Odpowiedź:
A = (- 2 , - 1 ) , B = (4 , 2 ) , C = (- 1 , 4 )
xa = - 2 , xb = 4 , xc = - 1 , ya = - 1 , yb = 2 , yc = 4
IABI = √[(xb - xa)² + (yb - ya)²] = √[(4 + 2)² + (2 + 1)²] = √(6² + 3²) =
= √(36 + 9) = √45 = √(9 * 5) = 3√5 [j] - bok najdłuższy
IACI = √[(xc - xa)² + (yc - ya)²] = √[( - 1 + 2)² + (4 + 1)²] = √(1² + 5²) =
= √(1 + 25) = √26 [j]
IBCI = √[xc - xb)² + (yc - yb)²] = √[(- 1 - 4)² + (4 - 2)²] = √[(- 5)² + 2²] =
= √(25 + 4) = √29 [j]
Odp: C
