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IL TEOREMA DI PITAGORA APPLICATO AL ROMBO E AL TRAPEZIO

IL TEOREMA DI PITAGORA APPLICATO AL ROMBO E AL TRAPEZIO. Di Susanna Lamberti 2°B. IL ROMBO. OB= CB2-CO2. CO= BC2-BO2. dM. CB= CO2+OB2. dm. CONSIDERATO IL TRIANGOLO CBO POSSIAMO TROVARE: LA META’ DELLA DIAGONALE MAGGIORE, DI QUELLA MINORE E UN LATO DEL ROMBO.

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IL TEOREMA DI PITAGORA APPLICATO AL ROMBO E AL TRAPEZIO

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  1. IL TEOREMA DI PITAGORA APPLICATO AL ROMBO E AL TRAPEZIO Di Susanna Lamberti 2°B

  2. IL ROMBO OB= CB2-CO2 CO= BC2-BO2 dM CB= CO2+OB2 dm CONSIDERATO IL TRIANGOLO CBO POSSIAMO TROVARE: LA META’ DELLA DIAGONALE MAGGIORE, DI QUELLA MINORE E UN LATO DEL ROMBO. COSI’ FACENDO POSSIAMO TROVARE L’AREA E IL PERIMETRO

  3. TRAPEZIO RETTANGOLO CB= CH2+HB2 CH= CB2-HB2 HB CB2-CH2 h DIFFERENZA TRA LE BASI AB-DC

  4. TRAPEZIO ISOSCELE AD= AH2+HD2 AB-DC 2 AH= DA2-DH2 h DH= DA2-AH2

  5. TRAPEZIO SCALENO DAH = CKB H K

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