SOLUTION: Side lengths are 16, 48, and 50. Is this a right triangle?

Question 796496: Side lengths are 16, 48, and 50. Is this a right triangle?
Found 2 solutions by Edwin McCravy, SRiz360:
Answer by Edwin McCravy(20081) (Show Source): You can put this solution on YOUR website!

Let's find out. If so, the Pythagorean theorm will hold So we have: 16� + 48� ≟ 50� 256 + 2304 ≟ 2500 2560 ≟ 2500 Nope, not quite! Drawn to scale it looks like this: That angle at the top looks like a 90� angle but it's only about 88�, just a couple of degrees short. Edwin

Answer by SRiz360(1) (Show Source): You can put this solution on YOUR website!
In this question you use Pythagorean theorem, a^2+b^2=c^2. So you plug the numbers in equation. *Remember the bigger length is always the hypothenuse*
>>Link>> https://www.google.com/search?q=hypotenuse&es_sm=122&source=lnms&tbm=isch&sa=X&ei=WkpTUt_bA5TI4AOdvIDADw&ved=0CAcQ_AUoAQ&biw=1024&bih=475&dpr=1#es_sm=122&q=hypotenuse+leg&tbm=isch&facrc=_&imgdii=_&imgrc=7D-ltG-4mhO9IM%3A%3B1_nbljP7ToRwEM%3Bhttp%253A%252F%252Fwww.wyzant.com%252FImages%252FHelp%252Fright_tri1.gif%3Bhttp%253A%252F%252Fwww.wyzant.com%252Fhelp%252Fmath%252Fgeometry%252Ftriangles%252Fright_triangle_congruence%3B299%3B308 <<, Link <<<
^^^^^^^^^use that link^^^^^^^^^ to see the hypothenuse the big length of a triangle. So the formula a^2+b^2 are legs and we use two small number compare to 50 are 16 and 48.
a^2 would be substitute for 16^2 + and b^2 would be substitute for b^2 for 48^2 and c^2 for 50^2.
16^2+48^2=50^2 now we solve
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16*16=256
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48*48=2304
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50*50=2500
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256+2304=2500
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2560=2500 DONE!!!!!
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Yes it is a right triangle because two legs when combined have to be greater than hypothenuse.
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