30.2
[tex]x=\frac{a\sqrt{2} }{2}\\ \\H^{2}=(2a)^{2}-(\frac{a\sqrt{2} }{2})^{2} \\ \\H^{2}=4a^{2}-\frac{1}{2}a^{2}\\ \\ H^{2}=\frac{7}{2}a^{2}\\ \\ H= \frac{\sqrt{7} }{\sqrt{2} }a=\frac{\sqrt{14} }{2} a[/tex]
ODP: B
30.3
R - promień koła opisanego na kwadracie
r - promień koła wpisanego w kwadrat
[tex]R=\frac{a\sqrt{2} }{2}\\ \\r=\frac{1}{2}a\\ \\P=\pi *R^{2}=\pi *(\frac{a\sqrt{2} }{2})^{2}=\frac{1}{2}a^{2}\pi \\ \\ p=\pi *r^{2}=\pi *(\frac{1}{2}a)^{2}=\frac{1}{4}a^{2}\pi \\ \\ \frac{P}{p}= \frac{\frac{1}{2}a^{2}\pi}{\frac{1}{4}a^{2}\pi} =2[/tex]
ODP: A
Odpowiedź:
odpowiedź w załączniku pozdrawiam :)
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Verified answer
30.2
[tex]x=\frac{a\sqrt{2} }{2}\\ \\H^{2}=(2a)^{2}-(\frac{a\sqrt{2} }{2})^{2} \\ \\H^{2}=4a^{2}-\frac{1}{2}a^{2}\\ \\ H^{2}=\frac{7}{2}a^{2}\\ \\ H= \frac{\sqrt{7} }{\sqrt{2} }a=\frac{\sqrt{14} }{2} a[/tex]
ODP: B
30.3
R - promień koła opisanego na kwadracie
r - promień koła wpisanego w kwadrat
[tex]R=\frac{a\sqrt{2} }{2}\\ \\r=\frac{1}{2}a\\ \\P=\pi *R^{2}=\pi *(\frac{a\sqrt{2} }{2})^{2}=\frac{1}{2}a^{2}\pi \\ \\ p=\pi *r^{2}=\pi *(\frac{1}{2}a)^{2}=\frac{1}{4}a^{2}\pi \\ \\ \frac{P}{p}= \frac{\frac{1}{2}a^{2}\pi}{\frac{1}{4}a^{2}\pi} =2[/tex]
ODP: A
Odpowiedź:
odpowiedź w załączniku pozdrawiam :)