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 = = ,