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GEOMETRIE-13
Suport teoretic:
Piramida patrulatera regulata, aria unei sectiuni paralela cu baza, volumul unui poliedru, distanta dintre doua plane.
Enunt:
Fie piramida patrulatera regulata [VABCD], in care baza are lungimea egala cu a, iar
inaltimea h. Se sectioneaza piramida cu doua plane paralele cu baza, astfel incat
volumele celor trei poliedre obtinute sa fie egale.
Sa se afle ariile sectiunilor si distanta dintre planele acestora.
Raspuns:
![\frac{{a^2}\sqrt[3]{3}}{3},\;\frac{{a^2}\sqrt[3]{12}}{3};\;{\frac{{h}\sqrt[3]{9}}{3}}\cdot{(\sqrt[3]{2}-1)}.](http://s.wordpress.com/latex.php?latex=%5Cfrac%7B%7Ba%5E2%7D%5Csqrt%5B3%5D%7B3%7D%7D%7B3%7D%2C%5C%3B%5Cfrac%7B%7Ba%5E2%7D%5Csqrt%5B3%5D%7B12%7D%7D%7B3%7D%3B%5C%3B%7B%5Cfrac%7B%7Bh%7D%5Csqrt%5B3%5D%7B9%7D%7D%7B3%7D%7D%5Ccdot%7B%28%5Csqrt%5B3%5D%7B2%7D-1%29%7D.&bg=FFFFFF&fg=000000&s=0 "\frac{{a^2}\sqrt[3]{3}}{3},\;\frac{{a^2}\sqrt[3]{12}}{3};\;{\frac{{h}\sqrt[3]{9}}{3}}\cdot{(\sqrt[3]{2}-1)}.")
Rezolvare:
Se formeaza astfel inca doua piramide, sau din alta perspectiva, o piramida si doua
trunchiuri de piramida. Conform teoremei potrivit careia raportul volumelor a doua
piramide asemenea este egal cu raportul cuburilor inaltimilor (apotemelor, laturilor
bazelor) se obtine inaltimea
![h_1=\frac{h}{\sqrt[3]{3}}](http://s.wordpress.com/latex.php?latex=h_1%3D%5Cfrac%7Bh%7D%7B%5Csqrt%5B3%5D%7B3%7D%7D&bg=FFFFFF&fg=000000&s=0 "h_1=\frac{h}{\sqrt[3]{3}}")
si baza piramidei mici:
![a_1=\frac{a}{\sqrt[3]{3}}.](http://s.wordpress.com/latex.php?latex=a_1%3D%5Cfrac%7Ba%7D%7B%5Csqrt%5B3%5D%7B3%7D%7D.&bg=FFFFFF&fg=000000&s=0 "a_1=\frac{a}{\sqrt[3]{3}}.")
Deducem aria sectiunii superioare (baza piramidei mici):
![{\mathcal{A}}_1={a_1}^2=\frac{a^2}{\sqrt[3]{9}}.](http://s.wordpress.com/latex.php?latex=%7B%5Cmathcal%7BA%7D%7D_1%3D%7Ba_1%7D%5E2%3D%5Cfrac%7Ba%5E2%7D%7B%5Csqrt%5B3%5D%7B9%7D%7D.&bg=FFFFFF&fg=000000&s=0 "{\mathcal{A}}_1={a_1}^2=\frac{a^2}{\sqrt[3]{9}}.")
La fel pentru piramida intermediara:
![h_2=h\sqrt[3]{\frac{2}{3}},](http://s.wordpress.com/latex.php?latex=h_2%3Dh%5Csqrt%5B3%5D%7B%5Cfrac%7B2%7D%7B3%7D%7D%2C&bg=FFFFFF&fg=000000&s=0 "h_2=h\sqrt[3]{\frac{2}{3}},")
![{a_2}=a\sqrt[3]{\frac{2}{3}},](http://s.wordpress.com/latex.php?latex=%7Ba_2%7D%3Da%5Csqrt%5B3%5D%7B%5Cfrac%7B2%7D%7B3%7D%7D%2C&bg=FFFFFF&fg=000000&s=0 "{a_2}=a\sqrt[3]{\frac{2}{3}},")
![{\mathcal{A}}_2={a^2}{\sqrt[3]{\frac{4}{9}}}.](http://s.wordpress.com/latex.php?latex=%7B%5Cmathcal%7BA%7D%7D_2%3D%7Ba%5E2%7D%7B%5Csqrt%5B3%5D%7B%5Cfrac%7B4%7D%7B9%7D%7D%7D.&bg=FFFFFF&fg=000000&s=0 "{\mathcal{A}}_2={a^2}{\sqrt[3]{\frac{4}{9}}}.")
In final, distanta dintre cele doua plane de sectiune este egala cu:
![h_2-h_1=...={\frac{h}{\sqrt[3]{3}}}\cdot{(\sqrt[3]{2}-1)}={\frac{{h}\sqrt[3]{9}}{3}}\cdot{(\sqrt[3]{2}-1)}.](http://s.wordpress.com/latex.php?latex=h_2-h_1%3D...%3D%7B%5Cfrac%7Bh%7D%7B%5Csqrt%5B3%5D%7B3%7D%7D%7D%5Ccdot%7B%28%5Csqrt%5B3%5D%7B2%7D-1%29%7D%3D%7B%5Cfrac%7B%7Bh%7D%5Csqrt%5B3%5D%7B9%7D%7D%7B3%7D%7D%5Ccdot%7B%28%5Csqrt%5B3%5D%7B2%7D-1%29%7D.&bg=FFFFFF&fg=000000&s=0 "h_2-h_1=...={\frac{h}{\sqrt[3]{3}}}\cdot{(\sqrt[3]{2}-1)}={\frac{{h}\sqrt[3]{9}}{3}}\cdot{(\sqrt[3]{2}-1)}.")
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