3 - 4 - 5 Triangle

[ted]

Greetings:

 

I am a first time user of this web site, so, Imay make a few mistakes. I want to re-learn the formula for determining the diagonal distance between the ends of a 3-4-5 right triangle. The the known measurements are 10 ft and 12 ft.

[gcol]

Wow, that is really difficult. Even Pythagoras was stumped...or was he?

[timo]

Assuming "measurement" is what I assume it to be (0.5*a*b, where a and b are the two shorter sides of the triangle), then additional assuming a<b (meaning that a is the shortest side), you can get the length of a by substituting b=4/3*a in above. Knowing the length of a, you know the lengths of the other sides (because you know the ratios).

 

EDIT: Forget what I said in above. If the "measurements" are the area as I supposed in above, then it should be square-feet, not feet. They can, however, also not be the lengths of two sides (because their ratios are 6:5 which is none of the two possible ratios 5:3 and 5:4). What is it?

[Ragib]

lol gcol why are you so cruel, maybe he just forgot..or maybe he just doesnt remember his pythagorean triads and thought 3,4,5 wudnt be right angled. either way, atheist is nice. btw ted, the side opposite the right angle, called the hypotenuse, is what you want. to get it with any right angled triangle, you add the squares of the smaller sides, then square root that. or, a^2 + b^2 = c^2 where c is the longest side, and a and b are the smaller ones.

[gcol]

lol gcol why are you so cruel,QUOTE]

Grovelling apologies. The first question of a mathematical nature in this forum I even understood. Could not resist it. I did manage to suppress all the convoluted punning ways of hinting at the answer, though

[timo]

Gimme a hint then: What is 10 ft and 12 ft ?

[the tree]

If I'm guessing correctly then all you need to know is:

the square of length of the longest side of a right angled triangle is equal to the sum of the squares of the two remaining sides

[Ragib]

yes of course you guessed correctly, i bet the person were trying to help doesnt even read this thread anymore

[justchosen]

hi guy's new to the forum, hey why don't you use one variable trigonometry.its simple,and it does away with all trig.never have to use pathgoreans theorum. x= 12h/20,4h/5, (.8h). y=12h/15, 3h/5 ,(.6h) cause .6^2+.8^2=h^2,.36+.64=1^2 see h is a constant to both x and y