Question 239329
The perimeter of an isosceles triangle is equal to 2*s + b where s equals one of the equal sides of the triangle and b equals the base.


If the base is equal to one of the sides, then you have an equilateral triangle.


The perimeter of this triangle has to be greater than 21 units and is not a prime number.


Possible perimeters are therefore:


22,23,24,25,26,27,28,29,30,...


The length of each side of the triangle has to be a prime number.   


I presume this includes the base of the triangle as well as the 2 equal sides.


Since 7*3 = 21, one of the sides of this triangle has to be greater than 7 at least.


We'll start with the next highest prime number which is 11.


If 11 is the base of the triangle, then the smallest side could be 7 which would make the total perimeter equal to 14 + 11 = 25.


25 is not a prime.


25 looks good.


If 11 was a side of the triangle, then the perimeter of the triangle could be 22 + 2 = 24


24 is not a prime.


24 looks good.


None of the other combinations produce a perimeter less than 24.


23 would be no good because it's a prime number.


The smallest possible perimeter that meets the requirements is 22 but no combinations I was able to find yielded a total of 22.


I'd go with 24 which is a combination of 2 times 11 plus 2.