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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28.1 | 4 | 1. |
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29.1 | 5 | [[image:geogebra-export.png||width=600]] |
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18.1 | 6 | |
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11.1 | 7 | 1. {{formula}}V_{Würfel}{{/formula}} = a³ = (10 cm)³ = 1.000 cm³ = 1dm³ = 1 Liter |
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16.1 | 8 | {{formula}}V_{Pyramide}{{/formula}} = 1/3 ∙ G ∙ h = 1/3 ∙ b² ∙ 10 cm = … |
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12.1 | 9 | |
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1.1 | 10 | Pythagoras: b² = (a/2)² + (a/2)² = 2 ∙ (a/2)² = 2 ∙ a²/4 = a²/2 = (10 cm)²/2 = 50 cm² |
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20.1 | 11 | |
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1.1 | 12 | … = 1/3 ∙ 50 cm² ∙ 10 cm = 500/3 cm³ ≈ 166,67 cm³ |
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3.1 | 13 | |
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12.1 | 14 | {{formula}}V_{Würfel}{{/formula}} : {{formula}}V_{Pyramide}{{/formula}} = 1.000 cm³ : 166,66… cm³ = 6 : 1 |
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| 16 | 1. {{formula}}O_{Würfel}{{/formula}} = 6a² = 6 ∙ (10 cm)² = 600 cm² | ||
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24.1 | 17 | {{formula}}O_{Pyramide}{{/formula}} = G + 4 ∙ b ∙ {{formula}}h_b{{/formula}} = 50 cm² + 4 ∙ √50 cm ∙ {{formula}}h_b{{/formula}} = … |
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12.1 | 18 | |
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14.1 | 19 | Pythagoras: {{formula}}h_b²{{/formula}} = (b/2)² + h² = (√50 cm / 2)² + (10 cm)² = 112,5 cm² | √ |
| 20 | {{formula}}h_b{{/formula}} = √(112,5 cm²) ≈ 10,61 cm | ||
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12.1 | 21 | |
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1.1 | 22 | … = 50 cm² + 4 ∙ √50 cm ∙ √(112,5 cm²) = 50 cm² + 4 ∙ 75 cm² = 350 cm² |
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3.1 | 23 | |
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21.1 | 24 | {{formula}}O_{Pyramide}{{/formula}} / {{formula}}O_{Würfel}{{/formula}} = 350 cm² / 600 cm² = 7/12, |
| 25 | 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 | 26 | |
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22.1 | 27 | 1. {{formula}}V_{Pyramide}{{/formula}} = 1/3 ∙ b² ∙ h = 1/3 ∙ 50 cm² ∙ h = 1.000 cm³ , |
| 28 | d.h. h = (1.000 cm³) / (1/3 ∙ 50 cm²) = 1.000 cm³ ∙ (3 / (50 cm²)) = 3.000 / 50 cm = 60 cm | ||
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17.1 | 29 | |
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12.1 | 30 | 1. {{formula}}V_{Würfel}{{/formula}} = a³ = 167,67 cm³, d.h. a = ∛167,67 cm ≈ 5,51 cm |
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1.1 | 31 | |
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3.1 | 32 | {{/aufgabe}} |
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