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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,270

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

Character is who you are when no one is looking.

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

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

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

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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**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 17,270

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

Character is who you are when no one is looking.

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

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

Or not ...

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

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

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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