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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?