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Hajm tushunchasi. Parallelepipedning hajmi

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Geometriya
11-sinf
Mavzu:
Hajm tushunchasi.
Parallelepipedning hajmi
Reja
οƒ˜ Hajm tushunchasi
οƒ˜ Parallelepipedning hajmi
οƒ˜ Kubning hajmi
οƒ˜ Mavzuga oid masalalar yechish
πŸπŸ•πŸ“-masala (157-sahifa)
Oktaedr bitta yog‘ining yuzi πŸ“, πŸ“ π’„π’ŽπŸ bo‘lsa, uning
to‘la sirtini toping.
Berilgan:
𝑺 = πŸ“, πŸ“ π’„π’ŽπŸ
𝒀=πŸ–
Topish kerak:
𝑺𝒕𝒐‘𝒍𝒂 = ?
Masalaning yechimi:
𝑺𝒕𝒐‘𝒍𝒂 = πŸ– βˆ™ 𝑺
𝑺𝒕𝒐‘𝒍𝒂 = πŸ– βˆ™ πŸ“, πŸ“ = πŸ’πŸ’
Javob: πŸ’πŸ’ π’„π’Ž
𝟐
πŸπŸ–πŸ”-masala (158-sahifa)
Agar kubning qirrasi 1 birlik orttirilsa, uning to‘la
sirti 54 birlikka ortadi. Kubning qirrasini toping.
Yechish:
Berilgan:
𝟐
𝑺 = πŸ”π’‚
𝑺
𝒂
𝟐
𝒂+𝟏
𝑺 + πŸ“πŸ’
𝑺 + πŸ“πŸ’ = πŸ” βˆ™ 𝒂 + 𝟏
Topish kerak:
πŸ”π’‚πŸ + πŸ“πŸ’ = πŸ” βˆ™ π’‚πŸ + πŸπ’‚ + 𝟏
𝟐
𝟐
𝒂=?
πŸ”π’‚ + πŸ“πŸ’ = πŸ”π’‚ + πŸπŸπ’‚ + πŸ”
πŸ’πŸ– = πŸπŸπ’‚ 𝒂 = πŸ’
Javob: πŸ’
πŸπŸ—πŸ’-masala (158-sahifa)
To‘rtburchakli muntazam prizma asosining tomoni
6 cm, prizmaning balandligi 5 cm. Prizmaning diagonal
kesimi yuzini toping.
Berilgan:
𝒂 = πŸ” π’„π’Ž
𝒉 = πŸ“ π’„π’Ž
Topish kerak:
𝑺𝒅 = ?
Yechish:
𝒅=𝒂 𝟐=πŸ” 𝟐
𝑺𝒅 = 𝒉 βˆ™ 𝒅
𝑺𝒅 = πŸ“ βˆ™ πŸ” 𝟐 = πŸ‘πŸŽ 𝟐
Javob: πŸ‘πŸŽ 𝟐 π’„π’Ž
𝟐
𝒉
𝒅
𝒂
𝒂
Hajm tushunchasi
Hajm tushunchasi
Hajm – fazoviy jismning quyidagi xossalarga ega
bo‘lgan miqdoriy ko‘rsatkichidir:
𝑽>𝟎
Hajm – fazoviy jismning quyidagi xossalarga ega
bo‘lgan miqdoriy ko‘rsatkichidir:
𝟏
𝟏
𝟏
Hajm o‘lchov birliklari
π’ŽπŸ‘ , π’…π’ŽπŸ‘ , π’”π’ŽπŸ‘ , π’Žπ’ŽπŸ‘ , 𝒍
𝑽=𝟏
Parallelepipedning hajmi
To‘g‘ri burchakli
parallelepipedning
hajmi uning uchta
o‘lchamlari ko‘paytmasiga teng:
𝒄
𝒂
𝒃
𝑽=π’‚βˆ™π’ƒβˆ™π’„
Parallelepipedning hajmi
𝒉
𝑺𝒂𝒔𝒐𝒔
To‘g‘ri burchakli
parallelepipedning
hajmi asosining yuzi
bilan balandligining
ko‘paytmasiga teng:
𝑽 = 𝑺𝒂𝒔𝒐𝒔 βˆ™ 𝒉
Parallelepipedning hajmi
Ixtiyoriy
parallelepipedning hajmi
asosining
yuzi bilan balandligining
ko‘paytmasiga teng:
𝑽 = 𝑺𝒂𝒔𝒐𝒔 βˆ™ 𝒉
𝒉
Kubning hajmi
𝑽=π’‚βˆ™π’‚βˆ™π’‚
πŸ‘
𝑽=𝒂
𝒂
𝒂
𝒂
1-masala
Rasmda berilgan yoyilmaga ko‘ra yasalgan
idishning hajmini toping.
πŸ’πŸ– π’„π’Ž
πŸ– π’„π’Ž
πŸ‘πŸ” π’„π’Ž
πŸ– π’„π’Ž
Masalaning yechimi
𝒂 = πŸ‘πŸ” − 𝟐 βˆ™ πŸ– = πŸ‘πŸ” − πŸπŸ” = 𝟐𝟎
𝒃 = πŸ’πŸ– − 𝟐 βˆ™ πŸ– = πŸ’πŸ– − πŸπŸ” = πŸ‘πŸ
πŸ‘
𝑽 = 𝒂 βˆ™ 𝒃 βˆ™ 𝒄 = 𝟐𝟎 βˆ™ πŸ‘πŸ βˆ™ πŸ– = πŸ“πŸπŸπŸŽ π’„π’Ž
𝒄=πŸ–
πŸ’πŸ– π’„π’Ž
πŸ– π’„π’Ž
πŸ‘πŸ” π’„π’Ž
𝒃
𝒂
πŸ– π’„π’Ž
𝒄
Javob: πŸ“πŸπŸπŸŽ π’„π’ŽπŸ‘
2-masala
To‘g‘ri burchakli parallelepiped yog‘ining yuzi
12 cm ga va unga perpendikulyar qirra uzunligi
18 cm ga teng. Parallelepipedning hajmini toping.
Berilgan:
Yechish:
𝑽 = 𝑺𝒂𝒔𝒐𝒔 βˆ™ 𝒉
𝒉 = πŸπŸ– π’„π’Ž
𝑺𝒂𝒔𝒐𝒔 = 𝟏𝟐 π’„π’Ž
𝑽 = πŸπŸ– βˆ™ 𝟏𝟐 = πŸπŸπŸ”
𝒉 = πŸπŸ–
Topish kerak:
Javob: πŸπŸπŸ” π’„π’ŽπŸ‘
𝑽=?
3-masala
To‘g‘ri burchakli parallelepiped bir uchidan
chiquvchi uchta qirralarining uzunliklari 4, 6 va 9 ga teng.
Unga tengdosh kub qirrasini toping.
Berilgan:
Yechish:
𝑽=π’‚βˆ™π’ƒβˆ™π’„
𝒂=πŸ’
𝒃=πŸ”
𝑽 = πŸ’ βˆ™ πŸ” βˆ™ πŸ— = πŸπŸπŸ”
𝒄=πŸ—
𝑽 = π’‚π’Œπ’–π’ƒ πŸ‘
Topish kerak:
π’‚π’Œπ’–π’ƒ = ?
𝒃
𝒂
𝒄
3-masala
To‘g‘ri burchakli parallelepiped bir uchidan
chiquvchi uchta qirralarining uzunliklari 4, 6 va 9 ga teng.
Unga tengdosh kub qirrasini toping.
Berilgan:
𝒂=πŸ’
𝒃=πŸ”
𝒄=πŸ—
Topish kerak:
π’‚π’Œπ’–π’ƒ = ?
Yechish:
𝑽=π’‚βˆ™π’ƒβˆ™π’„
𝑽 = πŸ’ βˆ™ πŸ” βˆ™ πŸ— = πŸπŸπŸ”
𝑽 = π’‚π’Œπ’–π’ƒ πŸ‘
π’‚π’Œπ’–π’ƒ πŸ‘ = πŸπŸπŸ” = πŸ”πŸ‘
π’‚π’Œπ’–π’ƒ = πŸ”
Javob: πŸ”
𝒂
𝒂
𝒂
4-masala
Agar kubning qirralarini 1 birlik orttirilsa, uning
hajmi 19 birlikka ortadi. Kubning qirrasini toping.
Berilgan:
𝑽
𝒂
𝒂+𝟏
𝑽 + πŸπŸ—
Topish kerak:
𝒂=?
Yechish:
𝑽 = π’‚πŸ‘
𝑽 + πŸπŸ— = 𝒂 + 𝟏 πŸ‘
π’‚πŸ‘ + πŸπŸ— = π’‚πŸ‘ + πŸ‘π’‚πŸ + πŸ‘π’‚ + 𝟏
πŸ‘π’‚πŸ + πŸ‘π’‚ − πŸπŸ– = 𝟎
π’‚πŸ + 𝒂 − πŸ” = 𝟎
π’‚πŸ = −πŸ‘
π’‚πŸ = 𝟐
Javob: 𝟐 π’„π’Ž
5-masala
Birinchi kubning hajmi ikkinchisinikidan 8 marta
katta. Birinchi kubning to‘la sirtining yuzi
ikkinchisinikidan necha marta katta?
Berilgan:
Yechish:
πŸ‘
πŸ‘
π‘½πŸ = π‘½πŸ βˆ™ πŸ–
π’‚πŸ = π’‚πŸ βˆ™ πŸ–
Topish kerak:
π‘ΊπŸ
=?
π‘ΊπŸ
πŸ‘
πŸ‘
π’‚πŸ = π’‚πŸ βˆ™ πŸπŸ‘
π’‚πŸ‘πŸ = π’‚πŸ βˆ™ 𝟐 πŸ‘
π’‚πŸ = π’‚πŸ βˆ™ 𝟐
π‘ΊπŸ πŸ” βˆ™ π’‚πŸπŸ π’‚πŸπŸ
=
= 𝟐=
𝟐
π‘ΊπŸ πŸ” βˆ™ π’‚πŸ π’‚πŸ
π’‚πŸ
π’‚πŸ
𝟐
π’‚πŸ βˆ™ 𝟐
=
π’‚πŸ
𝟐
=πŸ’
π‘ΊπŸ = π‘ΊπŸ βˆ™ πŸ’
Javob: πŸ’
π’‚πŸ
π’‚πŸ
π’‚πŸ
π’‚πŸ
π’‚πŸ
π’‚πŸ
Mustaqil bajarish uchun topshiriqlar
165-166-sahifalar
209
220
213
223
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