III gimnazjum

Liczy się matematyka 3

Matematyka

Korzystając z zależności między bokami w trójkątach o kątach 30∘,60∘,90∘ &nbsp; 45∘,45∘,90∘ &nbsp;obliczamy długości odpowiednich

Przekrój walca jest kwadratem. Oznacza to, że wysokość walca oraz średnica podstawy mają taką samą długość.&nbsp; d=H &nbsp; Pole przekroju wynosi 32. P=32 &nbsp; Obliczamy, ile wynosi

Przyjmijmy oznaczenia jak na poniższym rysunku.&nbsp; <img class="img-responsive img-thumbnail dropzone-img" src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/162006/qGeP/x2w3FSoSvZrIguruA%3D%3D_7752032c59c28e7bb541b85f2db82d0cb6303f477fb91bfbad81de3eeb808a57.jpg" width="231" height="257"> Przekrój walca jest prostokątem, którego przekątna ma długość 26 cm a jeden z boków

Mniejsze szklanki:&nbsp; Średnica podstawy tych szklanek ma długość 6 cm.&nbsp; d=6 cm &nbsp; Promień podstawy tych szklanek ma długość:&nbsp; r=21​d=21​⋅6 cm=3 cm &nbsp; Woda sięga do wysokości 8 cm, czyli: h=8 cm &nbsp; Obliczamy, ile wynosi

Przekrój osiowy stożka jest trójkątem równobocznym. Pole tego trójkąta wynosi 16√3.&nbsp; P=163<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>​ &nbsp; Obliczamy, ile wynosi długość boku (a) tego trójkąta. 163<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>​=4a23<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>​​         ∣⋅4 &nbsp; 643<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>​=a23<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>​        ∣: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
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M834 80h400000v40h-400000z'/></svg>​ &nbsp; 64=a2 &nbsp; a=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
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M834 80h400000v40h-400000z'/></svg>​=8 &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp; Tworzące stożka oraz średnica podstawy mają

Pole podstawy stożka wynosi 72π cm2. Pp​=72π cm2 &nbsp; Obliczamy, ile wynosi długość promienia podstawy. Pp​=πr2 &nbsp; 72π=πr2      ∣:π &nbsp; 72=r2 &nbsp;&nbsp;&nbsp; r=72<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>​=36⋅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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M834 80h400000v40h-400000z'/></svg>​=62<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
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​   [cm] &nbsp; Promień podstawy stożka ma długość 6√2 cm. Wysokość stożka ma długość 3 cm. H=3 cm &nbsp; Przyjmijmy oznaczenia jak na poniższym rysunku:

Pole podstawy stożka wynosi 18π.&nbsp; Pp​=18π &nbsp; Obliczamy, ile wynosi długość promienia podstawy. 18π=πr2      ∣:π &nbsp; 18=r2 &nbsp; r=18<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⋅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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M834 80h400000v40h-400000z'/></svg>​=32<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>​ &nbsp; Promień podstawy ma długość 3√2. Pole powierzchni bocznej wynosi 54π. Obliczamy, ile wynosi

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