Please can you help me with this question?
I've got 3 lengths
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! ^_^
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
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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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.
Sorry ganesh, you are a lot faster than me! lol.
Last edited by irspow (2006-02-18 03:29:06)