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While this might appear to be a 3-4-5 triangle, you can’t square each side of a 3-4-5 and expect the pattern to hold: a 3-4-5 triangle must be some triangle whose sides can be reduced to the ratio 3x:4x:5x. As such, we have to resort to the Pythagorean Theorem, which in this case gives us 9^2+b^2=25^2, or b^2=544. At this point you can approximate – 544−−−√ is greater than 400−−−√, or 20, and less than 625−−−√, or 25, so the answer must be between 20 and 25: (B) the only such option.

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The query here isnt the answer to the problem but the solution given by veritas.. the way i deduced the answer was using the properties of triangle third side lies between the difference & sum of other two sides hence: 25-9<BC< 25+9 16<BC<34

and the only option falling is B. The explanation here says answer is between 20 and 25.. who is going wrong and where??

While this might appear to be a 3-4-5 triangle, you can’t square each side of a 3-4-5 and expect the pattern to hold: a 3-4-5 triangle must be some triangle whose sides can be reduced to the ratio 3x:4x:5x. As such, we have to resort to the Pythagorean Theorem, which in this case gives us 92+b2=252, or b2=544. At this point you can approximate – 544−−−√ is greater than 400−−−√, or 20, and less than 625−−−√, or 25, so the answer must be between 20 and 25: (B) the only such option.

---xx----

The query here isnt the answer to the problem but the solution given by veritas.. the way i deduced the answer was using the properties of triangle third side lies between the difference & sum of other two sides hence: 25-9<BC< 25+9 16<BC<34

and the only option falling is B. The explanation here says answer is between 20 and 25.. who is going wrong and where??

any help would be appreciated

I assume there is a diagram attached with it which gives some more information (that the triangle is a right triangle) etc. Also, judging from the question, we need to give the actual length of BC, not give the length that BC CAN take. Hence, there has to be some more info.

Given the limited info, your method is fine. Since 25 > BC > 9, the only possible options are (A) and (B). But sum of any two sides of a triangle must be greater than the third side so BC cannot be 16. Answer (B)
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While this might appear to be a 3-4-5 triangle, you can’t square each side of a 3-4-5 and expect the pattern to hold: a 3-4-5 triangle must be some triangle whose sides can be reduced to the ratio 3x:4x:5x. As such, we have to resort to the Pythagorean Theorem, which in this case gives us 92+b2=252, or b2=544. At this point you can approximate – 544−−−√ is greater than 400−−−√, or 20, and less than 625−−−√, or 25, so the answer must be between 20 and 25: (B) the only such option.

---xx----

I really like this question, because a lot of students immediately think the answer must be 16 to maintain that 3-4-5 pattern they've heard so much about, but obviously that is the trap answer for those going too fast.

Once you've figured out that the Pythagorean Theorem will unlock the answer for you, the major hurdle is approximating square roots. This made me think of a blog I wrote on this topic a month or two back. It's actually perfect for exactly this question, so I figured I'd link it here in case it helped anyone:

Re: The side lengths of triangle ABC are such that AC > BC > AB. [#permalink]

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18 May 2013, 07:07

I'm still having difficulty deciding between answer choices B and D

So I understand that the third side BC should be 16<BC<34

If I try to evaluate answer (B) 4sqt34 it gives 23.3 which is a possible answer. Also when I evaluate (D) 4sqt52 it gives 28.8 which is still a possible answer. I seem to be missing a trick. Please help
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Re: The side lengths of triangle ABC are such that AC > BC > AB. [#permalink]

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19 May 2013, 05:10

There is a very simple way to solve this question. We know AC=25 , AB = 9 and assume BC = x

We also know that since it is a right triangle x^2 + 81 = 25^2

Now before embarking on lengthy calculations, it can easily be observed that for the equation x^2 should have the units digit as 4. By simple process of elimination and no lengthy calculations you can deduce that only option B fits the bill. (if x =4root34, x^2 units place = 4 since x^2 =16 x 34

Hope this makes sense
_________________

You've been walking the ocean's edge, holding up your robes to keep them dry. You must dive naked under, and deeper under, a thousand times deeper! - Rumi

http://www.manhattangmat.com/blog/index.php/author/cbermanmanhattanprep-com/ - This is worth its weight in gold

Economist GMAT Test - 730, Q50, V41 Aug 9th, 2013 Manhattan GMAT Test - 670, Q45, V36 Aug 11th, 2013 Manhattan GMAT Test - 680, Q47, V36 Aug 17th, 2013 GmatPrep CAT 1 - 770, Q50, V44 Aug 24th, 2013 Manhattan GMAT Test - 690, Q45, V39 Aug 30th, 2013 Manhattan GMAT Test - 710, Q48, V39 Sep 13th, 2013 GmatPrep CAT 2 - 740, Q49, V41 Oct 6th, 2013

GMAT - 770, Q50, V44, Oct 7th, 2013 My Debrief - http://gmatclub.com/forum/from-the-ashes-thou-shall-rise-770-q-50-v-44-awa-5-ir-162299.html#p1284542

I'm still having difficulty deciding between answer choices B and D

So I understand that the third side BC should be 16<BC<34

If I try to evaluate answer (B) 4sqt34 it gives 23.3 which is a possible answer. Also when I evaluate (D) 4sqt52 it gives 28.8 which is still a possible answer. I seem to be missing a trick. Please help

Notice that the word 'right' had been added to the question. If it is a right triangle, you can easily use Pythagorean theorem and get your answer.

The original poster had tried to solve it for any generic triangle (the question without the word 'right'). You understand 16 < BC < 34. Also, realize that you are given that AC > BC Since AC = 25, BC must be less than 25. So only (B) works.
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Re: The side lengths of triangle ABC are such that AC > BC > AB. [#permalink]

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28 Jul 2014, 22:26

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: The side lengths of triangle ABC are such that AC > BC > AB. [#permalink]

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30 Dec 2016, 06:05

Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________