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13 Oct 2021, 09:01
Kudos
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avigutman Please add your view in this Problem
avigutman
Expert
Tutor
Joined: 17 Jul 2019
Last visit: 30 Sep 2025
Posts: 1,285
Given Kudos: 66
Location: Canada
Schools: NYU Stern (WA)
GMAT 1: 780 Q51 V45
GMAT 2: 780 Q50 V47
GMAT 3: 770 Q50 V45
Expert reply
13 Oct 2021, 13:55
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kskumar
kskumar here's how I think through this problem.
Start with statement 2, as I don't know how it could possibly be enough to answer the question. You see, when I look at c+2, I ask myself what impact does the +2 have here? I don't even know whether it's 2cm or 2 feet or 2km... I have no way of knowing whether +2 is a lot or a little in the context of this triangle. The fact that a+b > c goes without saying, so the only new information provided by statement 2 is information for which I totally lack context and is therefore useless.
So, I've eliminated BD and I'm ready to evaluate statement 1.
Statement 1 provides the 2-dimensional ratio for the three sides (3:4:6), which I appreciate because that fits well with the Pythagorean theorem.
If the squares of the two smaller sides are 3 and 4 ratio units respectively, the Pythagorean theorem tells us that the square of the third side would have to be exactly 7 ratio units for this triangle to be a right triangle. Since it's less than 7 ratio units (it's 6 ratio units), I can conclude that the angle across from the longest side is less than 90 degrees, and the other angles have to be smaller still, so we get a definitive YES and I pick A.
08 Nov 2021, 02:02
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moonwalking
I didnt know how to go about solving it, so I made a quick guess.
I reasoned that since stat 1 gives us areas, we can find the length of the diameter, and thus the lengths of all three sides. So now the triangle has fixed length sides. I assumed that if we make one angle obtuse, the length of the opposite side will change. Thus, we cant really change the shape of the triangle. It is sort of locked. Thus, there's a good chance that the angles will be locked in too. SO A is suff
For stat 2 I tested values, c=2 then a=2 and b=1 or c=3 then a=2 and b=2
Thus, the lengths of the sides cannot be locked into one particular length, SO, there's a good chance that B is insuff.
Now, does this all make sense, or I just got really lucky?
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