The lengths of the sides of a triangle are integrals, and its area is also integer. One side is 21 and the perimeter is 48. Find the shortest side.

A = √ s(s-a)(s-b)(s-c)

If the area is an integer

Then [s(s-a)(s-b)(s-c)] should be proper square

If s = Then s = = 24

Hence ;

A = √ 24(24-a)(24-b)(24-c)

If side of triangle are

a = 21 and b + c = 27

let c be smallest side

then b = 27-c

∴ √ 24(24-21)(24-27 + c)(24-c)

⇒ √ 24 × 3 × (c-3)(24-c)

⇒ √ 2 × 2 × 2 × 3 × 3 × (c-3)(24-c)

⇒ 2 × 3√2(c-3)(24-c)

⇒ 6√2(c-3)(24-c)

∴ the value of [2(c-3)(24-c)] must be a perfect square for area to be a integer

For getting square 2(c-3) should be equal to (24-c)

2(c-3) = (24-c)

2c-6 = 24-c

2c + c = 24 + 6

3c = 10

c = 10; b = 27-c = 27-10 = 17

Hence the size of smallest size is 10.