Geometriya 11-sinf Mavzu: Hajm tushunchasi. Parallelepipedning hajmi Reja ο Hajm tushunchasi ο Parallelepipedning hajmi ο Kubning hajmi ο Mavzuga oid masalalar yechish πππ-masala (157-sahifa) Oktaedr bitta yog‘ining yuzi π, π πππ bo‘lsa, uning to‘la sirtini toping. Berilgan: πΊ = π, π πππ π=π Topish kerak: πΊππ‘ππ = ? Masalaning yechimi: πΊππ‘ππ = π β πΊ πΊππ‘ππ = π β π, π = ππ Javob: ππ ππ π πππ-masala (158-sahifa) Agar kubning qirrasi 1 birlik orttirilsa, uning to‘la sirti 54 birlikka ortadi. Kubning qirrasini toping. Yechish: Berilgan: π πΊ = ππ πΊ π π π+π πΊ + ππ πΊ + ππ = π β π + π Topish kerak: πππ + ππ = π β ππ + ππ + π π π π=? ππ + ππ = ππ + πππ + π ππ = πππ π = π Javob: π πππ-masala (158-sahifa) To‘rtburchakli muntazam prizma asosining tomoni 6 cm, prizmaning balandligi 5 cm. Prizmaning diagonal kesimi yuzini toping. Berilgan: π = π ππ π = π ππ Topish kerak: πΊπ = ? Yechish: π =π π=π π πΊπ = π β π πΊπ = π β π π = ππ π Javob: ππ π ππ π π π π π Hajm tushunchasi Hajm tushunchasi Hajm – fazoviy jismning quyidagi xossalarga ega bo‘lgan miqdoriy ko‘rsatkichidir: π½>π Hajm – fazoviy jismning quyidagi xossalarga ega bo‘lgan miqdoriy ko‘rsatkichidir: π π π Hajm o‘lchov birliklari ππ , π ππ , πππ , πππ , π π½=π Parallelepipedning hajmi To‘g‘ri burchakli parallelepipedning hajmi uning uchta o‘lchamlari ko‘paytmasiga teng: π π π π½=πβπβπ Parallelepipedning hajmi π πΊππππ To‘g‘ri burchakli parallelepipedning hajmi asosining yuzi bilan balandligining ko‘paytmasiga teng: π½ = πΊππππ β π Parallelepipedning hajmi Ixtiyoriy parallelepipedning hajmi asosining yuzi bilan balandligining ko‘paytmasiga teng: π½ = πΊππππ β π π Kubning hajmi π½=πβπβπ π π½=π π π π 1-masala Rasmda berilgan yoyilmaga ko‘ra yasalgan idishning hajmini toping. ππ ππ π ππ ππ ππ π ππ Masalaning yechimi π = ππ − π β π = ππ − ππ = ππ π = ππ − π β π = ππ − ππ = ππ π π½ = π β π β π = ππ β ππ β π = ππππ ππ π=π ππ ππ π ππ ππ ππ π π π ππ π Javob: ππππ πππ 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: π½ = πΊππππ β π π = ππ ππ πΊππππ = ππ ππ π½ = ππ β ππ = πππ π = ππ Topish kerak: Javob: πππ πππ π½=? 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: π½=πβπβπ π=π π=π π½ = π β π β π = πππ π=π π½ = ππππ π Topish kerak: ππππ = ? π π π 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: π=π π=π π=π Topish kerak: ππππ = ? Yechish: π½=πβπβπ π½ = π β π β π = πππ π½ = ππππ π ππππ π = πππ = ππ ππππ = π Javob: π π π π 4-masala Agar kubning qirralarini 1 birlik orttirilsa, uning hajmi 19 birlikka ortadi. Kubning qirrasini toping. Berilgan: π½ π π+π π½ + ππ Topish kerak: π=? Yechish: π½ = ππ π½ + ππ = π + π π ππ + ππ = ππ + πππ + ππ + π πππ + ππ − ππ = π ππ + π − π = π ππ = −π ππ = π Javob: π ππ 5-masala Birinchi kubning hajmi ikkinchisinikidan 8 marta katta. Birinchi kubning to‘la sirtining yuzi ikkinchisinikidan necha marta katta? Berilgan: Yechish: π π π½π = π½π β π ππ = ππ β π Topish kerak: πΊπ =? πΊπ π π ππ = ππ β ππ πππ = ππ β π π ππ = ππ β π πΊπ π β πππ πππ = = π= π πΊπ π β ππ ππ ππ ππ π ππ β π = ππ π =π πΊπ = πΊπ β π Javob: π ππ ππ ππ ππ ππ ππ Mustaqil bajarish uchun topshiriqlar 165-166-sahifalar 209 220 213 223