of AC and BC respectively.
We can easily prove Triangles ADO and ADG congruent.
so all triangles have equal area i.e.
Tri ADG,DGP,DOP,EOP,EPF,EFB,CDO,CEO.
(area)ABC=8*(area ADG)
Thanks. Post Kudos if you like.
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can we solve using symmetry of the triangle? then option would be sufficient.
--> x=2y

chetan2u, please help
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AnkithaSrinivas
can we solve using symmetry of the triangle? then option would be sufficient.
--> x=2y

chetan2u, please help


Of course we can use symmetry to solve questions => if ratio of sides is a:b, then ratio of area becomes a^2:b^2

But it is not clear what do x and y stand for.
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