Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible? (A) 7 (B) 12 (C) 9 (D) 13 (E) 11

Please answer with details in logic order. Thanks to all.

Any side of triangle is less than the sum of other two sides so in this case x < 22. Also any side of triangle is larger than positive difference of other two sides so x > 2. 2 < x < 22 Total of 19 possible values. As this is acute angled triangle x must be largest hence x > 12 and x < 22 so total of 9 possible values hence C.

acute angle traingle - all angles in the traingle has to be less than 90. Two possibilities lets take a or almost a right traingle(89 as one angle -> and the maximum value of the traingle will be somewhat closer to the hypotenue of the traingle.

let x be the other side .. if x is hypotenue it value is less than (12 ^ 2 + 10 ^ 2)^1/2 < 16 .. so max value is 15 let x be the other side .. if 12 is hypotenue, then value of x is greater than (12 ^ 2 - 10 ^ 2)^1/2 > 6 .. so min value is 7

idea behind this ... draw a traingle with 10 as base and 12 as the height and angle formed between these 2 is right traingle. other side is hypotenue and the max value will be 15. then slide the side 12 towards right slowly and at some point side 12 will become hypotense .. and if you move that afterwards towards rights, then angle between 10 and x will be >90

If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible? (A) 7 (B) 12 (C) 9 (D) 13 (E) 11

Please answer with details in logic order. Thanks to all.

Any side of triangle is less than the sum of other two sides so in this case x < 22. Also any side of triangle is larger than positive difference of other two sides so x > 2. 2 < x < 22 Total of 19 possible values. As this is acute angled triangle x must be largest hence x > 12 and x < 22 so total of 9 possible values hence C.

x need not have to the largest side - take an example where x=21 .. this would be an obtuse angled traingle.

If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible? (A) 7 (B) 12 (C) 9 (D) 13 (E) 11

Please answer with details in logic order. Thanks to all.

IMO C...

The catch in the acute angled triangle.

Consider the figures attached...

Case 1: \(x < sqrt{12^2 + 10^2}\) i.e. \(x < 16\)

Case 2: \(x > sqrt{12^2 - 10^2}\) i.e. \(x > 6\)

Therefore x ranges from 7 to 15 = 9 Values...

Attachments

1.png [ 21.93 KiB | Viewed 6186 times ]

_________________

Cheers! JT........... If u like my post..... payback in Kudos!!

|Do not post questions with OA|Please underline your SC questions while posting|Try posting the explanation along with your answer choice| |For CR refer Powerscore CR Bible|For SC refer Manhattan SC Guide|

first of using some of two sides greater than third side gives 2<x<22

now using \(Cos A = \frac{(b^2 + c^2 - a^2)}{2bc}\)

for cos A to be +ve sum of squares of 2 sides should be > square of third side ( equality will come when right angled, if its right angled then its not acute angled )

Actually, chix475ntu, jeetesh and gurpreet have all given the correct answer either using diagrams or formulas.. 6 < x < 16 is the same as values from 7 to 15 because only integral values are allowed.

Anyway, I am putting down the explanation. The question asks you for an acute triangle i.e. a triangle with all angles less than 90. An obtuse angled triangle has one angle more than 90. So the logic is that before one of the angles reaches 90, find out all the values that x can take. Starting from the first diagram where x is minimum and the angle is very close to 90, to the 2nd diagram where all angles are much less than 90 to the third diagram where the other angle is going towards 90.

Attachment:

Ques1.jpg [ 9.53 KiB | Viewed 5245 times ]

Note: The remaining angle cannot be 90 because that will make 10 the hypotenuse but hypotenuse is always the greatest side.

In the leftmost diagram \(x = \sqrt{(12^2 - 10^2)}\) x = root 44 which is 6.something x should be greater than 6.something because the angle cannot be 90.

In the rightmost diagram, \(x = \sqrt{(12^2 + 10^2)}\) x = root 244 which is 15.something x should be less than 15.something so that the angle is not 90.

Values that x can take range from 7 to 15 which is 15 - 7 = 8 + 1 = 9 values. (If the +1 above is not clear, check out the explanation on my blog http://gmatquant.blogspot.com/2010/11/four-prongs.html) _________________

If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible? (A) 7 (B) 12 (C) 9 (D) 13 (E) 11

Please answer with details in logic order. Thanks to all.

Any side of triangle is less than the sum of other two sides so in this case x < 22. Also any side of triangle is larger than positive difference of other two sides so x > 2. 2 < x < 22 Total of 19 possible values. As this is acute angled triangle x must be largest hence x > 12 and x < 22 so total of 9 possible values hence C.

x need not have to the largest side - take an example where x=21 .. this would be an obtuse angled traingle.

Your explanations is easy to assimilate , but i'm having a doubt (may be a silly one) - how do we deicide that x HAS to be the largest side ? Please enlighten. _________________

how do we deicide that x HAS to be the largest side ? Please enlighten.

I am not sure I understand your question very well... But let me try.

In an obtuse angled triangle, only one angle can be more than 90 (because the sum of all angles has to be 180). The side opposite that angle is largest. The side opposite the smallest angle is the smallest side. Similarly, in a right angled triangle, the side opposite 90 degree angle is greatest (the hypotenuse). In our question, we have to consider only acute triangles. Therefore, we are working with right triangles. The values of x we obtain from them will be very close to the smallest/greatest that are acceptable.

In the first figure, we are trying to calculate the smallest value of x. In that case, 12 will be the hypotenuse of the triangle. x will be less than 12. The angle opposite x is the smallest possible (if you look at the second figure, the angle opposite x is getting greater)

In the third figure, angle opposite to x has become the greatest possible i.e. 90 degrees (actually it should be slightly less than 90). Here x will be the hypotenuse and hence the greatest side of the triangle. x can't be any greater because then the triangle will become obtuse. _________________

how do we deicide that x HAS to be the largest side ? Please enlighten.

I am not sure I understand your question very well... But let me try.

In an obtuse angled triangle, only one angle can be more than 90 (because the sum of all angles has to be 180). The side opposite that angle is largest. The side opposite the smallest angle is the smallest side. Similarly, in a right angled triangle, the side opposite 90 degree angle is greatest (the hypotenuse). In our question, we have to consider only acute triangles. Therefore, we are working with right triangles. The values of x we obtain from them will be very close to the smallest/greatest that are acceptable.

In the first figure, we are trying to calculate the smallest value of x. In that case, 12 will be the hypotenuse of the triangle. x will be less than 12. The angle opposite x is the smallest possible (if you look at the second figure, the angle opposite x is getting greater)

In the third figure, angle opposite to x has become the greatest possible i.e. 90 degrees (actually it should be slightly less than 90). Here x will be the hypotenuse and hence the greatest side of the triangle. x can't be any greater because then the triangle will become obtuse.

HI Karishma , Thank you , i could understand the explanation as you told me . But Still have a doubt with 'bangalorian2000's statement "As this is acute angled triangle x must be largest hence x > 12 and x < 22 so total of 9 possible values hence C." To help me understand , pls tell me when we have 2 < x < 22 i.e. Total of 19 possible values. How many possible values of x when obtuse triangle ? How many values of x when right triangle. _________________

HI Karishma , Thank you , i could understand the explanation as you told me . But Still have a doubt with 'bangalorian2000's statement "As this is acute angled triangle x must be largest hence x > 12 and x < 22 so total of 9 possible values hence C." To help me understand , pls tell me when we have 2 < x < 22 i.e. Total of 19 possible values. How many possible values of x when obtuse triangle ? How many values of x when right triangle.

Ok. A side of a triangle is less than the sum of other two sides. Also a side of a triangle is greater than the difference of other two sides. Since the other two sides are 10 and 12, x should be less than 22 and more than 2.

Now we only have to consider acute angled triangles. Look at the figures below.

Attachment:

Ques.jpg [ 17.39 KiB | Viewed 4903 times ]

In the first figure when x is very small, till the angle goes to 90, the triangle is obtuse, which is not allowed. When x becomes \(\sqrt{(12^2 - 10^2)}\) = root 44 = 6.something, the angle is right angled. Now x greater than this value is allowed since we will get acute triangles. Values of 3, 4, 5 and 6 give obtuse angled triangles. Values of 7, 8, 9 ... give acute triangles so allowed.

Now look at the second figure. Finally x is \(\sqrt{(12^2 + 10^2)}\) = root 244 = 15.something So values of x till 15.something make acute triangles. Therefore 7, 8, 9, 10, 11, 12, 13, 14 and 15 (9 values) are allowed. When x is even bigger (16, 17, 18, 19, 20, 21), it will again make an obtuse angled triangle. And, as we saw, there are two values of x (6.something and 15.something) when we get a right angled triangle.

As for bangalorian2000's statement, he was probably mistaken. _________________

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. _________________

how do we deicide that x HAS to be the largest side ? Please enlighten.

I am not sure I understand your question very well... But let me try.

In an obtuse angled triangle, only one angle can be more than 90 (because the sum of all angles has to be 180). The side opposite that angle is largest. The side opposite the smallest angle is the smallest side. Similarly, in a right angled triangle, the side opposite 90 degree angle is greatest (the hypotenuse). In our question, we have to consider only acute triangles. Therefore, we are working with right triangles. The values of x we obtain from them will be very close to the smallest/greatest that are acceptable.

In the first figure, we are trying to calculate the smallest value of x. In that case, 12 will be the hypotenuse of the triangle. x will be less than 12. The angle opposite x is the smallest possible (if you look at the second figure, the angle opposite x is getting greater)

In the third figure, angle opposite to x has become the greatest possible i.e. 90 degrees (actually it should be slightly less than 90). Here x will be the hypotenuse and hence the greatest side of the triangle. x can't be any greater because then the triangle will become obtuse.

Some how can't we exploit the property to come to the conclusion

any side< sum of other two sides a<b+c any side> difference of two other sides a> |b-c| _________________

how do we deicide that x HAS to be the largest side ? Please enlighten.

I am not sure I understand your question very well... But let me try.

In an obtuse angled triangle, only one angle can be more than 90 (because the sum of all angles has to be 180). The side opposite that angle is largest. The side opposite the smallest angle is the smallest side. Similarly, in a right angled triangle, the side opposite 90 degree angle is greatest (the hypotenuse). In our question, we have to consider only acute triangles. Therefore, we are working with right triangles. The values of x we obtain from them will be very close to the smallest/greatest that are acceptable.

In the first figure, we are trying to calculate the smallest value of x. In that case, 12 will be the hypotenuse of the triangle. x will be less than 12. The angle opposite x is the smallest possible (if you look at the second figure, the angle opposite x is getting greater)

In the third figure, angle opposite to x has become the greatest possible i.e. 90 degrees (actually it should be slightly less than 90). Here x will be the hypotenuse and hence the greatest side of the triangle. x can't be any greater because then the triangle will become obtuse.

Some how can't we exploit the property to come to the conclusion

any side< sum of other two sides a<b+c any side> difference of two other sides a> |b-c|

Re: If 10, 12 and ‘x’ are sides of an acute angled triangle, ho [#permalink]

Show Tags

22 Nov 2013, 04:24

3

This post received KUDOS

Expert's post

2

This post was BOOKMARKED

honchos wrote:

I arrived at this 2 < x < 22 Total of 19 possible values.

Couldn't figure out how to minimize the range for acute angle?

If 10, 12 and 'x' are sides of an acute angled triangle, how many integer values of 'x' are possible? (A) 7 (B) 12 (C) 9 (D) 13 (E) 11

Say the lengths of the sides of a triangle are a, b, and c, where the largest side is c.

For a right triangle: \(a^2 +b^2= c^2\). For an acute triangle: \(a^2 +b^2>c^2\). For an obtuse triangle: \(a^2 +b^2<c^2\).

For a triangle with sides 10, 12 and x (\(2<x<22\)), the largest side could be 12 or x.

If the largest side is 12, then since it's an acute angle triangle: \(10^2+x^2>12^2\) --> \(x>\sqrt{44}=6.something\). If the largest side is x, then since it's an acute angle triangle: \(10^2+12^2>x^2\) --> \(x<15.something\).

\(6.something<x<15.something\) --> x can take 9 integer values: 7, 8, 9, 10, 11, 12, 13, 14, an 15.

Re: If 10, 12 and ‘x’ are sides of an acute angled triangle, ho [#permalink]

Show Tags

01 Feb 2015, 18:23

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. _________________

Re: If 10, 12 and ‘x’ are sides of an acute angled triangle, ho [#permalink]

Show Tags

11 Feb 2016, 15:23

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. _________________

http://blog.ryandumlao.com/wp-content/uploads/2016/05/IMG_20130807_232118.jpg The GMAT is the biggest point of worry for most aspiring applicants, and with good reason. It’s another standardized test when most of us...

With the limited financing options available to NZ citizens (especially those comme moi who aren't planning to return to work in NZ), I've really had to...

Strategy, innovation, marketing, finance... The second module has been pretty engaging. Though, no lack of memorable times. There is no lack of high profile guest speakers. One of the...