oblicz dlugosci odcinkow oznaczonych literami.
twierdzenie pitagorasa
a^2+b^2=c^2
a) 4^{2}+6^{2}=a^{2} \\
16+36=a^{2}\\
52=a^{2}\\
a=\sqrt{52} \\a^{2}=2\sqrt{13}\\
b) 3\sqrt{3} +b^{2}=3\sqrt{5} \\
27+b^{2}=45\\
b^{2}=18\\
b=\sqrt{18}\\
b=3\sqrt{2}\\
c) 3^{2}+h^{2}=5^{2}\\
9+h^{2}=25\\
h^{2}=16\\
h=4 \\
4^{2}+2^{2}=c^{2}\\
16+4=c^{2}\\
20=c^{2}\\
c=2\sqrt{5}
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a^2+b^2=c^2
a) 4^{2}+6^{2}=a^{2} \\
16+36=a^{2}\\
52=a^{2}\\
a=\sqrt{52} \\
a^{2}=2\sqrt{13}\\
b) 3\sqrt{3} +b^{2}=3\sqrt{5} \\
27+b^{2}=45\\
b^{2}=18\\
b=\sqrt{18}\\
b=3\sqrt{2}\\
c) 3^{2}+h^{2}=5^{2}\\
9+h^{2}=25\\
h^{2}=16\\
h=4 \\
4^{2}+2^{2}=c^{2}\\
16+4=c^{2}\\
20=c^{2}\\
c=2\sqrt{5}