The length of the sides of a triangle are integers, and its area is also integers. one side is 21 and the perimeter is 48. find the shortest side?
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The length of the sides of a triangle are integers, and its area is also integers. one side is 21 and the perimeter is 48. find the shortest side?
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Answer:
a + b + c = 48
Let a = 21,
b = 48 -a-c = 48 -21-c = 27 -c
b-3 = 24-c.
[tex] \bold{Area A = √ { s(s-a)(s-b)(s-c) }} [/tex]
s =(a+b+c)/2 =24
Area A = √ { 24(24–21)(24-b)(24-c)}
Area A = √ { 24(24–21)(24-b)(b-3).
Area A = √ { 72(24-b)(b-3)}
Area A =6 √ { 2(24-b)(b-3)}
For A to be an integer, 24 -b must be = 2(b-3)
Or b = 10.
The other possibility is : -
2(24 -b) = ( b-3) and in this case , we get b 17.
If b = 10 ,
Area A = 6×14 = 84 unit of area.
°•° a= 21, b =10, c = 17 units of length.a + b + c = 48