Wiki-Quellcode von Lösung Pyramide in Würfel
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| author | version | line-number | content |
|---|---|---|---|
| 1 | {{aufgabe id="Pyramide in Würfel" afb="III" kompetenzen="K2,K4,K5" quelle="Reinhard Ansorge" zeit="15"}} | ||
| 2 | |||
| 3 | (%class=abc%) | ||
| 4 | 1. | ||
| 5 | |||
| 6 | [[image:geogebra-export.png||width=600]] | ||
| 7 | |||
| 8 | 1. {{formula}}V_{Würfel}{{/formula}} = a³ = (10 cm)³ = 1.000 cm³ = 1dm³ = 1 Liter | ||
| 9 | {{formula}}V_{Pyramide}{{/formula}} = 1/3 ∙ G ∙ h = 1/3 ∙ b² ∙ 10 cm = … | ||
| 10 | |||
| 11 | Pythagoras: b² = (a/2)² + (a/2)² = 2 ∙ (a/2)² = 2 ∙ a²/4 = a²/2 = (10 cm)²/2 = 50 cm² | ||
| 12 | |||
| 13 | … = 1/3 ∙ 50 cm² ∙ 10 cm = 500/3 cm³ ≈ 166,67 cm³ | ||
| 14 | |||
| 15 | {{formula}}V_{Würfel}{{/formula}} : {{formula}}V_{Pyramide}{{/formula}} = 1.000 cm³ : 166,66… cm³ = 6 : 1 | ||
| 16 | |||
| 17 | 1. {{formula}}O_{Würfel}{{/formula}} = 6a² = 6 ∙ (10 cm)² = 600 cm² | ||
| 18 | {{formula}}O_{Pyramide}{{/formula}} = G + 4 ∙ b ∙ {{formula}}h_b{{/formula}} = 50 cm² + 4 ∙ √50 cm ∙ {{formula}}h_b{{/formula}} = … | ||
| 19 | |||
| 20 | Pythagoras: {{formula}}h_b²{{/formula}} = (b/2)² + h² = (√50 cm / 2)² + (10 cm)² = 112,5 cm² | √ | ||
| 21 | {{formula}}h_b{{/formula}} = √(112,5 cm²) ≈ 10,61 cm | ||
| 22 | |||
| 23 | … = 50 cm² + 4 ∙ √50 cm ∙ √(112,5 cm²) = 50 cm² + 4 ∙ 75 cm² = 350 cm² | ||
| 24 | |||
| 25 | {{formula}}O_{Pyramide}{{/formula}} / {{formula}}O_{Würfel}{{/formula}} = 350 cm² / 600 cm² = 7/12, | ||
| 26 | d.h. die O der Pyramide ist um 5/12 = 0,4166… ≈ 41,67 % kleiner als die des Würfels. | ||
| 27 | |||
| 28 | 1. {{formula}}V_{Pyramide}{{/formula}} = 1/3 ∙ b² ∙ h = 1/3 ∙ 50 cm² ∙ h = 1.000 cm³ , | ||
| 29 | d.h. h = (1.000 cm³) / (1/3 ∙ 50 cm²) = 1.000 cm³ ∙ (3 / (50 cm²)) = 3.000 / 50 cm = 60 cm | ||
| 30 | |||
| 31 | 1. {{formula}}V_{Würfel}{{/formula}} = a³ = 167,67 cm³, d.h. a = ∛167,67 cm ≈ 5,51 cm | ||
| 32 | |||
| 33 | {{/aufgabe}} |