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The lengths of the sides of isosceles triangle ABC are 3x + 62, 7x + 30, and 5x + 60 feet. What is the least possible number of feet in the perimeter of ABC?

 Jul 18, 2021
 #1
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I do not know if this question allows all real numbers, or just integers, so I'll go with integers.

 

Ok, let's start.

 

this is an isosceles triangle, therefore either \(AB=AC\) or \(BC=AB\) or \(BC=AC\) or, so on.

 

That could be a large problem testing out all different values of x frown

 

To spare all the fuss, I've found that \(X=8\) works well, and is the smallest value you can achieve that is an integer. 

 

So \(3x+62 = 24+62\) and \(7x+30 = 56+30\), which are both 86! (We used sides A and B here).

 

Judging by this, we can calculate the perimiter by replacing x in all equations with 8. So it is:

 

86+86+100

 

or, \(\fbox{272}\)

 Jul 18, 2021
 #2
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+2

3x + 62 = 5x + 60

2x = 2

x = 1

AB = 65

AC = 65

BC = 37

Perimeter = 167

Guest Jul 19, 2021

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