Odpowiedź:
a = 12
H = 10
a)
[tex]Pp = \frac{a^{2}\sqrt{3} }{4} = \frac{10^{2}*\sqrt{3} }{4} = \frac{100\sqrt{3} }{4} = 25\frac{\sqrt{3} }{4} = okolo 10.81 cm^{2}[/tex]
b) [tex]V = \frac{1}{3} * Pp * H = \frac{1}{3} * 25 \frac{\sqrt{3} }{4} * 10 = \frac{1000\sqrt{3} }{12} = \frac{250\sqrt{3} }{3}[/tex]
c) sin(60) = [tex]\frac{12}{x}[/tex]
[tex]\frac{\sqrt{3} }{2} = \frac{12}{x}[/tex]
x = [tex]\frac{24}{\sqrt{3} }[/tex] - przekatna podstawy
y = [tex]\frac{1}{3}x = \frac{1}{3} * \frac{24}{\sqrt{3} } = \frac{8}{\sqrt{3} }[/tex]
[tex]b^{2} = y^{2} + H^{2}\\b^{2} = (\frac{8}{\sqrt{2} }) ^{2} + 10^{2}\\b^{2} = \frac{64}{2} + 100\\b^{2} = 32+100 = 132\\b = \sqrt{132} = \sqrt{33*4} = 2\sqrt{33}\\[/tex]
d)
[tex]Pc = Pp + Pb\\Pp = 25\frac{\sqrt{3}}{4}\\Pb = 3*\frac{1}{2} a*h \\Pb = 3 * \frac{1}{2} * 12 * 2\sqrt{33} = 18 * 2\sqrt{33} = 36\sqrt{33}[/tex]
Szczegółowe wyjaśnienie:
