chetan2u wrote:

bhoot80 wrote:

sanahazari wrote:

I feel the answer should be (E) 140.

As the length of the third side of a triangle can't be more than the sum of the other two sides and can't be less than the difference of the other two sides, the other options are acceptable.

Hence the distance would be \({20} \leq distance \leq {120}\)

It's should be d. 120..law of triangles leg

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GMAT Club Forum mobile apphi.

the minimum distance will be WHEN D is in between O and Z then OZ is 70 and DZ is 50, so OD will be 70-20=50.. O-20-D-50-Z

MAX when again all three are in a straight line and D is on other side... O-70-Z-50-D OD is 50+70=120

140>120 not possible

E

chetan2u , I used your approach, because as far as I understand it, the triangle inequality theorem, and what is sometimes called its converse, are literally inequalities. One side cannot equal the sum of the lengths of two other sides, and one side cannot equal the difference between the two other sides.

You didn't mention that the \(\leq\) sign was impermissible, i.e., you did NOT say that it should be just a < sign.

I understand that there can be degenerate triangles with area zero. I also understand that the GMAT does not test them.

I used the straight-line method for values 20 and 120 (to allow them) because I thought that those two values would be impermissible with the triangle inequality theorem.

Is it true that the third side could be equal to the sum of the other two? Or equal to the difference between the other two? On the GMAT?

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