Question

the lengths of the sides of triangle is integer and perimeter also integer,one length of the side is 21, triangle perimeter is 48 the ln find out smallest side of triangle​

Answer 1

Step-by-step explanation:

a + b + c = 48

Let a = 21,

b = 48 -a-c = 48 -21-c = 27 -c

b-3 = 24-c.

Area A = √ { s(s-a)(s-b)(s-c) } &

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.

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