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# In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4

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In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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17 Aug 2007, 12:09
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In the triangle above, is x > 90?

(1) a^2 + b^2 <15
(2) c > 4

[Reveal] Spoiler:
Attachment:

triangle.GIF [ 1.66 KiB | Viewed 1396 times ]
[Reveal] Spoiler: OA

Last edited by Bunuel on 14 Aug 2017, 08:10, edited 3 times in total.

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In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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09 Feb 2012, 02:57
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In the triangle above, is x > 90?

(1) a^2 + b^2 <15
(2) c > 4

Each statement alone is clearly insufficient.

When taken together:
If angle x were 90 degrees than we would have $$a^2+b^2=c^2$$, since $$a^2+b^2<15<(16=c^2)$$ then angle x must be greater than 90 degrees (c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90).

P.S. If the lengths of the sides of a triangle are a, b, and c, where the largest side is c, then:

For a right triangle: $$a^2 +b^2= c^2$$.
For an acute (a triangle that has all angles less than 90°) triangle: $$a^2 +b^2>c^2$$.
For an obtuse (a triangle that has an angle greater than 90°) triangle: $$a^2 +b^2<c^2$$.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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17 Feb 2012, 01:42
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SergeNew wrote:
Quote:
c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90

Hi Guys,

Would anyone be able to explain why the angle is greater than 90 if c^2 is greater than a^2+b^2?

Serge.

If c^2 were equal to a^2+b^2 then we would have a^2+b^2=c^2, which would mean that angle x is 90 degrees. Now, since c^2 is more than a^2+b^2, then angle x, which is opposite c, must be more than 90 degrees: try to increase side c and you'll notice that angle x will increase too.

Hope it's clear.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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28 Jan 2013, 22:31
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Just a handy tool to prove that the answer is C.

c^2 = a^2+b^2-2abCos(c). In this case , Angle c = x.

Now, we know that c^2 >16

Also, a^+b^2<15.

Thus a^2+b^2<c^2 = a^2+b^2-c^2 <0

a^2+b^2-c^2 = 2abCos(x) ; 2abCos(x) <0. As ab!=0,Cos(x)<0. Thus, X>90 degree.

For c=90 degree, the above equality gives the famous theorem!
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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16 Feb 2012, 23:26
Quote:
c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90

Hi Guys,

Would anyone be able to explain why the angle is greater than 90 if c^2 is greater than a^2+b^2?

Serge.

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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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17 Feb 2012, 01:27
Whenever Asqr + Bsqr Is equal to Csqr than angle X is 90 degrees
Look at both options carefully ,

Asqr + Bsqr Is less than 15 and C greater than 4 means any value more than 4 for c and will automatically increase angle X bcoz as C increases angle 8 increases

Poor in making posts

Hope it will help.

Posted from my mobile device

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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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26 Jan 2013, 20:15
Bunuel wrote:
I attached the diagram, which was missing in initial post.

Attachment:
Trianlge.PNG
In the triangle above, is x > 90?

(1) a^2 + b^2 <15
(2) c > 4

Each statement alone is clearly insufficient. When taken together:
If angle x were 90 degrees than we would have a^2+b^2=4^2, since a^2+b^2<15<16 then angle x must be greater than 90 degrees (c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90).

Hope it helps.

What would happen is the statement was c>3? how would this question be framed such that the angle could be always below 90 degrees. Since in this particular question the solution will always be greater, what would be the opposite case?
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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26 Jan 2013, 20:50
fozzzy wrote:
Bunuel wrote:
I attached the diagram, which was missing in initial post.

Attachment:
Trianlge.PNG
In the triangle above, is x > 90?

(1) a^2 + b^2 <15
(2) c > 4

Each statement alone is clearly insufficient. When taken together:
If angle x were 90 degrees than we would have a^2+b^2=4^2, since a^2+b^2<15<16 then angle x must be greater than 90 degrees (c^2 is greater than a^2+b^2 then the angel opposite c must be greater than 90).

Hope it helps.

What would happen is the statement was c>3? how would this question be framed such that the angle could be always below 90 degrees. Since in this particular question the solution will always be greater, what would be the opposite case?

If you reverse the values, such as:
(1) a^2 + b^2 >16
(2) c < 4
You would get a case where angle x would be less than 90, provided a triangle is still formed. (Note that a+b will tend to be bigger and C tends to be smaller in this option)
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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24 Feb 2013, 20:46
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smily_buddy wrote:
Attachment:
Trianlge.PNG
In the triangle above, is x > 90?

(1) a^2 + b^2 <15
(2) c > 4

If a^2 + b^2 = c^2 then x= 90.
But , as given in st2, minimum value of C = 5, then C^2 = 25.
It means a^2 + b^2 < c^2.Thus, x <90. Therefore, C.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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24 Feb 2013, 22:14
abhi47 wrote:
E is the correct answer. Both statements are insufficient.

from one, possible values of (a,b) = (1,2) or (1,3) or (2,3) or (2,2)
from 2nd, given that c>4
Property of triangle is = sum of two sides > third side
therefor only possible values are a= 2, b=3 and c = 5 or a = 3 and b = 2 and c =5
since a^2 + b ^2 = c^2
hence it is a right angle triangle
hence both conditions required to answer
so option(c)

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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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01 Oct 2015, 16:41
a^2+b^2 = c^2 then angle opposite of c is right angle triangle
a^2+b^2 < c^2 then angle opposite of c is greater than 90
a^2+b^2 > c^2 then angle opposite of c is less than 90

Using the above logic, we can make of use of both the statements to answer the question.

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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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28 Jul 2017, 09:03
The inequality a^2+b^2,15 shows that the angle x is not equal to 90 so it can lesser or greater. So the statement 1 is sufficient. I know that i have some mistake with the concept, can someone please correct me. thank you

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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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29 Jul 2017, 00:10
jbisht wrote:
abhi47 wrote:
E is the correct answer. Both statements are insufficient.

from one, possible values of (a,b) = (1,2) or (1,3) or (2,3) or (2,2)
from 2nd, given that c>4
Property of triangle is = sum of two sides > third side
therefor only possible values are a= 2, b=3 and c = 5 or a = 3 and b = 2 and c =5
since a^2 + b ^2 = c^2
hence it is a right angle triangle
hence both conditions required to answer
so option(c)

How can sum of two sides be equal to third side in your argument above where a=2,b=3 and c=5, if this triangle is drawn will it not be a single line ?
IMO E is the correct answer

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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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29 Jul 2017, 00:49
longhaul123 wrote:
The inequality a^2+b^2,15 shows that the angle x is not equal to 90 so it can lesser or greater. So the statement 1 is sufficient. I know that i have some mistake with the concept, can someone please correct me. thank you

How does a^2 + b^2 < 15 imply that x is not 90 degrees? We are given that the sum of the square of the lengths of two sides is less than some number. We cannot deduce anything from this. If a = b = 1 and $$c = \sqrt{2}$$, then $$a^2 + b^2 = c^2$$, which will make x equal to 90 degrees but if a = b = c = 1, then x will be equal to 60 degrees.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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29 Jul 2017, 00:49
saurabhSuman wrote:
jbisht wrote:
abhi47 wrote:
E is the correct answer. Both statements are insufficient.

from one, possible values of (a,b) = (1,2) or (1,3) or (2,3) or (2,2)
from 2nd, given that c>4
Property of triangle is = sum of two sides > third side
therefor only possible values are a= 2, b=3 and c = 5 or a = 3 and b = 2 and c =5
since a^2 + b ^2 = c^2
hence it is a right angle triangle
hence both conditions required to answer
so option(c)

How can sum of two sides be equal to third side in your argument above where a=2,b=3 and c=5, if this triangle is drawn will it not be a single line ?
IMO E is the correct answer

The correct answer is C. Check here: https://gmatclub.com/forum/abc-with-a-t ... l#p1041826

Hope it helps.
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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30 Nov 2017, 15:46
Hello Quick question
Can the ans be A
Im a little confused

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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4 [#permalink]

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30 Nov 2017, 19:58
cocojatti92 wrote:
Hello Quick question
Can the ans be A
Im a little confused

The correct answer is C. Check here: https://gmatclub.com/forum/abc-with-a-t ... l#p1041826
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Re: In the triangle above, is x > 90? (1) a^2 + b^2 < 15 (2) c>4   [#permalink] 30 Nov 2017, 19:58
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