III gimnazjum

Matematyka wokół nas 3

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/25339/5jVNOb6gvOd1ahWqAeBVwQ%3D%3D_ff453f2128dd262f959d5f18d62ab7b553bc7233d0d38fe1b0c49ca8793b1781.jpg" width="456" height="232"> PABCD​=21​(12+24)⋅h PABCD​=21​(12+24)⋅h PABCD​=18h Figury AEHD, EFGH, FBCG mają równe pole, zatem każda z nich ma pole równe: 18h:3=6h PEFGH​=∣EF∣⋅h=6h |EF|=6 |AE|=|FB|=(24-6):2=9 6:9:6=2:3:2 Odp. Dłuższą podstawę należy podzielić w stosunku 2:3:2.

Rozwiązanie 1

a) 16:8=2     42​=x8​

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c-2,6,-10,9,-24,9
c-8,0,-12,-0.7,-12,-2z M1001 80
h400000v40h-400000z'/></svg>​=2a3<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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M834 80h400000v40h-400000z'/></svg>​​ h=2a3<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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l0 -0
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k=0,6 P1​+P2​=68cm2 P2​P1​​=0,62=0,36⇒P1​=10036​P2​

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/24968/iV4KEV0kH3NvPzN3Kc899Q%3D%3D_19c9c6cd24cc23cc015573c1517384f2b0fe7f7882a082744dfd2f01d7cea637.jpg" width="294" height="243"> ∣AC∣=172−152<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
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M1001 80h400000v40h-400000z'/></svg>​=289−225<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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l0 -0
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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>​=64<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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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>​=8cm

Ob1​=2⋅100m+2⋅80m=200m+160m=360m

Wiemy, że obwód prostokątnej działki na planie w skali 1:500 jest równy 104 cm Wprowadźmy oznaczenia: a  -długość jednego boku tej działki (w rzeczywistości) b  -długość drugiego boku tej działki (w rzeczywistości) Długość obwodu tej działki na planie jest 500 razy mniejsza niż w rzeczywistości, zatem rzeczywistą długość obwodu tej działki możemy zapisać za pomocą następującego wyrażenia algebraicznego: 2a+2b=104⋅500 [cm] 2a+2b=52000 [cm]

<img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/24969/1XKO8RsFj3hIFUKy9hcfWg%3D%3D_d57f0fe6c2f4f09b9f4ad1078fb97144afb05c95cb6e285f4f151ca638e6a1ea.jpg" width="393" height="264"> h=∣FE∣=52−(228−20​)2<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='3.08em' viewBox='0 0 400000 3240' preserveAspectRatio='xMinYMin slice'><path d='M473,2793
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606zM1001 80h400000v40H1017.7z'/></svg>​=25−16<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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M834 80h400000v40h-400000z'/></svg>​=9<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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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>​=3cm

∇ ABC ~ ∇ DBE          k = 2               (bbb) ∣DB∣=21​∣AB∣

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