Odpowiedź:
Twierdzenie Pitagorasa
[tex] {a}^{2} + {b}^{2} = {c}^{2} [/tex]
Wzór na pole równoległoboku
P=a*h
a)
[tex] {2}^{2} + {h}^{2} = {3}^{2} \\ 4 + {h}^{2} = 9 \\ {h}^{2} = 9 - 4 \\ {h}^{2} = 5 \: \: \: \: \: | \sqrt{} \\ h = \sqrt{5} [/tex]
2+5=7
P=[tex]7 \times \sqrt{5} = 7 \sqrt{5} [/tex]
b)
[tex] {3}^{2} + {h}^{2} = {4}^{2} \\ 9 + {h}^{2} = 16 \\ {h}^{2} = 16 - 9 \\ {h}^{2} = 7 \: \: \: \: \: | \sqrt{} \\ h = \sqrt{7} [/tex]
3+4=7
P=[tex]7 \times \sqrt{7} = 7 \sqrt{7} [/tex]
