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## #1 2005-08-09 16:48:32

NeoXtremeX
Member
Registered: 2005-08-09
Posts: 1

### help with Inequality

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?

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## #2 2005-08-09 17:19:28

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,812

### Re: help with Inequality

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

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #3 2005-08-09 20:04:40

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: help with Inequality

Ganesh is right, but as BC=4, it can be continued further:
AB < (4+8)/2
AB < 12/2
AB < 6

Why did the vector cross the road?
It wanted to be normal.

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## #4 2005-08-09 21:05:37

wcy
Member
Registered: 2005-08-04
Posts: 117

### Re: help with Inequality

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.

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## #5 2005-08-09 21:14:47

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

### Re: help with Inequality

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.

Why did the vector cross the road?
It wanted to be normal.

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## #6 2005-08-09 23:48:04

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,812

### Re: help with Inequality

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

It is no good to try to stop knowledge from going forward. Ignorance is never better than knowledge - Enrico Fermi.

Nothing is better than reading and gaining more and more knowledge - Stephen William Hawking.

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## #7 2005-08-10 00:06:08

MathsIsFun
Registered: 2005-01-21
Posts: 7,631

### Re: help with Inequality

Ooohh ... so you CAN have one side of a triangle longer than the other two, it just somehow slips into imaginary space ...

Or not ...

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..."  - Leon M. Lederman

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## #8 2005-08-10 00:06:20

wcy
Member
Registered: 2005-08-04
Posts: 117

### Re: help with Inequality

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

Last edited by wcy (2005-08-10 00:07:00)

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## #9 2005-08-10 12:35:14

wcy
Member
Registered: 2005-08-04
Posts: 117

### Re: help with Inequality

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.

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