pole i objętość ostrosłupa prawidłowego
pole i objętość ostrosłupa prawidłowego
Oblicz objętość i pole powierzchni całkowitej ostrosłupa prawidłowego czworokątnego, którego wysokość H na długość 8 cm, natomiast kąt nachylenia ścian bocznych do płaszczyzny podstawy ma miarę \(\displaystyle{ 45^{o}}\) .
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barakuda
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- Rejestracja: 22 paź 2009, o 19:37
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pole i objętość ostrosłupa prawidłowego
\(\displaystyle{ H=8}\)
\(\displaystyle{ tg45^o = \frac{H}{ \frac{1}{2}d }}\)
\(\displaystyle{ 1 = \frac{8}{ \frac{1}{2}d } \Rightarrow d=16}\)
\(\displaystyle{ d=a \sqrt{2} \Rightarrow 16=a \sqrt{2} \Rightarrow a=8 \sqrt{2}}\)
\(\displaystyle{ V= \frac{1}{3}a^2 \cdot H = \frac{1}{3} \cdot (8 \sqrt{2})^2 \cdot 8 = \frac{1024}{3} = 341 \frac{1}{3} \ cm^3}\)
\(\displaystyle{ sin45^o = \frac{H}{b}}\)
\(\displaystyle{ \frac{ \sqrt{2} }{2} = \frac{8}{b} \Rightarrow b=8 \sqrt{2}}\)
\(\displaystyle{ h_{b} = \sqrt{b^2 - ( \frac{1}{2}a)^2 } = \sqrt{(8 \sqrt{2})^2 + (4 \sqrt{2})^2 } = \sqrt{96} = 4 \sqrt{6}}\)
\(\displaystyle{ P_{c} = a^2 + 4 \cdot \frac{1}{2}a \cdot h_{b} = (8 \sqrt{2})^2 + 2 \cdot 8 \sqrt{2} \cdot 4 \sqrt{6} = 128+128 \sqrt{3} = 128(1+ \sqrt{3}) \ cm^2}\)
\(\displaystyle{ tg45^o = \frac{H}{ \frac{1}{2}d }}\)
\(\displaystyle{ 1 = \frac{8}{ \frac{1}{2}d } \Rightarrow d=16}\)
\(\displaystyle{ d=a \sqrt{2} \Rightarrow 16=a \sqrt{2} \Rightarrow a=8 \sqrt{2}}\)
\(\displaystyle{ V= \frac{1}{3}a^2 \cdot H = \frac{1}{3} \cdot (8 \sqrt{2})^2 \cdot 8 = \frac{1024}{3} = 341 \frac{1}{3} \ cm^3}\)
\(\displaystyle{ sin45^o = \frac{H}{b}}\)
\(\displaystyle{ \frac{ \sqrt{2} }{2} = \frac{8}{b} \Rightarrow b=8 \sqrt{2}}\)
\(\displaystyle{ h_{b} = \sqrt{b^2 - ( \frac{1}{2}a)^2 } = \sqrt{(8 \sqrt{2})^2 + (4 \sqrt{2})^2 } = \sqrt{96} = 4 \sqrt{6}}\)
\(\displaystyle{ P_{c} = a^2 + 4 \cdot \frac{1}{2}a \cdot h_{b} = (8 \sqrt{2})^2 + 2 \cdot 8 \sqrt{2} \cdot 4 \sqrt{6} = 128+128 \sqrt{3} = 128(1+ \sqrt{3}) \ cm^2}\)
pole i objętość ostrosłupa prawidłowego
ale ja mam chyba to źle.
H=8
tg \(\displaystyle{ 45^{o}}\) = \(\displaystyle{ \frac{8}{x}}\)
1=\(\displaystyle{ \frac{8}{x}}\) /*x
x=8
\(\displaystyle{ sin 45^{o}}\) = \(\displaystyle{ \frac{8}{h}}\)
\(\displaystyle{ \frac{ \sqrt{2} }{2}}\) = \(\displaystyle{ \frac{5}{h}}\) /*h
\(\displaystyle{ \frac{ \sqrt{2} }{2}}\) h = 5 /:\(\displaystyle{ \frac{ \sqrt{2} }{2}}\)
h = 8 \(\displaystyle{ \frac{ \sqrt{2} }{2}}\) = \(\displaystyle{ \frac{16}{ \sqrt{2} }}\) * \(\displaystyle{ \frac{ \sqrt{2} }{2}}\)
h= \(\displaystyle{ \frac{16 \sqrt{2} }{2}}\) = 8 \(\displaystyle{ \sqrt{2}}\)
dalej mam chyba źle... a nawet napewno
-- 18 lis 2009, o 19:36 --
\(\displaystyle{ P_{p}}\) = \(\displaystyle{ a^{2}}\)
\(\displaystyle{ P_{p}}\) = \(\displaystyle{ 16 ^{2}}\) = \(\displaystyle{ 256 cm ^{2}}\)
\(\displaystyle{ P_{sb}}\) = \(\displaystyle{ \frac{ a^{2} \sqrt{3} }{4}}\) = \(\displaystyle{ \frac{ 16^{2} \sqrt{3} }{4}}\) = \(\displaystyle{ \frac{ 256\sqrt{3} }{4}}\) = 64 \(\displaystyle{ \sqrt{3}}\)
\(\displaystyle{ P_{c}}\) = \(\displaystyle{ P_{p}}\) + 4 \(\displaystyle{ P_{sb}}\)
\(\displaystyle{ P_{c}}\) = 256 + 4 * 64 \(\displaystyle{ \sqrt{3}}\) = 256 + 256 \(\displaystyle{ \sqrt{3}}\)
V = \(\displaystyle{ \frac{1}{3}}\) \(\displaystyle{ P_{p}}\) * H = \(\displaystyle{ \frac{1}{3}}\) 256 * 8 = 682 \(\displaystyle{ \frac{2}{5}}\)
H=8
tg \(\displaystyle{ 45^{o}}\) = \(\displaystyle{ \frac{8}{x}}\)
1=\(\displaystyle{ \frac{8}{x}}\) /*x
x=8
\(\displaystyle{ sin 45^{o}}\) = \(\displaystyle{ \frac{8}{h}}\)
\(\displaystyle{ \frac{ \sqrt{2} }{2}}\) = \(\displaystyle{ \frac{5}{h}}\) /*h
\(\displaystyle{ \frac{ \sqrt{2} }{2}}\) h = 5 /:\(\displaystyle{ \frac{ \sqrt{2} }{2}}\)
h = 8 \(\displaystyle{ \frac{ \sqrt{2} }{2}}\) = \(\displaystyle{ \frac{16}{ \sqrt{2} }}\) * \(\displaystyle{ \frac{ \sqrt{2} }{2}}\)
h= \(\displaystyle{ \frac{16 \sqrt{2} }{2}}\) = 8 \(\displaystyle{ \sqrt{2}}\)
dalej mam chyba źle... a nawet napewno
-- 18 lis 2009, o 19:36 --
\(\displaystyle{ P_{p}}\) = \(\displaystyle{ a^{2}}\)
\(\displaystyle{ P_{p}}\) = \(\displaystyle{ 16 ^{2}}\) = \(\displaystyle{ 256 cm ^{2}}\)
\(\displaystyle{ P_{sb}}\) = \(\displaystyle{ \frac{ a^{2} \sqrt{3} }{4}}\) = \(\displaystyle{ \frac{ 16^{2} \sqrt{3} }{4}}\) = \(\displaystyle{ \frac{ 256\sqrt{3} }{4}}\) = 64 \(\displaystyle{ \sqrt{3}}\)
\(\displaystyle{ P_{c}}\) = \(\displaystyle{ P_{p}}\) + 4 \(\displaystyle{ P_{sb}}\)
\(\displaystyle{ P_{c}}\) = 256 + 4 * 64 \(\displaystyle{ \sqrt{3}}\) = 256 + 256 \(\displaystyle{ \sqrt{3}}\)
V = \(\displaystyle{ \frac{1}{3}}\) \(\displaystyle{ P_{p}}\) * H = \(\displaystyle{ \frac{1}{3}}\) 256 * 8 = 682 \(\displaystyle{ \frac{2}{5}}\)