I have a different answer to this. I find each of the information suffecient. Please hear me out
Let the sides be a, b and c.
2(ab+bc+ac) = 94
Therefore, ab+bc+ac = 47.
Now, statement 1 = lets assume ab, side is 12
therefore, bc+ac = 35
(b+a).c =35
Now either b+a = 7 and c = 5 or
b+a = 5, c = 7
If b+a= 7, possible values of abc are (5,2,5) (4,3,5) or (6,1,5). However, only 1 combination gives 1 side area as 12 which is the (4,3,5)
If b+a=5 possible values of abc are (3,2,7) ,(4,1,7) .
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Let the sides be a, b and c.
2(ab+bc+ac) = 94
Therefore, ab+bc+ac = 47.
Now, statement 1 = lets assume ab, side is 12
therefore, bc+ac = 35
(b+a).c =35
Now either b+a = 7 and c = 5 or
b+a = 5, c = 7
If b+a= 7, possible values of abc are (5,2,5) (4,3,5) or (6,1,5). However, only 1 combination gives 1 side area as 12 which is the (4,3,5)
If b+a=5 possible values of abc are (3,2,7) ,(4,1,7) .
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