III gimnazjum

Policzmy to razem 3

Sześcian: V=a3 V=(9cm)3=729cm3   Pc=6⋅a2

Rozwiązanie 1

Obliczamy długość krawędzie modelu sześciennego bloku: Pc=8,64cm2 Pc=6a2 6a2=8,64cm2

Sprowadzamy jego wymiary do jednakowej jednostki: a=3dm=30cm b=12cm

Wyznaczamy wzór na długość przekątnej sześcianu. <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/43884/TUi%2Bl/IAOAGrHPsrLnc61g%3D%3D_1f8cdc685aaf95601e04714dd7c655b9a2f7650733429ff9eb90342617afc104.png"> Z twierdzenia Pitagorasa: a2+(2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
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c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
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c69,-144,104.5,-217.7,106.5,-221
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c5.3,-9.3,12,-14,20,-14
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
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c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​ x=3<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​a

a)  Pp=150cm2 H=24cm V=Pp⋅H V=150cm2⋅24cm=3600cm2​​

a) V=31​Pp⋅H Aby prawidłowo obliczyć objętość, musimy wyrazić pole podstawy i wysokość w takich samych jednostkach: Pp=300cm2 H=0,2m=20cm V=31​⋅300cm2⋅20cm=2000cm3​​

Nanosimy podane wymiary wycinków na rysunek i odczytujemy z niego wymiary powstałej siatki prostopadłościanu.  <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/43896/bPP5s5gd/1T24UYeAwAmcQ%3D%3D_6249f479b75bf699f84e502a7a7c1543a95262af3407992621d4e66d022117d7.jpg"> Prostopadłościan którego siatka powstała ma wymiary: a=12cm