Odpowiedź:
RYSUNEK 1
P1 = 2×2=4
P2 = 1/2 × 2 × 3 = 3
Pc = 4+4×3=4+12=16
RYSUNEK 2
P1 = 2×4 = 8
P2 = 1/2 × 4 × 4 = 8
P3 = 1/2 × 2 × 4 = 4
P4 =
[tex] \frac{ {a}^{2} \sqrt{3} }{4} = \frac{ {4}^{2} \sqrt{3} }{4} = 4 \sqrt{3} [/tex]
Pc =
[tex]8 + 8 + 2 \times 4 + 4 \sqrt{3} = 24 + 4 \sqrt{3} [/tex]
RYSUNEK 3
P1 = 1/2 × 3 × 3 =9/2
liczę bok trójkąta P2 z twierdzenia Pitagorasa na podstawie trójkąta P1
[tex] {x}^{2} = {3}^{2} + {3}^{2} \\ {x}^{2} = 18 \\ x = \sqrt{18} = \sqrt{9 \times 2} = 3 \sqrt{2} [/tex]
P2 =
[tex] \frac{ {(3 \sqrt{2}) }^{2} \sqrt{3} }{4} = \frac{18 \sqrt{3} }{4} = \frac{9 \sqrt{3} }{2} [/tex]
Pc =
[tex]3 \times \frac{9}{2} + \frac{9 \sqrt{3} }{2} = \frac{27}{2} + \frac{9 \sqrt{ 3} }{2} = \frac{27 + 9 \sqrt{3} }{2} = \frac{9(3 + \sqrt{ 3}) }{2} [/tex]
