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Esercizio: semicirconferenze tangenti
Date le (semi)circonferenze in
figura si ha che:
๐ด๐น = ๐น๐ถ = ๐ถ๐บ = ๐บ๐ต = ๐‘Ž
Trovare il raggio della
circonferenza rossa (tangente
alle tre semicirconferenze) in
funzione di a.
Svolgimento
In queste condizioni a รจ il raggio
della semicirconferenza blue e
della semicirconferenza verde
(interne). Quindi 2a sarร  il
raggio della semicirconferenza
nera (grande).
Indichiamo con x il raggio della
circonferenza rossa.
Applicando il teorema di Pitagora al triangolo rettangolo FCD otteniamo:
๐น๐ท + ๐น๐ถ = ๐ถ๐ท
Sostituendo:
(๐‘Ž + ๐‘ฅ) + ๐‘Ž = (2๐‘Ž โˆ’ ๐‘ฅ)
Risolvendo rispetto a x si trova:
๐‘ฅ =
๐‘Ž
3

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Esercizio semicirconferenze tangenti [sc]

  • 1. Esercizio: semicirconferenze tangenti Date le (semi)circonferenze in figura si ha che: ๐ด๐น = ๐น๐ถ = ๐ถ๐บ = ๐บ๐ต = ๐‘Ž Trovare il raggio della circonferenza rossa (tangente alle tre semicirconferenze) in funzione di a. Svolgimento In queste condizioni a รจ il raggio della semicirconferenza blue e della semicirconferenza verde (interne). Quindi 2a sarร  il raggio della semicirconferenza nera (grande). Indichiamo con x il raggio della circonferenza rossa. Applicando il teorema di Pitagora al triangolo rettangolo FCD otteniamo: ๐น๐ท + ๐น๐ถ = ๐ถ๐ท Sostituendo: (๐‘Ž + ๐‘ฅ) + ๐‘Ž = (2๐‘Ž โˆ’ ๐‘ฅ) Risolvendo rispetto a x si trova: ๐‘ฅ = ๐‘Ž 3