Showing Page:
1/1
8-sinf, geometriya.
(2 soatlik bayonnoma)
«20-22» -dekabr, 2018-yil.
Mavzu: 3-nazorat ishi.
1. Darsning maqsadi: Yuz(a), tengdosh shakllar, yuz(a)ni o‘lchash, to‘g‘ri to‘rtburchakning yuzi,
parallelogrammning yuzi, uchburchakning yuzi, ularning xossalari haqida tushunchalar hosil
qilish, o‘quvchilarning BKM larini shakllantirish, rivojlantirish va mustahkamlash hamda
tekshirish.
2. Darsning usuli: yozma ish.
3. Darsning jihozi: DTS ko‘rgazmalari, tarqatma materiallar.
DARSNING BORISHI:
1. Tashkiliy qism: o‘quvchilar bilan salomlashish, tozalikni tekshirish, davomatni aniqlash,
o‘quvchilarning dars mashg‘ulotlariga ruhiy jihatda tayyorliklarini aniqlash.
2. Yangi mavzuning bayoni: (Ikki variantdagi yozma ish matni)
1-variant.
1) To‘g‘ri to‘rtburchakning ikki tomoni 60 sm va 5,8 sm. Uning yuzini toping.
Yechilishi:
  
; Javob: 
.
2) Kvadratning tomoni 1,3 sm. Kvadratning yuzini toping.
Yechilishi:
󰇛

󰇜

; Javob: 
.
3) To‘g‘ri burchakli uchburchakning katetlari 5 sm va 6 sm. Shu uchburchakning yuzini toping.
Yechilishi:
1)
    
; Javob: 
.
4) Parallelogrammning asosi 60 sm va balandligi 0,5 m bo‘lsa, uning yuzini toping.
Yechilishi:
  bo‘lgani uchun:   deb olamiz:
  
; Javob: 
.
2-variant.
1) To‘g‘ri to‘rtburchakning ikki tomoni 4 m va 1,4 m. Uning yuzini toping.
Yechilishi:
  
; Javob: 
.
2) Kvadratning tomoni 250 mm. Kvadratning yuzini toping.
Yechilishi:
󰇛

󰇜

; Javob: 
.
3) To‘g‘ri burchakli uchburchakning katetlari 12 sm va 18 sm. Shu uchburchakning yuzini toping.
Yechilishi:
1)
    
; Javob: 
.
4) Parallelogrammning asosi 0,25 m va balandligi 100 sm bo‘lsa, uning yuzini toping.
Yechilishi:
  bo‘lgani uchun:   deb olamiz:
  
; Javob: 
.
3. Uyga vazifa: Yuz(a), tengdosh shakllar, yuz(a)ni o‘lchash, to‘g‘ri to‘rtburchakning yuzi,
parallelogrammning yuzi, uchburchakning yuzi, ularning xossalari kabi tushunchalarni
mustahkamlash uchun o‘tilgan mavzularni takrorlash, o‘qib o‘rganish va masalalar yechish.

Unformatted Attachment Preview

8-sinf, geometriya. (2 soatlik bayonnoma) «20-22» -dekabr, 2018-yil. Mavzu: 3-nazorat ishi. 1. Darsning maqsadi: Yuz(a), tengdosh shakllar, yuz(a)ni o‘lchash, to‘g‘ri to‘rtburchakning yuzi, parallelogrammning yuzi, uchburchakning yuzi, ularning xossalari haqida tushunchalar hosil qilish, o‘quvchilarning BKM larini shakllantirish, rivojlantirish va mustahkamlash hamda tekshirish. 2. Darsning usuli: yozma ish. 3. Darsning jihozi: DTS ko‘rgazmalari, tarqatma materiallar. DARSNING BORISHI: 1. Tashkiliy qism: o‘quvchilar bilan salomlashish, tozalikni tekshirish, davomatni aniqlash, o‘quvchilarning dars mashg‘ulotlariga ruhiy jihatda tayyorliklarini aniqlash. 2. Yangi mavzuning bayoni: (Ikki variantdagi yozma ish matni) 1-variant. 1) To‘g‘ri to‘rtburchakning ikki tomoni 60 sm va 5,8 sm. Uning yuzini toping. Yechilishi: 𝑆 = 𝑎 ∙ 𝑏 = 60 𝑠𝑚 ∙ 5,8 𝑠𝑚 = 340 𝑠𝑚2 ; Javob: 𝑆 = 340 𝑠𝑚2 . 2) Kvadratning tomoni 1,3 sm. Kvadratning yuzini toping. Yechilishi: 𝑆 = 𝑎2 = (1,3 𝑠𝑚)2 = 1,69 𝑠𝑚2 ; Javob: 𝑆 = 1,69 𝑠𝑚2 . 3) To‘g‘ri burchakli uchburchakning katetlari 5 sm va 6 sm. Shu uchburchakning yuzini toping. Yechilishi: 1 1 1) 𝑆 = 𝑎 ∙ 𝑏 = ∙ 5 𝑠𝑚 ∙ 6 𝑠𝑚 = 5 𝑠𝑚 ∙ 3 𝑠𝑚 = 15 𝑠𝑚2 ; Javob: 𝑆 = 15 𝑠𝑚2 . 2 2 4) Parallelogrammning asosi 60 sm va balandligi 0,5 m bo‘lsa, uning yuzini toping. Yechilishi: 𝑎 = 60 𝑠𝑚 = 0,6 𝑚 bo‘lgani uchun: 𝑎 = 0,6 𝑚, ℎ = 0,5 𝑚 deb olamiz: 𝑆 = 𝑎 ∙ ℎ = 0,6 𝑚 ∙ 0,5 𝑚 = 0,3 𝑚2 ; Javob: 𝑆 = 0,3 𝑚2. 2-variant. 1) To‘g‘ri to‘rtburchakning ikki tomoni 4 m va 1,4 m. Uning yuzini toping. Yechilishi: 𝑆 = 𝑎 ∙ 𝑏 = 4 𝑚 ∙ 1,4 𝑚 = 5,6 𝑚2 ; Javob: 𝑆 = 5,6 𝑚2. 2) Kvadratning tomoni 250 mm. Kvadratning yuzini toping. Yechilishi: 𝑆 = 𝑎2 = (250 𝑚𝑚)2 = 62 500 𝑚𝑚2; Javob: 𝑆 = 62 500 𝑚𝑚2 . 3) To‘g‘ri burchakli uchburchakning katetlari 12 sm va 18 sm. Shu uchburchakning yuzini toping. Yechilishi: 1 1 1) 𝑆 = 2 𝑎 ∙ 𝑏 = 2 ∙ 12 𝑠𝑚 ∙ 18 𝑠𝑚 = 6 𝑠𝑚 ∙ 18 𝑠𝑚 = 108 𝑠𝑚2; Javob: 𝑆 = 108 𝑠𝑚2. 4) Parallelogrammning asosi 0,25 m va balandligi 100 sm bo‘lsa, uning yuzini toping. Yechilishi: 𝑎 = 0,25 𝑚 = 25 𝑠𝑚 bo‘lgani uchun: 𝑎 = 25 𝑠𝑚, ℎ = 100 𝑠𝑚 deb olamiz: 𝑆 = 𝑎 ∙ ℎ = 25 𝑠𝑚 ∙ 100 𝑠𝑚 = 2 500 𝑠𝑚2 ; Javob: 𝑆 = 2 500 𝑠𝑚2 . 3. Uyga vazifa: Yuz(a), tengdosh shakllar, yuz(a)ni o‘lchash, to‘g‘ri to‘rtburchakning yuzi, parallelogrammning yuzi, uchburchakning yuzi, ularning xossalari kabi tushunchalarni mustahkamlash uchun o‘tilgan mavzularni takrorlash, o‘qib o‘rganish va masalalar yechish. Name: Description: ...
User generated content is uploaded by users for the purposes of learning and should be used following Studypool's honor code & terms of service.
Studypool
4.7
Trustpilot
4.5
Sitejabber
4.4