Odpowiedź:
a) P = (/3 + /2):2
obw = /3 + /2 + /5
/ to pierwiastek
b)
[tex]{(2 \sqrt{3})}^{2} + {2}^{2} = {x}^{2} \\ 12 + 4 = {x}^{2} \\ x = \sqrt{16} = 4[/tex]
[tex]p = \frac{a \times h}{2} \\ p = 2 \times 2 \sqrt{3} \times \frac{1}{2} \\ p = 2 \sqrt{3} [/tex]
[tex]obw = 6 + 2 \sqrt{3} [/tex]
c)
[tex] {(5 \sqrt{2} )}^{2} + {(5 \sqrt{2)} }^{2} = {x}^{2} \\ 50 + 50 = {x}^{2} \\ x = \sqrt{100} \\ x = 10[/tex]
[tex]p = \frac{5 \sqrt{2} \times 5 \sqrt{2} }{2} = 50 : 2 = 25
[tex] obw = 10 + 10 \sqrt{2} [/tex]
proszę o niezwracanie uwagi na te litery i znaki w podpunkcie c (</p> <p> i [tex])
nie wiem jak je usunąć
