Mavzu: stereometriya kursini o‘qitish metodikasi aylanish figuralari

Javob: πL² ( – cosα).


7

– masala. Radiusi R ga teng bo‘lgan shardan o‘q kesimida α burchak tashkil qiluvchi shar sektorining hajmi va to‘la sirtini toping.
Yechilishi:
Shar radiusini R bilan, segment balandligini DC =h va
AD kesmani r bilan belgilaylik. Sektor hajmi V= πR²h ga
teng. ∆ACD uchburchakda ACD = , bundan h = rtg .
∆ADO uchburchakdan r= R sin . Ma’lumki,
V= πR²h= R³ sin² .
Sektorning to‘la sirti ABC segment yuzi 2πRh va AOB konus yon sirti πRr lardan tashkil topadi.
Demak, S = 2πRh + πRr = πR²sin (2tg +1).
Javob: V = R³sin² ; S =πR²sin (2tg +1).




8 – masala. Silindrning balandligi 5 ga, uning asosiga ichki
chizilgan muntazam uchburchakning tomoni 3 ga teng.
Silindrning hajmini toping.
Yechilishi:

Ma’lumki silindr hajmi V =S_asos*H = πR²H formula
bilan hisoblanadi. Bu yerda Н =5 shartda berilgan. R radiusni
ABC muntazam uchburchakdan foydalanib topamiz. Ya’ni

R = OC = = · = .

ekanligidan R =3 . Demak, V = 45π .

Javob: 45 π.


9

– masala. silindrning yon sirti yoyilganda uning diagonali asos tekisligi bilan 45⁰ burchak tashkil qiladi. silindrning yon sirti 144π² ga teng. silindr asosining radiusini toping.
Yechilishi:
Masala shartidan S_yon = 144π²
ekanligiga ko‘ra,

AB · BS = 144π²  AB =BC =12 π.

Silindr asosi aylanasi uzunligi

AB ga teng ekanligidan,
2πR =AB, 2πR = 12π
demak R =6.
Javob: 6


10 – masala. Radiusi 5 ga teng bo‘lgan sharga ichki chizilgan konusning balandligi 4 ga teng. Konusning hajmini toping.