Question

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?​

Answer 1

Answer:

a + b + c = 48

Let a = 21,

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

b-3 = 24-c.

$\bold{Area A = √ { s(s-a)(s-b)(s-c) }}$

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.a + b + c = 48