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Comment on Possible Right Angle
Can we say that since AC = 12
Unfortunately, that approach
Unfortunately, that approach is not valid. You're assuming that angle ACB is a right angle in order to prove that angle ACB is a right angle.
Here's what I mean. We know that AC = 12 and BC = 5. However, we cannot conclude that the length of side AB is 13, because we don't know for certain that triangle ABC is a right triangle (in fact, the goal of this question is to determine whether or not triangle ABC is a right triangle).
So, we cannot assume that triangle ABC is a right triangle and then use this assumption to conclude that triangle ABC is a right triangle.
well, we cannot assume that
That's right; the hypotenuse
That's right; the hypotenuse (AB) can vary, because ∠ACB can vary.
Please clarify where I am misunderstanding. I was using the same logic outlined above and deduced it was a right angle. I read your comment, but since we have sides we can use pythagoras to find the the hypotenuse which in this case is 13 - making it a right angle. Furthermore, since this is a DS question, if we have two of the sides we can find the third side and know it gives 13 or not and this would be enought to know if its a right angle. It is not possible for the hypotenuse be other than 13 and it be a right angle in a 5, 12, 13 triangle right?
You are using a bit of circular logic here.
We want to determine whether angle ACB is a right angle.
In order to apply the Pythagorean theorem, we must be certain that we have a right triangle.
So, we can't assume we have a right triangle, then apply the Pythagorean theorem, and then conclude that we have a right triangle.
However, even if we were told that triangle ACB is a right triangle, we still wouldn't know which angle is 90°.
In other words, if we know that two sides of a right triangle have lengths 5 and 12, we still wouldn't be able to determine the length of the third side, since we don't know whether 5 and 12 are the lengths of the two legs OR the length of one leg and the length of the hypotenuse.
Here's what I mean: https://imgur.com/g3Q3wGi
I see what you mean. We can
I think it's conceivable the
I think it's conceivable the test makers would do something like that.
When evaluating the statments
We can't say that the
We can't say that the triangles are similar, and we can't say that they're equal. This is because we can't show that the two triangles have any angles in common.
If BC divides AC in 2 equal parts, can't it be the bisector and then ACB be 90°?
Since BC divides AC in 2 equal parts, BC COULD be the bisector and ∠ACB COULD be 90°. However, that need not be the case.
IF it were true that AB = BD (making ∆ABD an isosceles triangle), then ∠ACB would definitely be 90°.
However, since we don't know whether AB = BD, we can't say for certain that ∠ACB = 90°.
Does that help?
Yes, thank you!
I selected B using the following reasoning. Please help what I missed out on:
Since the AC=CD, then the angle opposite them must also be equal i.e. angle(ABC)=angle(DBC), hence when we move the point B in either direction, the angles do not remain equal.
Be careful. If AC = CD, we
Be careful. If AC = CD, we can't necessarily conclude that ∠ABC = ∠DBC.
Doesn't it follow the rule that the angles opposite the sides with the length have the same angle?
That rule applies to 2 (of
That rule applies to 2 (of the 3 sides) of a triangle. More here: https://www.gmatprepnow.com/module/gmat-geometry/video/865
In this case, AC and CD are not two sides of the same triangle.
We can say that AC and CD COMBINE to make ONE side of triangle ABD, but this doesn't really help us.
The lock rule is a game
Thank you for that Brent!
I'm glad you like it, Erik!
I'm glad you like it, Erik!