IMO E.
I would choose A, if i am able to prove BAD and ABD are each x such that their sum equals 2x(exterior angle).
If anybody could prove the above, please post.
Triangle
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cubicle_bound_misfit
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i think angle BAD is X.
it is a BAD case of printing mistake.
let me know if that makes sense
regards,
it is a BAD case of printing mistake.
let me know if that makes sense
regards,
Cubicle Bound Misfit
IMO C.
Reasoning:
1. consider any triangle BDC. BDC is an isosceles triangle from question.
Fixing AD = 6, we can vary the height of the triangle BDC varying x. So we cant determine BC.
[Think of it like you can just pull point B up and down with sides as strings]
INSUFFICIENT.
2. Now, if you keep the angle 2x the same, i.e 72 degrees, then you can vary AD to any length and change the size of the triangle (Similar triangles keeping the ratio x:2x:2x).
INSUFFICIENT.
Together, if you fix the angle x and AD, then you cant vary the height of the triangle hence you will be able to find BC.
Please correct me if i am wrong
Reasoning:
1. consider any triangle BDC. BDC is an isosceles triangle from question.
Fixing AD = 6, we can vary the height of the triangle BDC varying x. So we cant determine BC.
[Think of it like you can just pull point B up and down with sides as strings]
INSUFFICIENT.
2. Now, if you keep the angle 2x the same, i.e 72 degrees, then you can vary AD to any length and change the size of the triangle (Similar triangles keeping the ratio x:2x:2x).
INSUFFICIENT.
Together, if you fix the angle x and AD, then you cant vary the height of the triangle hence you will be able to find BC.
Please correct me if i am wrong
