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

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

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New post 26 Mar 2017, 23:17
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Re: If 10, 12 and ‘x’ are sides of an acute angled triangle, ho [#permalink]

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New post 27 Mar 2017, 02:24
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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New post 27 Mar 2017, 10:06
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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New post 03 Sep 2017, 05:54
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, 05:54

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

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