III gimnazjum

Matematyka wokół nas 3

Siatka walca wygląda następująco: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/23131/bNxgv686jhC3UrAA7LzJNg%3D%3D_5327597832b8e68fb3ef064575db2006daec7822f2099f19a59b87db0bc0e56f.png" width="343" height="363"> a)

Prostokąt w skali 1:2 ma wymiary 2 cm × 4 cm.

Długosˊcˊ boku kwadratu w skali 1:3 wynosi 4 cm.

Z warunkoˊw zadania widzimy, z˙e prostokąt ma dwa boki - jeden o długosˊci 43<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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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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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>​ cm, a drugi o długosˊci 4⋅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>​⋅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>​=12 cm. Wynika to z tego, z˙e przekątna prostokąta tworzy troˊjkąt prostokątny o kątach 30o i 60o z dwoma bokami tego prostokąta.

Boki prostokąta powierzchni bocznej walca wynoszą następująco:

Bok powierzchni bocznej (czyli bok kwadratu) z wzoru na obwoˊd koła (podstawy walca) będzie roˊwny 2⋅π⋅4=8π cm.

Bok kwadratu (oznaczmy przez a) o polu 1,44 m2 wynosi a=1,44 m2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.28em' viewBox='0 0 400000 1296' preserveAspectRatio='xMinYMin slice'><path d='M263,681c0.7,0,18,39.7,52,119
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c340,-704.7,510.7,-1060.3,512,-1067
l0 -0
c4.7,-7.3,11,-11,19,-11
H40000v40H1012.3
s-271.3,567,-271.3,567c-38.7,80.7,-84,175,-136,283c-52,108,-89.167,185.3,-111.5,232
c-22.3,46.7,-33.8,70.3,-34.5,71c-4.7,4.7,-12.3,7,-23,7s-12,-1,-12,-1
s-109,-253,-109,-253c-72.7,-168,-109.3,-252,-110,-252c-10.7,8,-22,16.7,-34,26
c-22,17.3,-33.3,26,-34,26s-26,-26,-26,-26s76,-59,76,-59s76,-60,76,-60z
M1001 80h400000v40h-400000z'/></svg>​=1,2 m. &nbsp; Oznacza to, z˙e obwoˊd koła będący podstawą tego walca tez˙ wynosi 1,2 m.

Obwoˊd podstawy wynosi 10π cm.

Przekroˊj osiowy walca jest kwadratem o polu 256 cm2. Bok kwadratu (wysokosˊcˊ walca oraz sˊrednica podstawy walca) jest więc roˊwny a=256<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>​=16 cm.

Z warunkoˊw zadania widzimy, z˙e prostokąt będący przekrojem osiowym walca ma dwa boki - 8 cm i 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
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.

Pole podstawy walca wynosi 36π cm2, zatem promienˊ podstawy walca wynosi r=π36π​<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>​=6 cm.

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