Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**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?

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,497

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.

Offline

**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.

Offline

**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.

Offline

**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.

Offline

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 21,497

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.

Offline

**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,626

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

Offline

**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)*

Offline

**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.

Offline

Pages: **1**