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## #1 2006-02-18 03:06:25

Math Student
Guest

### Tricky Pythagoras' Theorem Question

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! ^_^

## #2 2006-02-18 03:17:39

ganesh
Moderator
Registered: 2005-06-28
Posts: 21,730

### Re: Tricky Pythagoras' Theorem Question

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.

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## #3 2006-02-18 03:27:44

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

### Re: Tricky Pythagoras' Theorem Question

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)

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