Aylanish figuralari

Y

echilishi:
Masala shartiga ko`ra CO<sub>1</sub>=4 O<sub>1</sub>O =1
chunki R =5. To`g`ri burchakli ∆O<sub>1</sub>OB uchbur-
chakdan O<sub>1</sub>B = = .
Konus asosining yuzi:
S<sub>acoc</sub>= π *O<sub>1</sub>B<sup>2</sup> = 24π.
Demak konusni hajmi:
V = π*O<sub>1</sub>B<sup>2</sup> π*CO<sub>1</sub> = 32π.
Javob: 32π.
11 – masala. Sharning radiusi ga teng. Radiusning oxiridan u bilan 60<sup>0</sup> 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<sub>sh</sub> = 4π R<sup>2</sup> = 16π, konus
sirti S =π · AD · AB + π · AD<sup>2</sup>
AD masofa to`g`ri burchakli ∆ADO dan
topiladi: AD = = ,