Moderne Wiskunde (13e editie) - 2 vmbo k(gt)
'Stelling van Pythagoras'.
| 2 vmbo k(gt) | 7.3 Langste zijde berekenen |
opgave 1Gegeven is \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) met \(B\kern{-.8pt}C = 18 \text{,}\) \(A\kern{-.8pt}C = 47\) en \(\angle \text{C} = 90\degree \text{.}\) 3p Bereken de lengte van zijde \(A\kern{-.8pt}B \text{.}\) Pythagoras (1) 007c - Stelling van Pythagoras - basis - 0ms ○ Pythagoras in \(\triangle A\kern{-.8pt}B\kern{-.8pt}C\) geeft \(B\kern{-.8pt}C^{2} + A\kern{-.8pt}C^{2} = A\kern{-.8pt}B^{2} \text{.}\) 1p ○ \(A\kern{-.8pt}B^{2} = 18^{2} + 47^{2} = 2\,533 \text{.}\) 1p ○ \(A\kern{-.8pt}B = \sqrt{2\,533} ≈ 50{,}3 \text{.}\) 1p |
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| 2 vmbo k(gt) | 7.4 Rechthoekszijde berekenen |
opgave 1Gegeven is \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) met \(P\kern{-.8pt}Q = 20 \text{,}\) \(P\kern{-.8pt}R = 50\) en \(\angle \text{Q} = 90\degree \text{.}\) 3p Bereken de lengte van zijde \(Q\kern{-.8pt}R \text{.}\) Pythagoras (2) 007d - Stelling van Pythagoras - basis - 0ms ○ Pythagoras in \(\triangle P\kern{-.8pt}Q\kern{-.8pt}R\) geeft \(P\kern{-.8pt}Q^{2} + Q\kern{-.8pt}R^{2} = P\kern{-.8pt}R^{2}\) ofwel \(20^{2} + Q\kern{-.8pt}R^{2} = 50^{2} \text{.}\) 1p ○ \(Q\kern{-.8pt}R^{2} = 50^{2} - 20^{2} = 2\,100 \text{.}\) 1p ○ \(Q\kern{-.8pt}R = \sqrt{2\,100} ≈ 45{,}8 \text{.}\) 1p |