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. {{formula]} V_{Würfel} {{/formula}} = a³ = (10 cm)³ = 1.000 cm³ = 1dm³ = 1 Liter | ||
| 5 | V_{Pyramide} = 1/3 ∙ G ∙ h = 1/3 ∙ b² ∙ 10 cm = … | ||
| 6 | Pythagoras: b² = (a/2)² + (a/2)² = 2 ∙ (a/2)² = 2 ∙ a²/4 = a²/2 = (10 cm)²/2 = 50 cm² | ||
| 7 | … = 1/3 ∙ 50 cm² ∙ 10 cm = 500/3 cm³ ≈ 166,67 cm³ | ||
| 8 | V_{Würfel} : V_{Pyramide} = 1.000 cm³ : 166,66… cm³ = 6 : 1 | ||
| 9 | |||
| 10 | 1. O_{Würfel} = 6a² = 6 ∙ (10 cm)² = 600 cm² | ||
| 11 | O_{Pyramide} = G + 4 ∙ b ∙ hb = 50 cm² + 4 ∙ √50 cm ∙ hb = … | ||
| 12 | Pythagoras: hb² = (b/2)² + h² = (√50 cm / 2)² + (10 cm)² = 112,5 cm² | √ | ||
| 13 | hb = √(112,5 cm²) ≈ 10,61 cm | ||
| 14 | … = 50 cm² + 4 ∙ √50 cm ∙ √(112,5 cm²) = 50 cm² + 4 ∙ 75 cm² = 350 cm² | ||
| 15 | |||
| 16 | O_{Pyramide} / O_{Würfel} = 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. | ||
| 17 | |||
| 18 | 1. V_{Pyramide} = 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 | ||
| 19 | 1. V_{Würfel} = a³ = 167,67 cm³, d.h. a = ∛167,67 cm ≈ 5,51 cm | ||
| 20 | |||
| 21 | {{/aufgabe}} |