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**

**Math Student****Guest**

Please can you help me with this question?

I've got 3 lengths

x

x+7

and x+8

x+8 is the hypotenuse.

I need to find the length of each side, I've tried but still can't figure it out, please help! Thanks in advance! ^_^

**ganesh****Moderator**- Registered: 2005-06-28
- Posts: 20,909

Since Pythogoras theorem is applicable, the triangle is right-angled. As (x+8) is the hypotenuse,

x²+ (x+7)²= (x+8)²

x²+x²+14x + 49 = x²+16x + 64

2x²+14x+49-x²-16x-64=0

x²-2x-15=0

x = [2 ± √(4+60)]/2 = (2 ± 8)/2 = 5 or -3.

The lenghts of the sides are 5, 12 and 13, and these form a Pythogorean triple.

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

**irspow****Member**- Registered: 2005-11-24
- Posts: 457

Using the theorem;

c = √(a² + b²) gives;

x+8 = √(x² + (x+7)²);

x+8 = √(x² + x² + 14x + 49);

x+8 = √(2x² + 14x + 49);

Squaring both sides gives;

x² + 16x + 64 = 2x² + 14x + 49;

0 = x² - 2x - 15;

0 = (x-5)(x+3);

This is only true when x = 5 or -3;

Since x represents one of the sides it cannot be negative;

Therefore x = 5;

So the triangle has sides of 5, 12, and 13.

edit*

Sorry ganesh, you are a lot faster than me! lol.

*Last edited by irspow (2006-02-18 03:29:06)*

Offline

Pages: **1**