Math Is Fun Forum
  Discussion about math, puzzles, games and fun.   Useful symbols: √ ∞ ≠ ≤ ≥ ≈ ⇒ ∈ Δ θ ∴ ∑ ∫ π -

Login

Username

Password

Not registered yet?

Post a reply

Go back

Write your message and submit
:) :| :( :D :o ;) :/ :P :lol: :mad: :rolleyes: :cool: | :dizzy :eek :kiss :roflol :rolleyes :shame :down :up :touched :sleep :wave :swear :tongue :what :faint :dunno
Options

Go back

Topic review (newest first)

irspow
2006-02-19 02:27:44

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.

ganesh
2006-02-19 02:17:39

Since Pythogoras theorem is applicable, the triangle is right-angled. As (x+8) is the hypotenuse,
x+ (x+7)= (x+8)
x+x+14x + 49 = x+16x + 64
2x+14x+49-x-16x-64=0
x-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. smile smile smile

Math Student
2006-02-19 02:06:25

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! ^_^up

Board footer

Powered by FluxBB