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Consider an obtuseangled triangles with sides 8 cm, 15 cm a
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04 Oct 2011, 07:33
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Consider an obtuseangled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist? A. 5 B. 21 C. 10 D. 15 E. 14
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Re: number of traingles
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04 Oct 2011, 08:06
One of the angles must be greater than 90 a^2 + b^2 < c^2 z is the side opposite the obtuse angle & hence the longest. If 15 is the big side, then: 8^2 + x^2 < 15^2 64 + x^2 < 225 x^2 < 161 x < 12.jfdhfdh So, keeping with our other range, x could be 8, 9, 10, 11 or 12. If we make x our large side, we have: 8^2 + 15^2 < x^2 64 + 225 < x^2 289 < x^2 17 < x x could be 18, 19, 20, 21 or 22. 10 possible values for x. C
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Re: number of traingles
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04 Oct 2011, 08:15
sorry typo i meant c be the longest side.



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Re: Consider an obtuseangled triangles
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06 Oct 2011, 05:05
How did you get 10 possible values We know that X value should be greater that 8 and since in a triangle sum of two sides should be greater than third side that value of X can vary starting from 8 so we can have infinite values but how did you justify that 10 values possible.
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Re: Consider an obtuseangled triangles
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06 Oct 2011, 05:24
kotela wrote: How did you get 10 possible values
We know that X value should be greater that 8 and since in a triangle sum of two sides should be greater than third side that value of X can vary starting from 8 so we can have infinite values but how did you justify that 10 values possible. If two sides are 8 and 15, the third side will have to be between (158)=7 AND (15+8)=23. So, x is a value such that 7<x<23.
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Re: Consider an obtuseangled triangles
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08 Oct 2011, 02:55
by the property of triangle, we know that sum of two sides must be greater than the third side, therefore , 15+8>x also x must be greater than 7 because if x=7, then it will go against the property of triangle and 7+8 will be 15, which should be greater than 15. hence 7<x<23, therefore x=15... thanx in advance for the kudos
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Re: Consider an obtuseangled triangles
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10 Nov 2013, 16:33
I think the answer is 5. Here's why. Firstly you need to realise that the possible values for the third side lie between 7 and 23 that is 15 possible values. Secondly you need to understand that 8, 15 and 17 represents a Pythagorean triplet. Hence, the angle between 8 and 15 is 90 degrees. Now, to ensure that the angle between 8 and 15 is >90 degrees that it is obtuse the third side MUST be greater than 17 because if it is less than 17, the triangle will become ACUTE angled triangle. Previously the range was 7<third side<23. Now, the range of the third side is contracted to 17<third side< 23. Hence, third side can take 5 possible values and consequently 5 such obtuse angled triangles can be formed.



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Re: Consider an obtuseangled triangles
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10 Nov 2013, 23:54
madn800 wrote: Consider an obtuseangled triangles with sides 8 cm, 15 cm and x cm. If x is an integer then how many such triangles exist?
A. 5 B. 21 C. 10 D. 15 E. 14
I think the answer is 5. Here's why. Firstly you need to realise that the possible values for the third side lie between 7 and 23 that is 15 possible values. Secondly you need to understand that 8, 15 and 17 represents a Pythagorean triplet. Hence, the angle between 8 and 15 is 90 degrees. Now, to ensure that the angle between 8 and 15 is >90 degrees that it is obtuse the third side MUST be greater than 17 because if it is less than 17, the triangle will become ACUTE angled triangle. Previously the range was 7<third side<23. Now, the range of the third side is contracted to 17<third side< 23. Hence, third side can take 5 possible values and consequently 5 such obtuse angled triangles can be formed. The problem with your solution is that you assumed that x must be the largest side. But it's possible that 15 is the largest side and in this case you'd have: \(8^2 + x^2 < 15^2\) > \(x^2 < 161\) > \(x < 12.something\). Since we know that \(7 < x\), then for this case you'd have 5 more integer values for x: 8, 9, 10, 11, and 12. Therefore x can take total of 10 integer values. Answer: C.
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Re: Consider an obtuseangled triangles with sides 8 cm, 15 cm a
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13 Feb 2014, 06:15
The case where 15 could be the longest side is what I missed. this gives another 5 integer values for x. Making it total of 10. good question!
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Re: Consider an obtuseangled triangles with sides 8 cm, 15 cm a
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03 Apr 2016, 15:44



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Re: Consider an obtuseangled triangles with sides 8 cm, 15 cm a
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03 Apr 2016, 18:50
Chiragjordan wrote: Hey chetan2u Can you please explain this one.. None of the above solution works for me .. and What exactly are these acute obtuse cases? i know that x can take values between (7,23) Hi Obtuse angled triangle is a triangle with one angle>90 degree.. Because of this there are few restriction on the legth of sides that you can have ..A simpler example.. let the sides be 6, 8 and 10... this is a right angled triangle with sides 6 , 8 and hypotenuse as 10.. If it is an obtuse angled triangle, 6,8,10 cannot be sides as no angle >90 here too we can have two case 1) 15 is the biggest side..if it is 90 degree, x will be \(\sqrt{15^28^2}\) = \(\sqrt{161}\)= between 12 and 13.. if we take 13, the angle will become less than 90.. so 12 and values below it are correct.. but min value = 158 +1= 7+1 =8 so values are8 to 12= 5 values..2) let x be the largest sidesimilarly for it to be 90 degree, x should be \(\sqrt{8^2+15^2}\)= 17.. so it has to be >17.. max value = 5 +181 =22 so values = 18,19,20,21,22 5 more valuestotal  5+5=10 values
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Re: Consider an obtuseangled triangles with sides 8 cm, 15 cm a
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11 Aug 2018, 07:19
The sides of an obtuse triangle are 7 cm , 24 cm ,x cm. If x is an integer, how many values can it take?
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Re: Consider an obtuseangled triangles with sides 8 cm, 15 cm a
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11 Aug 2018, 07:28
Why x can not take the value more than 22 or less than 8?
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Re: Consider an obtuseangled triangles with sides 8 cm, 15 cm a &nbs
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11 Aug 2018, 07:28






