Z sześcianu o krawędzi 6 cm odcięto czworościan, jak na rysunku.
Punkty A, B, C są środkami odpowiednich krawędzi sześcianu.
Oblicz ile razy pole powierzchni otrzymanej bryły jest większe od pola powierzchni odciętego czworościanu.
Wynik podaj z dokładnością do 0,1 cm kwadratowego
Wynik:około 9,9 razy.
Rysunek w załączniku.
z góry wielkie dzięki ;)
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Prawy górny wierzchołek sześcianu oznacz literką D.
![a=6cm a=6cm](https://tex.z-dn.net/?f=a%3D6cm)
![|DA|=|DB|=|DC|=3cm |DA|=|DB|=|DC|=3cm](https://tex.z-dn.net/?f=%7CDA%7C%3D%7CDB%7C%3D%7CDC%7C%3D3cm)
![|AC|^2|=|DA|^2+|DC|^2 |AC|^2|=|DA|^2+|DC|^2](https://tex.z-dn.net/?f=%7CAC%7C%5E2%7C%3D%7CDA%7C%5E2%2B%7CDC%7C%5E2)
![|AC|^2|=3^2+3^2 |AC|^2|=3^2+3^2](https://tex.z-dn.net/?f=%7CAC%7C%5E2%7C%3D3%5E2%2B3%5E2)
![|AC|^2|=2\cdot3^2 |AC|^2|=2\cdot3^2](https://tex.z-dn.net/?f=%7CAC%7C%5E2%7C%3D2%5Ccdot3%5E2)
![|AC|=3 \sqrt{2}cm |AC|=3 \sqrt{2}cm](https://tex.z-dn.net/?f=%7CAC%7C%3D3+%5Csqrt%7B2%7Dcm)
![P_1=\frac{|AC|^2 \sqrt{3}}{4}+3\cdot \frac{1}{2}|DC|\cdot|DA| P_1=\frac{|AC|^2 \sqrt{3}}{4}+3\cdot \frac{1}{2}|DC|\cdot|DA|](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7B%7CAC%7C%5E2+%5Csqrt%7B3%7D%7D%7B4%7D%2B3%5Ccdot+%5Cfrac%7B1%7D%7B2%7D%7CDC%7C%5Ccdot%7CDA%7C)
![P_1=\frac{(3 \sqrt{2})^2 \sqrt{3}}{4}+\frac{3}{2}\cdot 3\cdot 3 P_1=\frac{(3 \sqrt{2})^2 \sqrt{3}}{4}+\frac{3}{2}\cdot 3\cdot 3](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7B%283+%5Csqrt%7B2%7D%29%5E2+%5Csqrt%7B3%7D%7D%7B4%7D%2B%5Cfrac%7B3%7D%7B2%7D%5Ccdot+3%5Ccdot+3)
![P_1=\frac{18\sqrt{3}}{4}+\frac{27}{2} P_1=\frac{18\sqrt{3}}{4}+\frac{27}{2}](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7B18%5Csqrt%7B3%7D%7D%7B4%7D%2B%5Cfrac%7B27%7D%7B2%7D)
![P_1=\frac{18\sqrt{3}}{4}+\frac{54}{4} P_1=\frac{18\sqrt{3}}{4}+\frac{54}{4}](https://tex.z-dn.net/?f=P_1%3D%5Cfrac%7B18%5Csqrt%7B3%7D%7D%7B4%7D%2B%5Cfrac%7B54%7D%7B4%7D)
![P_1=21,3cm^2 P_1=21,3cm^2](https://tex.z-dn.net/?f=P_1%3D21%2C3cm%5E2)
![P_2=3a^2+3(a^2-\frac{|DA|\cdot|DC|}{2})+\frac{|AC|^2 \sqrt{3}}{4} P_2=3a^2+3(a^2-\frac{|DA|\cdot|DC|}{2})+\frac{|AC|^2 \sqrt{3}}{4}](https://tex.z-dn.net/?f=P_2%3D3a%5E2%2B3%28a%5E2-%5Cfrac%7B%7CDA%7C%5Ccdot%7CDC%7C%7D%7B2%7D%29%2B%5Cfrac%7B%7CAC%7C%5E2+%5Csqrt%7B3%7D%7D%7B4%7D)
![P_2=3\cdot 6^2+3\cdot (6^2-\frac{3\cdot3}{2})+\frac{18\sqrt{3}}{4} P_2=3\cdot 6^2+3\cdot (6^2-\frac{3\cdot3}{2})+\frac{18\sqrt{3}}{4}](https://tex.z-dn.net/?f=P_2%3D3%5Ccdot+6%5E2%2B3%5Ccdot+%286%5E2-%5Cfrac%7B3%5Ccdot3%7D%7B2%7D%29%2B%5Cfrac%7B18%5Csqrt%7B3%7D%7D%7B4%7D)
![P_2=3\cdot 36+3\cdot (36-\frac{9}{2})+\frac{9\sqrt{3}}{2} P_2=3\cdot 36+3\cdot (36-\frac{9}{2})+\frac{9\sqrt{3}}{2}](https://tex.z-dn.net/?f=P_2%3D3%5Ccdot+36%2B3%5Ccdot+%2836-%5Cfrac%7B9%7D%7B2%7D%29%2B%5Cfrac%7B9%5Csqrt%7B3%7D%7D%7B2%7D)
![P_2=108+3\cdot (36-4,5)+\frac{9\sqrt{3}}{2} P_2=108+3\cdot (36-4,5)+\frac{9\sqrt{3}}{2}](https://tex.z-dn.net/?f=P_2%3D108%2B3%5Ccdot+%2836-4%2C5%29%2B%5Cfrac%7B9%5Csqrt%7B3%7D%7D%7B2%7D)
![P_2=108+3\cdot 31,5+\frac{9\sqrt{3}}{2} P_2=108+3\cdot 31,5+\frac{9\sqrt{3}}{2}](https://tex.z-dn.net/?f=P_2%3D108%2B3%5Ccdot+31%2C5%2B%5Cfrac%7B9%5Csqrt%7B3%7D%7D%7B2%7D)
![P_2=108+94,5+\frac{9\sqrt{3}}{2} P_2=108+94,5+\frac{9\sqrt{3}}{2}](https://tex.z-dn.net/?f=P_2%3D108%2B94%2C5%2B%5Cfrac%7B9%5Csqrt%7B3%7D%7D%7B2%7D)
![P_2\approx210,3cm^2 P_2\approx210,3cm^2](https://tex.z-dn.net/?f=P_2%5Capprox210%2C3cm%5E2)
![\frac{P_2}{P_1}= \frac{202,5}{\frac{18\sqrt{3}+54}{4}} \frac{P_2}{P_1}= \frac{202,5}{\frac{18\sqrt{3}+54}{4}}](https://tex.z-dn.net/?f=%5Cfrac%7BP_2%7D%7BP_1%7D%3D+%5Cfrac%7B202%2C5%7D%7B%5Cfrac%7B18%5Csqrt%7B3%7D%2B54%7D%7B4%7D%7D)
![\frac{P_2}{P_1}\approx\frac{210,3}{21,3} \frac{P_2}{P_1}\approx\frac{210,3}{21,3}](https://tex.z-dn.net/?f=%5Cfrac%7BP_2%7D%7BP_1%7D%5Capprox%5Cfrac%7B210%2C3%7D%7B21%2C3%7D)
![\frac{P_2}{P_1}\approx9,9 \frac{P_2}{P_1}\approx9,9](https://tex.z-dn.net/?f=%5Cfrac%7BP_2%7D%7BP_1%7D%5Capprox9%2C9)
Obliczam |AC|
Obliczam pole powierzchni czworościanu
Obliczam pole powierzchni sześcianu
Obliczam stosunek pól