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Topic review (newest first)
- wcy
- 2005-08-11 10:35:14
The sum of the lengths of any two sides is greater than the lenght of the third side.
For all triangles,
by cosine rule,
c²=a²+b²-2abcosC
(a+b)²=a²+2ab+b²
a+b=√(a²+2ab+b²)
c=√(a²+b²-2abcosC)
now, a+b is definitely larger than c, as the maximum possible value of -2abcosC is less than 2ab, as angle C is smaller than 180 but larger than 0.
- wcy
- 2005-08-10 22:06:20
here is a proof for right angled triangle:
by pythagoras theorem,
a²+b²=c²
c=√(a²+b²) ----(1)
(a+b)²=a²+2ab+b²
a+b=√(a²+2ab+b²) ----(2)
comparing (1) and (2),
one can see that
a+b>c
- MathsIsFun
- 2005-08-10 22:06:08
Ooohh ... so you CAN have one side of a triangle longer than the other two, it just somehow slips into imaginary space ... 
Or not ... 
- ganesh
- 2005-08-10 21:48:04
Lets take the length of sides Mathsy has given.
The Hero's formula for area of a triangle is
Area = √(s(s-a)(s-b)(s-c)
where a, b, c are the sides and s = (a+b+c)/2
Therefore, Area = √[(11/2)(11/2 - 2)(11/2 - 3) (11/2 - 6)]
11/2 - 6 is a negative number,
So the Area of the triangle is an imaginary number 
- mathsyperson
- 2005-08-10 19:14:47
Try disproving it and fail. Just try drawing a triangle with lengths of 2, 3 and 6. The 2 and 3 are too short and too far apart to be able to join up.
- wcy
- 2005-08-10 19:05:37
How do we prove that the sum of the lengths of any two sides is greater than the lenght of the third side in the first place?
Just interested.
- mathsyperson
- 2005-08-10 18:04:40
Ganesh is right, but as BC=4, it can be continued further:
AB < (4+8)/2
AB < 12/2
AB < 6
- ganesh
- 2005-08-10 15:19:28
The Triangle inequality is
AB + BC > AC
AC + BC > AB
AB + AC > BC
You have given BC=4 and AC = 8 - AB
Therefore, the required inequality is
AB < BC + AC
AB < BC + 8 - AB
2 AB < BC + 8
AB < (BC + 8)/2
- NeoXtremeX
- 2005-08-10 14:48:32
The sum of the lengths of any two sides is greater than the lenght of the third side. in triangle ABC, BC = 4 and AC = 8 - AB. Write an inequality for AB.
Help?