Hejka
zad.1
T1 - bok c = 25, bok a = 24
bok b = [tex]\sqrt{25^2 -24^2 }=\sqrt{49} =7[/tex]
T2 - bok f = 25, bok d = 15
bok e = [tex]\sqrt{25^2 - 15^2 } = \sqrt{400} = 20[/tex]
P1 = 24*7/2 = 12*7 = 84
P2 = 15*20/2 = 150
P2 > P1
O1 = 24+ 7 +25 = 56
02 = 15 + 20 + 25 = 60
01 < 02
Odp : F , P
zad.2
E = 1dm
f = [tex]\sqrt{3}[/tex]dm
bok rombu = [tex]\sqrt{(\frac{e}{2})^2 +(\frac{f}{2})^2}= \sqrt{\frac{1+3}{4} } =\sqrt{1} =1[/tex]dm
Odp: O = 4dm, bo bok = 1dm
Pozdrawiam, miłego dnia ;)))
