Odpowiedź:
to w 6 zadaniu będzie kolejno FFP
w 7 zadaniu
a) L=
[tex]10 + 7 \sqrt{2} [/tex]
P=
[tex] \frac{3 \sqrt{2} + 5 }{2} \times \sqrt{7} = \frac{3 \sqrt{14} + 5 \sqrt{7} }{2} [/tex]
b) L=
[tex] \sqrt{24} = 2 \sqrt{6} [/tex]
[tex]l = 4 + 14 \sqrt{6} [/tex]
P=
[tex]5 \sqrt{6} \times (2 \sqrt{6} + 2) = 60 + 10 \sqrt{6} [/tex]
