Wiki-Quellcode von Lösung Pyramide in Würfel
Verstecke letzte Bearbeiter
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2.1 | 1 | {{aufgabe id="Pyramide in Würfel" afb="III" kompetenzen="K2,K4,K5" quelle="Reinhard Ansorge" zeit="15"}} |
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1.1 | 2 | |
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2.1 | 3 | (%class=abc%) |
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11.1 | 4 | 1. {{formula}}V_{Würfel}{{/formula}} = a³ = (10 cm)³ = 1.000 cm³ = 1dm³ = 1 Liter |
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12.1 | 5 | {{formula}}V_{Pyramide}{{/formula}} = 1/3 ∙ G ∙ h = 1/3 ∙ b² ∙ 10 cm = … |
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1.1 | 7 | Pythagoras: b² = (a/2)² + (a/2)² = 2 ∙ (a/2)² = 2 ∙ a²/4 = a²/2 = (10 cm)²/2 = 50 cm² |
| 8 | … = 1/3 ∙ 50 cm² ∙ 10 cm = 500/3 cm³ ≈ 166,67 cm³ | ||
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3.1 | 9 | |
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12.1 | 10 | {{formula}}V_{Würfel}{{/formula}} : {{formula}}V_{Pyramide}{{/formula}} = 1.000 cm³ : 166,66… cm³ = 6 : 1 |
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| 12 | 1. {{formula}}O_{Würfel}{{/formula}} = 6a² = 6 ∙ (10 cm)² = 600 cm² | ||
| 13 | {{formula}}O_{Pyramide}{{/formula}} = G + 4 ∙ b ∙ hb = 50 cm² + 4 ∙ √50 cm ∙ hb = … | ||
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1.1 | 15 | Pythagoras: hb² = (b/2)² + h² = (√50 cm / 2)² + (10 cm)² = 112,5 cm² | √ |
| 16 | hb = √(112,5 cm²) ≈ 10,61 cm | ||
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12.1 | 17 | |
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1.1 | 18 | … = 50 cm² + 4 ∙ √50 cm ∙ √(112,5 cm²) = 50 cm² + 4 ∙ 75 cm² = 350 cm² |
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3.1 | 19 | |
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12.1 | 20 | {{formula}}O_{Pyramide}{{/formula}} / {{formula}}O_{Würfel}{{/formula}} = 350 cm² / 600 cm² = 7/12, d.h. die O der Pyramide ist um 5/12 = 0,4166… ≈ 41,67 % kleiner als die des Würfels. |
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3.1 | 21 | |
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12.1 | 22 | 1. {{formula}}V_{Pyramide}{{/formula}} = 1/3 ∙ b² ∙ h = 1/3 ∙ 50 cm² ∙ h = 1.000 cm³ , d.h. h = (1.000 cm³) / (1/3 ∙ 50 cm²) = 1.000 cm³ ∙ (3 / (50 cm²)) = 3.000 / 50 cm = 60 cm |
| 23 | 1. {{formula}}V_{Würfel}{{/formula}} = a³ = 167,67 cm³, d.h. a = ∛167,67 cm ≈ 5,51 cm | ||
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1.1 | 24 | |
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3.1 | 25 | {{/aufgabe}} |
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