Odpowiedź:
a. | KL| = 3 - (-5) = 3 + 5 = 8
b. | PR | = 7 - (-4) = 7 + 4 = 11
c. | AB | =5 - (-1) = 5 + 1 = 6
| CD| = 4 - ( - 1) = 4 + 1 = 5
[tex]P_{ABC} = \frac{6 ( :2= 3) * 5 }{2(:2=1)} = 3*5 = 15 [j^{2} }[/tex]
d. | EF | = 1 - (-2) = 1 + 2 = 3
| HG | = 3 - ( - 4)= 3 + 4 = 7
[tex]P_{EFG }=\frac{3*7}{2}= \frac{21}{2} = 10\frac{1}{2} [j^{2} ][/tex]
