# GRE Math practice question. Possible wrong answer? So I was going through questions in a GRE prep book and I'm just convinced this answer is wrong but I understand the likelihood is very slim. Can someone check my work and explain why I am wrong? So I am given this isosceles triangle and I am told the measure of the angles are expressed by 2x+y=180. Since the triangle is isosceles I know the perimeter would either be represented by 7+7+4=18 or 7+4+4=15. Now from here can't I drop an altitude from R and figure out which perimeter correctly represents the triangle?

I showed my work however the answer key says the relationship between Quantity A and Quantity B cannot be determined.

• Can you post an image of the original question, what are you given to start with? Just that the triangle is isoceles, it has two angles x that are equal, y that isn't... where is the 7 in your (7+7+4) coming from? In your question, at the end put down a "-------" and underneath it, write exactly what is given in the question – frogeyedpeas Jul 10 '16 at 2:26
• yeah let me add the image I just scribbled all around it lol – Lil Jul 10 '16 at 2:28
• So, which answer was your answer? – Thomas Andrews Jul 10 '16 at 2:31
• I think its quantity A is greater but the book says the relationship cannot be determined – Lil Jul 10 '16 at 2:32
• The problem is that the picture implies that $7$ is the duplicated value, but nothing in the body of the question implies that. So yes, the triangle could be $4,4,7$ or $4,7,7$, so it could have perimeter $15$ or $18$. Basically, this is a trick question, reminding you that you should not draw any inference from the picture of the problem, except the specific lengths (4,7) listed and the fact given in the text. – Thomas Andrews Jul 10 '16 at 2:34

## 1 Answer

The problem is that the picture implies that $7$ is the duplicated value, but nothing in the body of the question implies that. So yes, the triangle could be $4,4,7$ or $4,7,7$, so it could have perimeter $15$ or $18$.

Basically, this is a trick question, reminding you that you should not draw any inference from the picture of the problem, except the specific lengths (4,7) listed and the fact given in the text.

The picture is not of a $4,4,7$ triangle, because it is not obtuse, but the picture is not something from which you should use when making geometric inferences.

• I understand the trap but is my process of dropping an altitude to find the length of the third side invalid? – Lil Jul 10 '16 at 2:38
• @Lil Dropping the perpendicular does not necessarily bisect the $4$ side - if it did, then you could conclude the third side is $7$ – Thomas Andrews Jul 10 '16 at 2:41
• hmm so I can't draw in a perpendicular to decide? – Lil Jul 10 '16 at 2:42
• Dropping a perpendicular from the vertex where the equal sides meet resolves nothing. If the equal sides have length $4$ then the length of the perpendicular is $\dfrac{\sqrt{15}}{2}$, if the equal sides have length $7$ then the length of the perpendicular is $3\sqrt{5}$. – John Wayland Bales Jul 10 '16 at 2:55
• You don't have any freaking idea where the altitude will land or even if it will land inside the triangle. The best you could do is you'd get $w \pm v =4$ and $7=\sqrt {x^2 +w^2}$ and the third $4 or 7=\sqrt {x^2+v^2}$. Which will give you two possible results that will not help you. (Oh, it will tell you if the third side is 4 whether the altitude will be in or out of the triangle-- but that's all). You can not resolve from this. – fleablood Jul 10 '16 at 6:53