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If 10, 12 and ‘x’ are sides of an acute angled triangle, ho

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

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Updated on: 14 Nov 2013, 06:24
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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

Originally posted by amirdubai1982 on 16 Feb 2010, 10:43.
Last edited by Bunuel on 14 Nov 2013, 06:24, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: interesting question in Geometry  [#permalink]

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11 Nov 2010, 04:36
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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 31317 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)
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Re: interesting question in Geometry  [#permalink]

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16 Feb 2010, 11:20
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amirdubai1982 wrote:
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

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 32279 times ]

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

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22 Nov 2013, 03:24
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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.

Hope it's clear.
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Re: interesting question in Geometry  [#permalink]

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16 Feb 2010, 11:06
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amirdubai1982 wrote:
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

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.
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Re: interesting question in Geometry  [#permalink]

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16 Feb 2010, 11:44
2
Ans C -9

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 )

this gives 6 < x < 16
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Re: interesting question in Geometry  [#permalink]

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16 Feb 2010, 11:10
1
bangalorian2000 wrote:
amirdubai1982 wrote:
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

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.
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Re: interesting question in Geometry  [#permalink]

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21 Nov 2010, 07:37
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girisshhh84 wrote:

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 30670 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.
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Re: If 10, 12 and ‘x’ are sides of an acute angled triangle, ho  [#permalink]

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22 Nov 2013, 02:59
1
I arrived at this
2 < x < 22
Total of 19 possible values.

Couldn't figure out how to minimize the range for acute angle?
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Re: interesting question in Geometry  [#permalink]

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16 Feb 2010, 11:06
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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

so possibilities = 9

C
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Re: interesting question in Geometry  [#permalink]

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11 Nov 2010, 02:13
There are like 3 different solutions that all give the same answer but different values for x. Could someone please look at it once again?
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Re: interesting question in Geometry  [#permalink]

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16 Nov 2010, 22:43
Hi,

Doubt Clrd but just a point in case, if it traingle would have been an obtused triangle

The points would have been b/w 2<x<6 and 15<x<22

Regards
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Re: interesting question in Geometry  [#permalink]

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20 Nov 2010, 14:12
chix475ntu wrote:
bangalorian2000 wrote:
amirdubai1982 wrote:
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

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 ?
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Re: interesting question in Geometry  [#permalink]

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20 Nov 2010, 17:37
girisshhh84 wrote:
how do we deicide that x HAS to be the largest side ?

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.
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Re: interesting question in Geometry  [#permalink]

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20 Nov 2010, 22:23
VeritasPrepKarishma wrote:
girisshhh84 wrote:
how do we deicide that x HAS to be the largest side ?

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.
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Re: interesting question in Geometry  [#permalink]

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22 Nov 2013, 02:08
VeritasPrepKarishma wrote:
girisshhh84 wrote:
how do we deicide that x HAS to be the largest side ?

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|
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Re: interesting question in Geometry  [#permalink]

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22 Nov 2013, 02:36
honchos wrote:
VeritasPrepKarishma wrote:
girisshhh84 wrote:
how do we deicide that x HAS to be the largest side ?

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|

Almost all solutions above use this property.

Similar question to practice: consider-an-obtuse-angled-triangles-with-sides-8-cm-15-cm-a-121500.html
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Re: If 10, 12 and ‘x’ are sides of an acute angled triangle, ho  [#permalink]

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27 Mar 2017, 01:24
sum of two sides greater than third side

12+10= 22 > x

x < 12-10 = 2

2 < x < 22

since this is acute angle triangle hence x > 12 therfore 12<x<22 => 9 values

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

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27 Mar 2017, 09:06
praveenmittal95605 wrote:
sum of two sides greater than third side

12+10= 22 > x

x < 12-10 = 2

2 < x < 22

since this is acute angle triangle hence x > 12 therfore 12<x<22 => 9 values

Hence C

This is incorrect reasoning that by fluke happens to get the right answer. The correct range of 9 values is 7<=x<=15, and it is obtained by using the Pythagorean theorem. x = 21, for instance, would NOT give an acute triangle. Nor is it necessary for x to be the largest side. I hope this clarifies; there were a number of confused comments earlier in the thread. to reiterate, this answer is correct, but this solution, along with the x values it gives, is wrong.
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If 10, 12 and ‘x’ are sides of an acute angled triangle, ho  [#permalink]

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03 Sep 2017, 04:54
Ans is Clearly C

49<= x^2<=225
7<=x<=15
x can have 15-7+1 integer values = 9
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If 10, 12 and ‘x’ are sides of an acute angled triangle, ho   [#permalink] 03 Sep 2017, 04:54

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