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= πR<sup>2</sup>h 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<sub>asos*</sub>H = πR<sup>2</sup>H 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.