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:
S_acoc= 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, S_kesim= =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 V_f = V_s – V_k.
V_s =π*AC²*OC = 36π,
V_k = πOB²*OC = 12π
Demak, V_f = 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 S_sh = 4π R² = 16π, konus
sirti S =π · AD · AB + π · AD²
AD masofa to‘g‘ri burchakli ∆ADO dan
topiladi: AD = = ,