Aylanish figuralari


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Aylanish figuralari

Y

echilishi:

Masala shartiga ko`ra CO1=4 O1O =1
chunki R =5. To`g`ri burchakli ∆O1OB uchbur-
chakdan O1B = = .
Konus asosining yuzi:
Sacoc= π *O1B2 = 24π.
Demak konusni hajmi:
V = π*O1B2 π*CO1 = 32π.
Javob: 32π.
11 – masala. Sharning radiusi ga teng. Radiusning oxiridan u bilan 600 li burchak tashkil etadigan kesuvchi tekislik o`tkazilgan. Kesimning yuzini toping.
Y

echilishi:

Masala shartiga ko`ra OB =OA = ,
0. Bundan ∆ABO ni teng tomonli
uchburchak ekanligini aytish mumkin.
O1A = = .
Demak, Skesim= =16. Javob: 16


1

2 - masala.
Ko`rsatilgan uchburchakning to`g`ri chiziq atrofida aylanishidan hosil bo`lgan jismning hajmini toping.
Yechilishi:
Masala shartida keltirilgan

aylanish figurasining hajmi ABB1A1
silindr hajmidan BSB1 konus hajmi-
ni ayirmasiga teng. YA`ni Vf = Vs – Vk.
Vs =π*AC2*OC = 36π,
Vk = πOB2*OC = 12π
Demak, Vf = 24 π
Javob: 24 π.


1

3 – masala.
Konusning balandligi 3, unga tashqi chizilgan sharning radisi 2 bo`lsa, konus sirtining shar sirtiga nisbatini toping.
Yechilishi:
Masala shartiga e`tibor bersak,
BO=2, BD =3 ekanligi ∆ABC ni teng

tomonli ekanligini bildiradi. Ma`-
lumki shar sirti Ssh = 4π R2 = 16π, konus
sirti S =π · AD · AB + π · AD2
AD masofa to`g`ri burchakli ∆ADO dan
topiladi: AD = = ,

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