Matematika. 4-sinf

O’tilgan mavzuni takrorlash uchun 5 ta kartochkada yozilgan savollarni o’quvchilarga tarqatiladi va ularga javoblar olinadi .

Foziljon Anapiyayev

2. Ko’rgazmali qurollar,tarqatma materiallar , bo’r va yozuv taxtasi va to’g’ri burchakli parallelepiped va kub maketlari.

IV. Darsning rejasi:
Darsning bosqichlari
Foydalanishga tavsiya etiladigan usul
Vaqt taqsimoti
Tashkiliy qism
O’tilgan mavzuni takrorlash
Qisqa test yoki og’zaki so’rov
Yangi mavzuni yoritish:
a) mavzuni yoritishga tayyorgarlik va motivatsiya
b) yangi mavzuni yoritish;
Savol-javob
Yangi mavzuni mustaxkamlash:
a) guruhlarda ishlash;
c) yakka tartibdagi mustaqil ish.
Guruhlarda ishlash
Aqliy hujum
Mustaqil ish
Darsga yakun yasash va baholash
Uyga vazifa
V. Darsning borishi:

a) Tashkiliy davr. O’quvchilarning davomatini aniqlab ,sinf va o’quvchilar tozaligiga e’tibor beriladi;

b) O’tilgan mavzuni takrorlash uchun tarqatma materialdan foydalaniladi.
v) Yangi mavzuni tafsilotlari

O’tilgan mavzuni takrorlash uchun 5 ta kartochkada yozilgan savollarni o’quvchilarga tarqatiladi va ularga javoblar olinadi .

Mavzuni o’rganish natijasida o’quvchilar egallashlari lozim bo’lgan
Bilimlar:
– hajm birliklari orasidagi bog’lanishlarni;
– to’g’ri burchakli parallelepiped hajmini hisoblash formulasini;
– kub hajmi formulasini bilish.
Ko’nikmalar :
– o’lchamlari(bo’yi, eni, balandligi) berilgan to’g’ri burchakli parallelepipedning;
– qirrasi berilgan kubning hajmini hisoblay olish;
– kub va to’g’ri burchakli parallelepiped hajmiga doir formulalarni amaliyotga qo’llay olish.
Malakalar:
– to’g’ri burchakli parallelepiped va kubning hajmiga doir sodda mashqlarni mustaqil yecha olish
Kompetensiya turi:
– o’quv – o’rganish
O’tilgan mavzuni takrorlash uchun test topshiriqlari:
1. To’g’ri burchakli parallelepipedning nechta uchi bor?
A) 4 ta B) 6 ta C) 8 ta D) 12 ta E) 2 ta
2. To’g’ri burchakli parallelepipedning nechta qirrasi bor?
A) 4 ta B) 6 ta C) 8 ta D) 12 ta E) 2 ta
3. Berilgan ma’lumotlarga ko’ra kub sirtining yuzini toping.Kubning qirrasi 3 dm.
A) 36 dm 2 B) 48 dm 2 C) 54 dm 2 D) 64 dm 2 108 dm 2

4. Berilgan ma’lumotlarga to’g’ri burchakli parallelepipedning sirtining yuzini toping.a = 2 sm, b = 3 sm , c = 8 sm.

A) 92 sm 2 B) 24 sm 2 C) 48 sm 2 D) 64 sm 2 E) 52 sm 2
Yangi mavzuni borishi:

Hajm o’lchov birligi sifatida birlik kub hajmi olinadi. Qirrasi bir uzunlik birligiga teng bo’lgan kub birlik kub deyiladi.Berilgan jismning hajmini topish u nechta birlik kubdan tashkil topganini bilish demakdir.To’g’ri burchakli parallelepipedni n dona birlik kubga ajratish mumkin bo’lsa , uning hajmi n birlik kub hajmiga teng deyiladi.O’lchamlari (qirralari uzunliklari) a,b,c bo’lgan to’g’ri burchakli parallelepipedning hajmi V=a QUOTE b QUOTE c (kub birlik) ga teng bo’ladi. Odatda to’g’ri burchakli parallelepipedning a-bo’yi, b- eni, c-balandligi deyiladi. a.b – to’g’ri burchakli parallelepipedning asosi yuzi bo’lganidan quyidagi qoida kelib chiqadi:

T o’g’ri burchakli parallelepipedning hajmi uning asosi yuzi bilan balandligining ko’paytmasiga teng.

Kub uchun barcha qirralar o’zaro teng:a=b=c. Kubning hajmi V=a.a.a=a 3 (kub birlik)ga tengdir.
Shundan keyin taxtada 1 va 2- masalalar yechib ko’rsatiladi.

To’g’ri burchakli parallelepipedning bir uchidan chiquvchi qirralari yig’indisi bir xil (o’zaro teng) bo’lsa-da, ularning hajmlari , sirtlari turlicha bo’lishi mumkin.

To’g’ri burchakli parallelelpiped hajmi formulasidagi a QUOTE ko’paytma asosinig yuzi S ni , c=qirra esa balandlik h ni ifodalasa V= SH formula hosil bo’ladi, ya’ni to’g’ri burchakli parallelepipedning hajmi asosi yuzi bilan balandligi ko’paytmasiga teng.

V = a b c formuladan a = V : (bc); b = V : (ac) ; c = V : (ab ) hamda V= SH formuladan S = V : H va H = V : S degan formulalar kelib chiqadi.

Matematika. 4-sinf

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