Lesson: Triangles - Part I

Comment on Triangles - Part I


In the above question,

(x+3) < third side < (5(x+1))

since x > 0, i took x = 1, then
4 < third side < 10

option A yields 9 => this is possible
option B yields 3 => not possible
option C yields 7 => possible
option D yields 11 => not possible
option E yields 19 => not possible

answer must be A and C right. how come in revel spoiler they have given E is also one of the answer? please correct me if i am wrong.
greenlight-admin's picture

Link: http://gre.myprepclub.com/forum/if-x-0-and-two-sides-of-a-certain-triang...

The problem is that we don't know the value of x. More importantly, the question asks "Which of the following COULD be the length of the third side?"

So, is it POSSIBLE that answer choice E is such that...
(x+3) < 2x+17 < (5(x+1))?

Yes, it is POSSIBLE.

Notice that, if x = 5, the inequality works.

So, 2x+17 (aka answer choice E) COULD be the length of the third side

If x=5 then there are most options possible

Sir, In the question x+3<third side<5x+5
What if we take x=100.. then 6x+1<5x+5.. In that case C will not be an option?
I am a bit confused with this question.
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-x-0-and-two-sides-of-a-certain-trian...

You're correct to say: x+3 < third side < 5x+5

The question asks us to find answer choices that COULD represent the length of the third side.
This is much different from finding answer choices that MUST represent the length of the third side

If answer choice C (6x+1) represents a possible length of the third side, then we need only find one value of x so that x+3 < 6x+1 < 5x+5

So, for example, when x = 1, the above inequality holds true.
This means (6x+1) COULD represent the length of the third side

The fact that the inequality x+3 < 6x+1 < 5x+5 does NOT hold true when x = 100 has no bearing here, since we've already found one value of x that makes the inequality possible.

Does that help?


I have a question regarding this :

The third side of the triangle is of course:
x + 3 < third side < 5x + 5

If we assume that x = 1, then we get the the range of values 4 < third side < 5x + 5. Now going to the answer choices

A) 4(1) + 5 = 9
b) (1) + 2 = 3
C) 6(1) + 1 = 7
D) 5(1) + 6 = 11
E) 2(1) + 17 = 19

Therefore, the answer choice should be A & C and NOT E.
What am I doing wrong here?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-x-0-and-two-sides-of-a-certain-trian...

The question asks, "Which of the following COULD be the length of the third side of the triangle?"

You have chosen only one possible x-value (x = 1).
Is there another possible x-value in which answer choice E COULD be the length of the third side?
You already noted that: x + 3 < third side < 5x + 5
E) 2x+17
Is there an x-value such that x + 3 < 2x+17 < 5x + 5 ?
If x = 10, then the inequality works.
So, answer choice E COULD be the length of the third side


So if you have to keep plugging values in for x until you're satisfied no value holds to be true then eliminate
greenlight-admin's picture

You COULD use that strategy, but it will take a long time.
It's better to look for a value that DOES satisfy the inequality (a much shorter approach).


You eliminated 42 based on y+z=20 therefore x is less than 20,so the answer can't be 42. But x+y=30 and y+z= 20 so the third side z can be 12 which is less than 30
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-the-sides-of-a-triangle-have-lengths...

You have to take BOTH equations (x + y = 30, and y + z = 20) into account.

You're suggesting that it's possible for z = 12
Let's see what that means for the values of x and y.

GIVEN: y + z = 20
If z = 12, then y = 8

GIVEN: x + y = 30
If y = 8, then y = 22

So, the three sides have lengths 8, 12 and 22 (for a perimeter of 42)
There's a big problem with these lengths.
The longest side (22) is longer than the sum of the other two sides (8 and 12)
This breaks the triangle property that says "length of third side < SUM of other two sides"

So, a triangle with lengths 8, 12 and 22 CANNOT exist.

Does that help?


After i sent this mail. I perused the question properly and saw this fact.

Isn't this question better phrased as which "MUST" be the perimeter of the triangle? becuause in "MUST BE" questions,when one solution yields false we totally eliminate it.
For eg
The perimeter can be 28 becuause y+z=20 and x(third side) must be less than the two sides.So x can be 8. But if x+y=30 z has to be -2 before the sum can be 28 and it's ruled out. But in this case must be true requires both solutions to check out,so 28 as perimeter is eliminated. But "COULD" requires just one solution to be true and since y+z=20 and x =8 is true and x+y=30 and z=-2 isn't. 28 could be d perimeter. Please help me understand my confusing
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/if-the-sides-of-a-triangle-have-lengths...

I agree that MUST questions are easier, since we can easily eliminate options that are false. However, on the GMAT, you will see both MUST questions and COULD question.

Also notice that changing the COULD to MUST would dramatically change the question.

Consider the following question:

If x is an integer, and k = x², then which of the following COULD be the value of k?
A) 4
B) 7
C) 25
D) 36
E) 50
F) 81
In this case, the correct answer is A, C, D and F, since those values are all squares of integers.

Compare that question with this one:

If x is an integer, and k = x², then which of the following MUST be the value of k?
A) 4
B) 7
C) 25
D) 36
E) 50
F) 81
In this case, there is no correct answer, because in addition to possibly equaling 4, 25, 36 or 81, k COULD also equal 9 or 16 or 49 or 64 or 100 or 121 or....

Likewise, if the linked question asked "Which of the following MUST be the perimeter of the triangle?" there would be infinitely many possible values for the perimeter. That is, there is no ONE perimeter that the triangle MUST have.

Does that help?


When the line was extended why was it assumed to be a right angle triangle? Was it because that's the only way to solve the question
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/qotd-6-which-is-greater-c-or-d-2655.html

The poster intentionally created a right angle there for 2 reasons:
1) By drawing a HORIZONTAL line, we create a right angle at the y-axis, which means we can determine that the remaining angle is 46°

2) By drawing a HORIZONTAL line (parallel to the x-axis), we can easily determine that the horizontal side has length |c|

3) By drawing a HORIZONTAL line, we can easily determine that the vertical side has length |d|

That said, we could have also solve the question by drawing a VERTICAL line up to x-axis. The logic for that solution would be pretty much identical to the logic in the original solution.


In the solution both were added to give a+b+c+d+x+y=300
But i set them equal to each other since they both equal 300 and got
a+c+x=b+d+y and couldn't solve further
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/what-is-the-value-of-a-b-c-d-12084.html

You've hit upon two crucial aspects of solving math questions:
1) be bold and try a variety of approaches.
2)While trying different approaches, keep asking yourself "Am I making any progress towards the solution?"

Here, the question asks "What is the value of a + b + c + d?"

So, when you derived the equation a + c + x = b + d + y, you probably found yourself stuck, because you weren't able to get the SUM a + b + c + d on one side of the equation (and our goal is to find that sum).

After attempting to get a + b + c + d on one side of the equation (and failing to do so), you should recognize that this route is going nowhere and try something else.

Alternatively, when we ADD the equations we get: a + b + c + d + x + y = 300

This certainly seems like a good step, since we now have a + b + c + d on one side of the equation.

So, we might continue with this approach.

Does that help?


Is 2x + 2 - x + 2 = 2x - x + 2 + 2 = x + 4 wrong
(2x + 2) - (x + 2) = 2x + 2 - x - 2 = x correct
greenlight-admin's picture

Both of your simplifications are correct.

That is,
2x + 2 - x + 2 = 2x - x + 2 + 2 = x + 4 (perfect)
(2x + 2) - (x + 2) = 2x + 2 - x - 2 = x (perfect)

More here: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...


But they gave different answers
greenlight-admin's picture

Can you tell me what "they" is referring to?
It will help if you provide a link.



In the solution provided by you, how can we let the third angle be y? what is the reasoning behind this?
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/in-the-figure-if-y-60-then-what-is-the-...

Good question! I actually have no idea why I did that, especially since another angle was already labeled as y.
After reviewing my (awful) solution, I decided to change my approach. Please see my revised solution.


Hi Brent,


The answer is A why is it showing B though?

greenlight-admin's picture

My bad!
Thanks for the heads up!!
I've edited my solution here: https://gre.myprepclub.com/forum/o-is-the-circle-s-center-and-por-is-a-r...

Cheers and thanks,

Hi Brent,


I can reach to the end but I picked the wrong answer.

I know the length of C < D, then the answer should be B.
(but the answer is A) because of the negative logic.

In my understanding "length of the vertical leg" never ever be negative, am I right?
How the length can be negative? I consider them like height/weight that should not be negative. However, it seemed like I misunderstood in this point, can you elaborate more that in this case that the length can be negative.

greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/qotd-6-which-is-greater-c-or-d-2655.html

We can say that the MAGNITUDE of C is less than the MAGNITUDE of D.
This is similar to saying "the length of C < D"

You're also correct to say that lengths are never negative.
However, in this case, we are referring to coordinates in the x-y plane, and coordinates can be either positive or negative.
In Quadrant III, both coordinates are negative.
When comparing NEGATIVE values, the value with the greatest MAGNITUDE is the lesser number.

For example, the MAGNITUDE of -10 is greater than the MAGNITUDE of -4
This tells us that -10 is the lesser number.
That is -10 < -4

You might want to review the video on the coordinate plane: https://www.greenlighttestprep.com/module/gre-algebra-and-equation-solvi...



Hello Brent, in this case besides assigning values, is there any other method to solve it.
thank you
greenlight-admin's picture

Link: https://gre.myprepclub.com/forum/angle-bad-y-angle-bac-z-angle-abc-x-and...

Here's another (more abstract) approach:
Notice that we can take point D and move it very very close to vertex C.
Doing so will make angle z very very small.
In fact, we can make angle z so small that it's practically zero (let's say it's 0.0000000001 degrees)

At this point, we have:
QUANTITY B: y + (almost zero)

From here, we can see that it COULD be the case that angle y is greater then, less than, or even equal to the sum of angles x and w.

As such, the answer is D


Hi Brent, I was working on a few practice problems involving triangle properties and came across this question: https://gre.myprepclub.com/forum/if-the-perimeter-of-the-isosceles-right-triangle-shown-is-16497.html?fl=similar.

I am having a hard time solving it and was wondering if you could provide insight on how you’d go about find the answer to this question.
greenlight-admin's picture

Tricky question! The solution involves a lot of algebra at the end!
Here's my fault illusion: https://gre.myprepclub.com/forum/if-the-perimeter-of-the-isosceles-right...

When trying to solve this problem:


My answer to this was: C,D, and E, which I found utilizing the same logic you used in your explanation; however I am not quite understanding how you are determining that B and F are also valid options.
greenlight-admin's picture

Link to my solution: https://gre.myprepclub.com/forum/points-q-r-and-s-lie-in-a-plane-if-the-...

If we assume QRS is a triangle then we get: 7 < SQ < 29
However, we need to recognize that QRS could also be a straight line.

So, we could have this situation:
In this case, QS = 29

Or we could have this situation:
In this case, QS = 7

Since QS can equal 7 and 29, the complete range of values for QS are as follows:
7 ≤ SQ ≤ 29

Does that help?


this is a very interesting problem I didnt get your solution though. can we think of this as a triangle or not? Originally I thought of it as a line and selected 7 and 29. How would we know that we can think of it as a triangle?
greenlight-admin's picture

Question Link: https://gre.myprepclub.com/forum/points-q-r-and-s-lie-in-a-plane-if-the-...

Tough question (41% success rate)!!
If three points lie in the coordinate plane, then those three points can form a triangle, or they can be collinear.

If we start by assuming the three points form a triangle, then we can use the noted triangle property to conclude that: 7 < x < 29
However, since we're specifically not told the three points create a triangle, we must consider the possibility that three points are collinear, in which case we must include 7 and 29 in our inequality to get: 7 ≤ x ≤ 29.

Does that help?


Hey Brent is a here showing like in co ordinate the magnitude of value of x from origin and b the vertical climb i.e. y
greenlight-admin's picture

Question link: https://gre.myprepclub.com/forum/which-is-greater-a-or-b-17268.html

That's correct.
Since the coordinate (a, b) lies in quadrant I, a and b are both positive, which means a (the x-coordinate) tells us the point's horizontal distance from the origin, and b (the y-coordinate) tells us the point's vertical distance from the origin.

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