III gimnazjum

Matematyka wokół nas 3

Pole powierzchni kuli wynosi Pp​=4πr2. Mamy zatem:

Rozwiązanie 1

Promień największego koła i zarazem promień kuli wynosi: 2π40π​ dm=20 dm

Pole przekroju wynosi 128π cm2. Z tego wynika, z˙e promienˊ tego przekroju jest roˊwny: r=128ππ​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
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l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​=128<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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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>​=82<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>​ cm

Policzmy promień kuli z wzoru na pole powierzchni: Pp​=4πr2

Promień koła wyznaczymy z objętości: V=34​πr3 r3=43​πV​=43​(288ππ​)=216

Policzmy promienie obydwu kul: r1​=4π12​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
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s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​=π3​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
c-1.3,-0.7,-38.5,-172,-111.5,-514c-73,-342,-109.8,-513.3,-110.5,-514
c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20
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s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​=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
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c69,-144,104.5,-217.7,106.5,-221
l0 -0
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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>​⋅π1​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
c-1.3,-0.7,-38.5,-172,-111.5,-514c-73,-342,-109.8,-513.3,-110.5,-514
c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20
s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13,-13s76,-122,76,-122s77,-121,77,-121
s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​ dm r2​=4π768​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
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c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20
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s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​=π192​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
c-1.3,-0.7,-38.5,-172,-111.5,-514c-73,-342,-109.8,-513.3,-110.5,-514
c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20
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s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​=83<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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l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​⋅π1​<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='2.48em' viewBox='0 0 400000 2592' preserveAspectRatio='xMinYMin slice'><path d='M424,2478
c-1.3,-0.7,-38.5,-172,-111.5,-514c-73,-342,-109.8,-513.3,-110.5,-514
c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5,25c-5.7,9.3,-9.8,16,-12.5,20
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s209,968,209,968c0,-2,84.7,-361.7,254,-1079c169.3,-717.3,254.7,-1077.7,256,-1081
l0 -0c4,-6.7,10,-10,18,-10 H400000
v40H1014.6
s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185
c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​ cm

Wystarczy przyroˊwnacˊ wzoˊr na pole powierzchni kuli i objętosˊcˊ kuli: Pp​=V

Objętosˊcˊ pomaranˊczy wynosi: Vp​=34​⋅π⋅43≈3256​π

Objętość kropli deszczu (o kształcie kuli) wynosi: Vk​=34​π⋅13=34​π mm3

Promienˊ koła wielkiego o obwodzie 24π cm wynosi: r=2π24π​=12 cm

Planeta Jowisz (r = 71 400 km): P1​=4πr2=4π⋅(71 400)2=4π⋅(714)2⋅(100)2 =2 039 184π⋅(100)2 km2

Promienˊ koła wpisanego w przekroˊj osiowy stoz˙ka wynosi 23<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
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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>​ cm. Z tego wynika, z˙e wysokosˊcˊ troˊjkąta roˊwnobocznego (i stoz˙ka) wynosi 63<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>​ cm, a bok troˊjkąta przekroju osiowego (i sˊrednica oraz tworząca) stoz˙ka 12 cm.