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AB + AC + BC = 48 AB + (AB +9) + (AB + 9 + 3) = 48

Solve and AB = 9

So AB = 9, AC = 18 and BC = 21

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Q: In triangle ABC side AB is 9 cm shorter than side AC while bc is 3cm longer than side AC if the perimeter is 48 cm find the lenghts of the three sides?
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What is the possible value if perimeter of a triangle is 42 cm and it's largest side is a fundamental number?

I am not sure what you mean by a "fundamental" number (I've never heard of that term being used with reference to the numbers themselves); I guess you mean an "integer". For a triangle to exist the shorter two sides must be longer than the longest side. Thus there is an upper limit to the length of the longest side of a triangle. For a given perimeter, the longest side must be less than half the perimeter. For a perimeter of 42cm this means that the longest side is less than 42 cm ÷ 2 = 21 cm. If we focus on the longest side of a triangle, as it becomes shorter, one or both of the other two sides must increase in length, they can equal but never be longer than this longest side. Thus there is also a lower limit below which the longest side cannot be; this is when all three sides are equal and the triangle is an equilateral triangle. For a perimeter of 42cm the longest side is greater than or equal to 42 cm ÷ 3 = 14 cm So with a perimeter of 42 cm we have: 14 cm ≤ longest side < 21 cm Which means for an integer length, the longest side can be 14 cm, 15 cm, 16 cm, 17 cm, 18 cm, 19 cm or 20 cm.

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