Odpowiedź :
2.
a)
[tex] {h}^{2} + {2}^{2} = {3}^{2} \\ {h}^{2} + 4 = 9 \\ {h}^{2} = 5 \\ h = \sqrt{5} [/tex]
[tex]P = 7 \times \sqrt{5} = 7 \sqrt{5} [/tex]
b)
[tex] {h}^{2} + {3}^{2} = {4}^{2} \\ {h}^{2} + 9 = 16 \\ {h}^{2} = 7 \\ h = \sqrt{7} [/tex]
[tex]P = 7 \times \sqrt{7} = 7 \sqrt{7} [/tex]
3.
a)
[tex] {x}^{2} + {6}^{2} = {10}^{2} \\ {x}^{2} + 36 = 100 \\ {x}^{2} = 64 \\ x = \sqrt{64} \\ x = 8[/tex]
[tex]P = 6 \times 8 = 48[/tex]
b)
[tex]x = 5 \sqrt{2} [/tex]
[tex]P = 5 \sqrt{2} \times 5 = 25 \sqrt{2} [/tex]
4.
a)
[tex] {a}^{2} = {8}^{2} + {3}^{2} \\ {a}^{2} = 64 + 9 \\ {a}^{2} = 73 \\ a = \sqrt{73} cm[/tex]
Obwód
[tex]4 \times \sqrt{73} = 4 \sqrt{73} cm[/tex]
b)
[tex] {x}^{2} + (3x {)}^{2} = {20}^{2} \\ {x}^{2} + 9{x}^{2} = {20}^{2} \\ 10 {x}^{2} = 400 | \div 10 \\ {x}^{2} = 40 \\ x = \sqrt{40} \\ x = \sqrt{4 \times 10} \\ x = 2 \sqrt{10} cm[/tex]
[tex]3x = 3 \times 2 \sqrt{10} = 6 \sqrt{10} cm[/tex]
Obwód
[tex]2(x + 3x) = 2(2 \sqrt{10} + 6 \sqrt{10} ) = 2 \times 8 \sqrt{10} = 16 \sqrt{10} cm[/tex]
Odpowiedź:
Zad. 2
a) 2² + h² = 3²
4 + h² = 9
h² = 5
h = pierwiastek z 5
pole:
5* pierwiastek z 5 = 5 pierwiastków z 5
b) h = 5, ponieważ trójka pitagorejska
pole: 4 * 5 = 20
zad. 3
a) 6² + x² = 10²
36 + x² = 100
x² = 64
x = 8
pole: 8 * 6 = 48
b) nie podam Ci odpowiedzi, ponieważ nie jestem pewny
zad. 4
a) 8² + 3² = x²
64 + 9 = x²
x² = 73
x = pierwiastek z 73
obwód: pierwiastek z 73 * 4 = pierwiastek z 292
b)
obwód: 20 + 20 pierwiastków z 3