1.
Odp.D
2.
[tex] {6}^{2} + {x}^{2} = {10}^{2} \\ 36 + {x}^{2} = 100 \\ {x}^{2} = 64 \\ x = 8cm[/tex]
Odp.B
3.
Wysokość podstawy
[tex] \frac{6 \sqrt{3} }{2} = 3 \sqrt{3} cm[/tex]
[tex] \frac{1}{3} \times 3 \sqrt{3} = \sqrt{3} cm[/tex]
Czyli
[tex] {2}^{2} + ( \sqrt{3} {)}^{2} = {x}^{2} \\ 4 + 3 = {x}^{2} \\ x = \sqrt{7} cm[/tex]
Odp.B
