Question 1047889: in a triangle ABC, AB=x+2, BC=2x+3,AC=3x-5. Find all possible values of x
Found 2 solutions by vyshu544, ikleyn:
Answer by vyshu544(3)   (Show Source):
You can put this solution on YOUR website! As one side of a triangle must be less than the sum of other two sides and the same side of the triangle should be more than the difference of othere two sides
so lets start ;
AB = x+2
BC = 2x+3
AC = 3x-5
for example if we take x = 2
then AB = 4 , BC = 7, AC = 1
7+4>1 ( SATISFIES )
7-4>1 so it doesnt satisfies
like wise do until you get the corrct formation
finally i got:
AB = 4 BC = 11 AC = 7
Answer by ikleyn(52756)  (Show Source):
You can put this solution on YOUR website! .
in a triangle ABC, AB=x+2, BC=2x+3,AC=3x-5. Find all possible values of x
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To solve the problem you need to solve all three "triangle inequalities":
AB + BC > AC, i.e. (x+2) + (2x+3) > 3x-5, (1)
AB + AC > BC, i.e. (x+2) + (3x-5) > 2x+3, (2) and
BC + AC > AB, i.e. (2x+3) + (3x-5) > x+2. (3)
(1) is equivalent to 5 > -5, which is always true and doesn't carry useful information;
(2) is equivalent to 4x-3 > 2x+3, i.e. 2x > 6, which means x > 3.
(3) is equivalent to 5x-2 > x+2, i.e. 4x > 4, which means x > 1.
So, you solved the problem. All possible values of x are x > 3.
It is the necessary and sufficient condition.
Answer. All possible values of "x" are determined by only one condition x > 3.
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