silindr prizma Hajm

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10 sm
V
2Silindr ko’rinishdagi idishga 1200 sm 3
suv to’ldirildi . Silindrga quyilgan
suv balandligi 12 sm . Unga biror jism tahslandi . Bundan suv 10 sm ga
ko’tarildi . Jism hajmini toping ? Javobini sm 3
larda ifodalangh S V a s 

2 1
V
V



2 1
h S
h S
o o
2 1
h
h

2 1
V
V
2 1
h
h1200 12
1012 sm1200 sm 3
V
1fizikadan ma’lumki , jism hajmi
ko’tarilgan suv hajmiga teng .
Hajmlar nisbatini ko’ramiz
3 х
1
0 хВ 9
0
1
0 0

Silindr ko’rinishdagi idishdagi suv balandligi 27 sm . Boshqa diametri 3marta
katta bo’gan ikkinchi idishga quyilsa u idishdagi suvning balandligini toping
Javobni santimetrlarda ifodalang

2 1
V V
2 2
1 1
h S
h S
2
2
1
2
2
3
2
h
d
h
d















2 1
V
V 2727 smV
3 х
1
0 хВ 9
3h
2
V d
3d2
rS
o
  22 12 4
9
4
h
d
h
d

21
9 hh

21
9 h
h1
1Hajmlar nisbatini topamiz


V
VSuyuqlik hajmi
o’zgarishi . V
1 =V
2
271
1
9 h=

To’g’ri burchakli muntazzam prizmaning katetlari 6 va 8 bo’lsa,
Yon qirralai teng . Tashqi chizilgan silindr hajmini toping
3 х
1
0 хВ 9
1
2 5
5

568
105
h S V a s. 
h r V
2
s.  









5
5
2
s V


5
25  

 


2
2
2
s  V Silindrga ichki chizilgan prizmaning asosi tomoni 2ga teng
Yon qirralari . Tashqi chizilgan silindr hajmini toping
3 х
1
0 хВ 9
4

2

2
h S V a s. 
h r V
2
s.  
4 2 2
2 2
2 2 2
2 2   d
8
2
 d
8  d
2 2  d
2  r
d

h S V a s.
3
1
Silindr va konusning asosi va balandligi bir xil bo’lib, Konusning
hajmi 27 ga teng bo’lsa Silindrning hajmini toping
3 х
1
0 хВ 9
8
1
sV V
k
Hajmalr orasidagi munosabat

h S
h S
a
a
3
1
h S V a s 
3
1
31
s. .

V V
k27
1
27
V
s . 3=

9 a
3 х
1
0 хВ 9
7
2 9
2 1
V VHajmlar orasidagi munosabat 
 3
3
9 a
a

33
729 aa
729
1
a V
2
V
1 Agar k ubning qirrasini 2 mart a ort irilsa,
k ubning hajmi necha mart a ort adi ?

Kub diagonali ga teng bo’lsa Uning hajmini
toping
3 х
1
0 хВ 9
8 22
312 a
12
12
a aa To’g’ri burchakli
parallelepiped uchund 2
= a 2
+ b 2
+ c 2
d 2
= 3 a 2
Kub uchun
2
3 12 a 
2
4 a 
2   a 2  a
3 2 
куб V3
kub. aV 

Kub hajmi 24 ga teng bo’lsa, uning diagonalini toping
3 х
1
0 хВ 9
63a aa To’g’ri burchakli
parallelepiped uchund 2
= a 2
+ b 2
+ c 2
d 2
= 3 a 2
Kub uchun
3
3 24 a 
 
33 3 2 a  
3 2  a
  2
2 3 2 3  d
3 4 3
2
   d
3 4 3    d
8
3

3
2  
2
3 
3
3 a 
2 3   d 3
куб. aV 

x 4Ikkita qirrasi 2, 4 teng bo’lgan to’g’ri burchakli parallelopipedning
diagonali 6 ga teng . Parallelopipedning hajmini toping
3 х
1
0 хВ 9
3
24 2 To’g’ri burchakli
parallelepiped uchund 2
= a 2
+ b 2
+ c 2
2222
426 x
6
416362
x
162
x
4x4 2 4    V
abc V  пар.

S = a 2
sinaA a
D B
b C
a
aA B
C
D
 parallelogramm
Romb S = a b sinaC a

A B
b S = a b sina
21

d
1
d
2 B
C
DA parallelogramm
romb S = d
1 d
2 sina
21
A d
2
D B
d
1 C
S = d
1 d
2 sin 90 0
21
A B C
D
dd
S = d 2
sina
21To’g’ri to’rtburchak 1

10 sm V 2Silindr ko’rinishdagi idishga 1200 sm 3 suv to’ldirildi . Silindrga quyilgan suv balandligi 12 sm . Unga biror jism tahslandi . Bundan suv 10 sm ga ko’tarildi . Jism hajmini toping ? Javobini sm 3 larda ifodalangh S V a s   2 1 V V    2 1 h S h S o o 2 1 h h  2 1 V V 2 1 h h1200 12 1012 sm1200 sm 3 V 1fizikadan ma’lumki , jism hajmi ko’tarilgan suv hajmiga teng . Hajmlar nisbatini ko’ramiz 3 х 1 0 хВ 9 0 1 0 0

Silindr ko’rinishdagi idishdagi suv balandligi 27 sm . Boshqa diametri 3marta katta bo’gan ikkinchi idishga quyilsa u idishdagi suvning balandligini toping Javobni santimetrlarda ifodalang  2 1 V V 2 2 1 1 h S h S 2 2 1 2 2 3 2 h d h d                2 1 V V 2727 smV 3 х 1 0 хВ 9 3h 2 V d 3d2 rS o   22 12 4 9 4 h d h d  21 9 hh  21 9 h h1 1Hajmlar nisbatini topamiz  V VSuyuqlik hajmi o’zgarishi . V 1 =V 2 271 1 9 h=

To’g’ri burchakli muntazzam prizmaning katetlari 6 va 8 bo’lsa, Yon qirralai teng . Tashqi chizilgan silindr hajmini toping 3 х 1 0 хВ 9 1 2 5 5  568 105 h S V a s.  h r V 2 s.            5 5 2 s V   5 25  

    2 2 2 s  V Silindrga ichki chizilgan prizmaning asosi tomoni 2ga teng Yon qirralari . Tashqi chizilgan silindr hajmini toping 3 х 1 0 хВ 9 4  2  2 h S V a s.  h r V 2 s.   4 2 2 2 2 2 2 2 2 2   d 8 2  d 8  d 2 2  d 2  r d

silindr prizma Hajm

10.08.2023

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553.6171875 KB

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