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04 Oct 2011, 08:33
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C
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Consider an obtuse-angled 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
A. 5
B. 21
C. 10
D. 15
E. 14
04 Oct 2011, 09:06
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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
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
11 Nov 2013, 00:54
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madn800
Consider an obtuse-angled 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.
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.
