Question 309769: The lengths of the sides of an isosceles triangle are 20, n, and n. If n is an integer, what is the smallest possible permiter of the triangle. Solution options are 40, 41, 42, 44, 60. Thank you.
Found 2 solutions by palanisamy, Edwin McCravy:
Answer by palanisamy(496)   (Show Source):
You can put this solution on YOUR website! The lengths of the sides of an isosceles triangle are 20, n, and n
Perimeter = 20+n+n = 20+2n, an even integer
The smallest possible permiter of the triangle is 40
Answer by Edwin McCravy(20054)  (Show Source):
You can put this solution on YOUR website!
The other tutor's solution is incorrect.
While it is true that the perimeter must be even, that
is not enough information.
The sides are 20, n, and n.
Every triangle must be such that
(its first side) + (its second side) > (its third side)
(its first side) + (its third side) > (its second side)
(its second side) + (its third side) > (its first side)
Let its first side = 20, its second side be n and its third side be n
Then we must have:
20 + n > n
20 + n > n
n + n > 20
or
20 > 0
20 > 0
2n > 20
The first two inequalities are always true, and
the last inequality
2n > 20
is true whenever
n > 10
The smallest integer n could be such that n > 10 is n = 11
So the sides of the triangle with smallest possible perimeter
are 20, 11 and 11, so the smallest perimeter is
20+11+11=42
Edwin
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