The lengths of the sides of the triangle are integrals and its area is also integer one side is 21 and the perimeter is 48 find the shortest side
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Answer:
sorry............................
A = √(24(24 - 21)(24 - b)(24 - c))
A = √(2332(24 - b)(24 - c))
A = 6√(2(24 - b)(24 - c))
Let's see if we can figure out what b and c would have to be from here:
>b and c have to be integers
>b and c have to add to 27 (because the perimeter is 48)
>2(24 - b)(24 - c) must be a perfect square (because A has to be an integer)
Since we're not working with very big numbers, guessing and checking isn't a bad idea.
I can't make b 0, 1, 2, or 3 or 24, 25, 26 or 27 because then c would be 27, 26, 25, or 24 or 3, 2, 1, or 0.
These choices of b and c would either give us a negative under the root or a non-triangle.
If we choose b to be 4 or 23 then we'll get 2(24 - 4)(24 - 23) = 2(4)(1) = 8 under the root. This isn't a perfect square.
If we choose b to be 5 or 22 then we'll get 2(24 - 5)(24 - 22) = 2(19)(2) = 76 under the root. This isn't a perfect square either.
If we choose b to be 6 or 21 then we'll get 2(24 - 6)(24 - 21) = 2(18)(3) = 108 under the root. Doesn't work.
If we choose b to be 7 or 20 then we'll get 2(24 - 7)(24 - 20) = 2(17)(4) = 136 under the root. Doesn't work.
If we choose b to be 8 or 19 we'll get 2(24 - 8)(24 - 19) = 2(16)(5) = 160 under the root. Not a square.
If we choose b to be 9 or 18 we'll get 2(24 - 9)(24 - 18) = 2(15)(6) = not a square because only one factor of 5 (I'm going to stop multiplying them out if I don't have to).
If we choose b to be 10 or 17 we get 2(24 - 10)(24 - 17) = 2(14)(7) = 14(14) and when we get to this point, we say "yay" because this is a number that we can take the square root of and get an integer.
Unless there are other choices for sides that work (and you can check the handful remaining doing the same thing), the sides of our triangle will be 21, 17, and 10.
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I'm going to see if there isn't an easier way to phrase this that allows us to get this result without checking every possibility ourselves.
We need 2(24 - b)(24 - c) to be a perfect square and we need b + c = 27.
Let's see if we can do this better...
2(24 - b)(24 - c) = p2 where p is a perfect square
2(576 - 24b - 24c + bc) = p2
4(288 - 12b - 12c + bc/2) = p2
This should be equivalent to saying that the expression the parentheses itself must be a perfect square:
bc/2 - 12b - 12c + 288 = p2 I'm using the same p, but I don't care what p is, so it's fine.
We have an equation relating what the sum of b and c is, so let's rewrite this again:
bc/2 - 12(b + c) + 288 = p2
bc/2 - 12(27) + 288 = p2
bc/2 - 324 + 288 = p2
bc/2 = p2 + 36
bc = 2(p2 + 36) ...
I'm still going to end up doing a guess and check process. I'll try replacing c with 27 - b though
b(27 - b) = 2(p2 + 36)
When I consider pairs of numbers that add to give me 27, the pair with the biggest product will be closest to the middle: 13 and 14 = 182. So b(27 - b) won't exceed this number. This means that the right hand side won't either, or that p2 + 36 won't exceed 91 which means that p2 won't exceed 55, which means that p itself won't exceed 7.
Thus, I'm going to look at no more than 8 cases in which p (I'm starting to think this is no simpler...) is 0, 1, 2, 3, 4, 5, 6, or 7.
What numbers do I add to get 27 but multiply to get any of the following (plugging 0 through 7 into 2(p2 + 36):
72, 74, 80, 90, 104, 122, 144, 170
Factors of 72 are 1 and 72, 2 and 36, 3 and 24, 4 and 18, 6 and 12, 8 and 9
One pair of these add to give me 27: 3 and 24, but thinking about our triangle, that would give us as our sides: 3, 21, and 24 (which can't be a triangle because the sum of two sides should exceed the length of the third side).
Factors of 74 are 1 and 74, 2 and 37
Neither of these pairs adds to 27.
Factors of 80 are: 1 and 80, 2 and 40, 4 and 20, 5 and 16, 8 and 10
None of these pairs sum to 27.
Factors of 90 are: 1 and 90, 2 and 45, 3 and 30, 5 and 18, 6 and 15, 9 and10
None of these pairs adds to 27.
Factors of 104 are: 1 and 104, 2 and 52, 4 and 26, 8 and 13
None of these work either.
Factors of 122 are: 1 and 122, 2 and 61
Neither of these work.
Factors of 144 are: 1 and 144, 2 and 72, 3 and 48, 4 and 36, 6 and 24, 8 and 18, 9 and 16, 12 and 12
None of these work either.
Finally, factors of 170 are: 1 and 170, 2 and 85, 5 and 34, 10 and 17
Oh, look, 10 and 17 work.
The first approach was definitely better...
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The sides of your triangle have to be 10, 17, and 21.
This is probably way over complicated.
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