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Expert replies
by resilient » Sun Feb 17, 2008 11:20 pm
Which of the following is a possible length for side AB of triangle ABC if AC = 6 and BC = 9?

I. 3
II. 9
III. 13.5


I only
II only
III only
II and III
I, II and III


qa is c..i chose a
Appetite for 700 and I scraped my plate!
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Source: — Problem Solving |

D is the correct answer.

by sunildutt_b » Mon Feb 18, 2008 12:53 am
In a triangle, sum of two sides should be greater than the third side at any point.
Given sides, AC = 6, BC= 9

Options:
1) 3 (AC+3 = 6+3 = 9 = BC). This is incorrect as the sum of 2 sides is equal to the third side.

options II and III are possible . So i think answer should be (d).
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by AleksandrM » Wed Feb 20, 2008 10:26 am
I agree. The answer should be D.

9-6 = 3 and 6+9 = 15, both of which would end up giving you a straight line, which is not a triangle. In order to keep the triangle, the third side must be 3<AB<15.
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by khurram » Thu Feb 21, 2008 12:05 pm
Why would D be correct

If two sides are equal, then right triange, using 9^2 +9^2 does not equal=6^2.

That is why C is only choice?
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by khurram » Thu Feb 21, 2008 12:09 pm
Why would D be correct

If two sides are equal, then right triange, using 9^2 +9^2 does not equal=6^2.

That is why C is only choice?
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by Stuart@KaplanGMAT » Thu Feb 21, 2008 12:48 pm
The shortest distance between any two points is a straight line. Therefore, every side of a triangle has to be LESS than the sum of the other two sides.

In other words, going straight from point A to point B MUST be shorter than going from A to C and then C to B.

For this rule to hold true for all 3 sides of a triangle, every side must also be GREATER than the difference between the other two sides.

So, the general rule is:

|Side 2 - Side 3| < Side 1 < Side 2 + Side 3

and this rule holds true for every side.
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by bikerguy.gmat » Thu Oct 08, 2009 6:07 am
Stuart Kovinsky wrote:The shortest distance between any two points is a straight line. Therefore, every side of a triangle has to be LESS than the sum of the other two sides.

In other words, going straight from point A to point B MUST be shorter than going from A to C and then C to B.

For this rule to hold true for all 3 sides of a triangle, every side must also be GREATER than the difference between the other two sides.

So, the general rule is:

|Side 2 - Side 3| < Side 1 < Side 2 + Side 3

and this rule holds true for every side.
Hi Stuart, I do not understand this. Can you please explain in more detail.

Never heard about, or seen this kind of rule anywhere else. And, I could not do it using geometry principles too.

Thanks in advance!
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by Stuart@KaplanGMAT » Thu Oct 08, 2009 10:21 am
bikerguy.gmat wrote:
Stuart Kovinsky wrote:The shortest distance between any two points is a straight line. Therefore, every side of a triangle has to be LESS than the sum of the other two sides.

In other words, going straight from point A to point B MUST be shorter than going from A to C and then C to B.

For this rule to hold true for all 3 sides of a triangle, every side must also be GREATER than the difference between the other two sides.

So, the general rule is:

|Side 2 - Side 3| < Side 1 < Side 2 + Side 3

and this rule holds true for every side.
Hi Stuart, I do not understand this. Can you please explain in more detail.

Never heard about, or seen this kind of rule anywhere else. And, I could not do it using geometry principles too.

Thanks in advance!
Sure!

Let's start with two houses, A and B.

Let's build a road in a straight line directly from A to B. That's the shortest possible road that we can build to connect the two houses.

Now let's build a third house, C, and build direct roads from A to C and from C to B.

We now have two possible routes to get from A to B. We can go directly from A to B, or we can go from A to C to B.

Another way we can think of AC and CB is as a detour from points A to B. Since the shortest distance between A and B is our original straight road, we can see that:

AB < AC + CB

Now for the second part of the relationship. What we just determined about AB must also be true for AC and CB. In other words:

AC < AB + CB

and

CB < AB + AC

We simply rearrange these inequalities:

AB < AC + CB
provides:
AB - CB < AC
AB - AC < CB

AC < AB + CB
provides:
AC - AB < CB
AC - CB < AC

CB < AB + AC
provides:
CB - AB < AC
CB - AC < AB

Putting those 9 inequalities together, we get:

|AB - AC| < CB < AB + AC
|AC - CB| < AB < AC + CB
|CB - AB| < AC < CB + AB
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by bikerguy.gmat » Thu Oct 08, 2009 6:55 pm
ohh brilliant!!!!

thanks Stuart, your explanation is amazing.

all your posts are really helpful. keep the good work going :)

Ciao!
-Avinash
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by sanjana » Fri Oct 09, 2009 1:55 am
My fav way of doing these problems..
You do not have the required permissions to view the files attached to this post.
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by NikolayZ » Fri Oct 09, 2009 8:20 am
Hey guys, it is just the basic rule for triangles' sides.
No side can be longer than the sum of other two sides.
Then you have to solve an inequality and find out what actually can be the side of a triangle. Rule is applicable to triangles of all types.
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