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Bob

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What set of numbers below can be the lengths of the sides of a triangle?

13,10,16
1,2,3
5.2, 11, 4.9
208, 9, 219
There must be a rule for determining but I don\\\\\\\'t recall it.

733 day(s) ago

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Answers (4)

matherapist 		This is just the application of the previous solution. The easiest way to do this is take the 2 longest sides, and see if the 3rd side satisfies (a-b)<side c<(a+b) so for the last set

219-208<9<219+8 is not true so it's not a triangle

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samzappala 		rt triangle: a=shortest side, b=mid side, c=hypotenuse. a^2+b^2=c^2. So 100+169=256. 256=16^2 no. 1+4=9 no. 4.9^2+5.2^2=11^2. no. 9^2+208^2=219^2. no.

oblique triangle: a^2+b^2-2ab*cos (gamma). Since there's no angle, one of the formulas you want is gamma=acos((a^2+b^2-c^2)/(2ab)). acos((100+169-256)/(2*10*13))=acos(13/260)=arcos .05=87.13=gamma.

Instead of going through this look at 13,10,16. 10+13=23, which is greater than 16. yes
1,2,3, no. 5.2,11,4.9. yes 11.4 is greater than10.1. 208,9,219. yes
As for a rule, a+b is greater than c. In this case, a doesn't have to be the shortest. c doesn't have to be the hypotenuse.

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Bob 		The questioner wants to know which of the above four combinations of three sides can POTENTIALLY form a triangle. I don't know where to begin.

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YummyCookies 		do u know the formula?
u need to be specific in wat u r trying to find?
to find the area u use A=0.5(b)(h)
to find the perimeter u use P = a + b + c

hope it helps! :)

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