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nazariya 2 (1)

Y echish: DK = h desak, h= 2 r bo‘ladi, bu yerda r – ichki chizilgan doira radiusi. ADK uchburchakdan AK = h ∙ tg300 =
ni topamiz.
DC= AB – 2AK = a
AD = BC = =
Trapetsiyaga aylana ichki chizilganligidan AB+DC=AD+BC, h ni aniqlash tenglamasini hosil qilamiz:
2a – =
Demak, h = , u holda r = . Izlanayotgan yuza .
Javob: .
2 – misol. Asoslari 2 va 14, yon tomoni 10 bo‘lgan teng yonli trapetsiyaga tashqi chizilgan aylana radiusini toping.
Yechish: Shartga ko‘ra AB = 14, DC = 2, BD = AD = 10. Ko‘rinib turibdiki,
AM =NB = (AB – DC) = 6,
AN = AB – NB = 8.
u holda Pifagor teoremasiga ko‘ra
CN = = 8, AC = = 8 .
ABCD trapetsiyaga tashqi chizilgan aylana o‘z navbatida ABC uchburchakka ham tashqi chizilganligini e’tiborga olamiz. R – aylana radiusi bo‘lsin.
S∆ABC = AB ∙ CN = 56
ekanligidan, R ni osongina topish mumkin.
R= = 5
Javob:_5__3_–_misol.'>Javob: 5
3 – misol. Tomoni a ga teng kvadratga aylana tashqi chizilgan. Kvadrat tomoni va aylana hosil qilgan segment yuzini toping.
Y echish:
Rasmdan ko‘rinadiki, AC = 2R = = a bundan R=
U holda segment yuzi quyidagi formula
bilan topiladi:
Ssegm = =
Javob:
4 – misol. Rombni yuzi Q Va ichki chizilgan doira yuzi S bo‘lsa, romb burchaklarini toping. Natijani Q =8, S = π da toping.
Yechish: Rasmdan quyidagiga ega bo‘lamiz
Sin < BAD =
Shartga ko‘ra MN ∙ AD =Q, bundan
2ra =Q va S= πr2. Bu tenglamalardan
r va a ni topish mumkin. Biz nisbatni topishimiz kerak. dan demak,< BAD = arcsin , Q = 8, S = π bo‘lganda < BAD = arcsin = 300

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