Matematika

8-sinf · Geometriya · 25 daqiqa

Pifagor teoremasi

Yuzalar: mavzular

Bu nima va nega kerak

Qurilish va me'morchilikda burchaklarning to'g'ri (90∘90^\circ) bo'lishi eng muhim shartdir. Agar bino poydevori yoki devorlari to'g'ri burchak ostida qurilmasa, inshoot mustahkam bo'lmaydi. Qadimgi Misrda muhandislar 12 ta teng bo'lakka tugilgan arqondan foydalanib, tomonlari 3, 4 va 5 bo'lgan uchburchak yasaganlar va to'g'ri burchak hosil qilganlar.

Buning sababi: 32+42=9+16=25=523^2 + 4^2 = 9 + 16 = 25 = 5^2.

Eramizdan avvalgi VI asrda yunon olimi Pifagor tomonidan isbotlangan bu qonuniyat Pifagor teoremasi deb ataladi. U to'g'ri burchakli uchburchak tomonlari orasidagi bog'liqlikni ifodalovchi asosiy qonundir.


Asosiy g'oya va ta'riflar

To'g'ri burchakli uchburchakda:

  • To'g'ri burchak hosil qiluvchi tomonlar — katetlar (aa va bb);
  • To'g'ri burchak qarshisidagi tomon — gipotenuza (cc).
          B
         /|
        / |
     c /  | a
      /   |
     /____|
    A  b  C

Chizmada: ∠C=90∘\angle C = 90^\circ, a,ba, b — katetlar, cc — gipotenuza.

1. Pifagor teoremasi

[!IMPORTANT] Pifagor teoremasi. To'g'ri burchakli uchburchakda gipotenuzaning kvadrati katetlar kvadratlarining yig'indisiga teng: a2+b2=c2a^2 + b^2 = c^2

Noma'lum tomonlarni topish:

  • Gipotenuza: c=a2+b2c = \sqrt{a^2 + b^2}
  • Katetlar: a=c2−b2a = \sqrt{c^2 - b^2}, b=c2−a2b = \sqrt{c^2 - a^2}

Isbot g'oyasi (yuzalar orqali)

Tomoni a+ba + b bo'lgan kvadrat chizib, uning burchaklariga 4 ta to'g'ri burchakli uchburchak joylashtirilsa, o'rtada tomoni cc bo'lgan kvadrat hosil bo'ladi.

  • Katta kvadrat yuzi: S=(a+b)2=a2+2ab+b2S = (a + b)^2 = a^2 + 2ab + b^2
  • Shakllar yig'indisi: S=4⋅(12ab)+c2=2ab+c2S = 4 \cdot (\frac{1}{2}ab) + c^2 = 2ab + c^2
  • Tenglashtirib: a2+2ab+b2=2ab+c2  ⟹  a2+b2=c2a^2 + 2ab + b^2 = 2ab + c^2 \implies a^2 + b^2 = c^2.

2. Pifagor uchliklari

Tomonlari butun sonlardan iborat to'g'ri burchakli uchburchaklar:

  • (3,4,5)(3, 4, 5) — Misr uchburchagi;
  • (5,12,13)(5, 12, 13);
  • (8,15,17)(8, 15, 17);
  • (7,24,25)(7, 24, 25).

Ularning karralilari ham to'g'ri burchakli uchburchak beradi: masalan, (6,8,10)(6, 8, 10), (9,12,15)(9, 12, 15).

3. Teskari teorema

[!IMPORTANT] Teskari teorema. Agar uchburchak tomonlari uchun a2+b2=c2a^2 + b^2 = c^2 bajarilsa, u to'g'ri burchakli uchburchak bo'ladi.

Burchak turini aniqlash (cc — eng katta tomon):

  1. a2+b2=c2  ⟹  a^2 + b^2 = c^2 \implies to'g'ri burchakli;
  2. a2+b2>c2  ⟹  a^2 + b^2 > c^2 \implies o'tkir burchakli;
  3. a2+b2<c2  ⟹  a^2 + b^2 < c^2 \implies o'tmas burchakli.

4. Metrik munosabatlar

To'g'ri burchakli uchburchakda gipotenuzaga hch_c balandlik tushirilsa:

  • hc2=ac⋅bc  ⟹  hc=ac⋅bch_c^2 = a_c \cdot b_c \implies h_c = \sqrt{a_c \cdot b_c}
  • a2=c⋅aca^2 = c \cdot a_c, b2=c⋅bcb^2 = c \cdot b_c

Yechilgan misollar

1-misol. Gipotenuzani hisoblash

To'g'ri burchakli uchburchakning katetlari 12 sm12\text{ sm} va 16 sm16\text{ sm}. Gipotenuzani toping.

  • c2=122+162=144+256=400c^2 = 12^2 + 16^2 = 144 + 256 = 400.
  • c=400=20 smc = \sqrt{400} = 20\text{ sm}.

Javob: 20 sm20\text{ sm}.

2-misol. Katet va yuzani hisoblash

Gipotenuzasi 25 sm25\text{ sm}, bitta kateti 7 sm7\text{ sm} bo'lgan to'g'ri burchakli uchburchakning ikkinchi katetini va yuzini toping.

  • Katet: b=252−72=625−49=576=24 smb = \sqrt{25^2 - 7^2} = \sqrt{625 - 49} = \sqrt{576} = 24\text{ sm}.
  • Yuza: S=12⋅7⋅24=84 sm2S = \frac{1}{2} \cdot 7 \cdot 24 = 84\text{ sm}^2.

Javob: Katet 24 sm24\text{ sm}, yuza 84 sm284\text{ sm}^2.

3-misol. Teng yonli uchburchak balandligi

Teng yonli uchburchakning yon tomoni 13 sm13\text{ sm}, asosi 10 sm10\text{ sm}. Asosga tushirilgan balandlikni toping.

  • Asosning yarmi: 10:2=5 sm10 : 2 = 5\text{ sm}.
  • Balandlik: h=132−52=169−25=144=12 smh = \sqrt{13^2 - 5^2} = \sqrt{169 - 25} = \sqrt{144} = 12\text{ sm}.

Javob: 12 sm12\text{ sm}.

4-misol. Rombning tomoni va perimetri

Rombning diagonallari 12 sm12\text{ sm} va 16 sm16\text{ sm}. Uning tomoni va perimetrini toping.

  • Yarim diagonallar: 6 sm6\text{ sm} va 8 sm8\text{ sm}.
  • Romb tomoni gipotenuza bo'ladi: a=62+82=100=10 sma = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ sm}.
  • Perimetr: P=4⋅10=40 smP = 4 \cdot 10 = 40\text{ sm}.

Javob: Tomoni 10 sm10\text{ sm}, perimetri 40 sm40\text{ sm}.


Mashq

  1. Katetlari 6 sm6\text{ sm} va 8 sm8\text{ sm} bo'lgan to'g'ri burchakli uchburchakning gipotenuzasini toping.
  2. Gipotenuzasi 17 sm17\text{ sm}, kateti 15 sm15\text{ sm} bo'lgan uchburchakning ikkinchi katetini hisoblang.
  3. To'g'ri to'rtburchakning tomonlari 5 sm5\text{ sm} va 12 sm12\text{ sm}. Uning diagonalini toping.
  4. Rombning diagonallari 16 sm16\text{ sm} va 30 sm30\text{ sm}. Uning perimetrini toping.
  5. Teng yonli uchburchakning asosi 16 sm16\text{ sm}, yon tomoni 10 sm10\text{ sm}. Asosga tushirilgan balandlikni hisoblang.
  6. Tomonlari 9 sm,12 sm,15 sm9\text{ sm}, 12\text{ sm}, 15\text{ sm} bo'lgan uchburchak to'g'ri burchakli bo'ladimi?
  7. Tomonlari 7 sm,8 sm7\text{ sm}, 8\text{ sm} va 11 sm11\text{ sm} bo'lgan uchburchak burchaklariga ko'ra qanday uchburchak?
  8. To'g'ri burchakli uchburchakda gipotenuzaga tushirilgan balandlik gipotenuzani 4 sm4\text{ sm} va 9 sm9\text{ sm} bo'laklarga ajratadi. Shu balandlikni toping.
  9. Kvadratning diagonali 82 sm8\sqrt{2}\text{ sm}. Uning tomoni va yuzini toping.
  10. Zinapoyaning uzunligi 10 m10\text{ m}, u devor tagidan 6 m6\text{ m} uzoqlikda yerga tirab qo'yilgan. Zinapoyaning yuqori uchi devorning qanday balandligiga yetadi?
Javoblar va yechim yo'llari
  1. c=62+82=36+64=10 smc = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = 10\text{ sm}. Javob: 10 sm.
  2. a=172−152=289−225=8 sma = \sqrt{17^2 - 15^2} = \sqrt{289 - 225} = 8\text{ sm}. Javob: 8 sm.
  3. d=52+122=25+144=13 smd = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = 13\text{ sm}. Javob: 13 sm.
  4. Yarim diagonallar 88 va 15 sm15\text{ sm}. Tomoni a=82+152=17 sma = \sqrt{8^2 + 15^2} = 17\text{ sm}. P=4⋅17=68 smP = 4 \cdot 17 = 68\text{ sm}. Javob: 68 sm.
  5. Asosning yarmi 8 sm8\text{ sm}. h=102−82=6 smh = \sqrt{10^2 - 8^2} = 6\text{ sm}. Javob: 6 sm.
  6. 92+122=81+144=225=1529^2 + 12^2 = 81 + 144 = 225 = 15^2. Javob: Ha, to'g'ri burchakli.
  7. Katta tomon 1111. 72+82=113<112=1217^2 + 8^2 = 113 < 11^2 = 121. Javob: O'tmas burchakli.
  8. h=4⋅9=6 smh = \sqrt{4 \cdot 9} = 6\text{ sm}. Javob: 6 sm.
  9. d=a2  ⟹  a=8 smd = a\sqrt{2} \implies a = 8\text{ sm}, S=82=64 sm2S = 8^2 = 64\text{ sm}^2. Javob: Tomoni 8 sm, yuzi 64 sm².
  10. h=102−62=100−36=8 mh = \sqrt{10^2 - 6^2} = \sqrt{100 - 36} = 8\text{ m}. Javob: 8 m.

Keng tarqalgan xatolar

1. Katetni topishda kvadratlarni qo'shib yuborish. ✗ a=c2+b2a = \sqrt{c^2 + b^2} deb hisoblash. ✓ Katet doimo gipotenuzadan qisqa, shuning uchun ayirish lozim: a=c2−b2a = \sqrt{c^2 - b^2}.

2. Pifagorni to'g'ri burchakli bo'lmagan uchburchaklarga qo'llash. ✗ Ixtiyoriy uchburchakda a2+b2=c2a^2 + b^2 = c^2 munosabatni ishlatish. ✓ Pifagor teoremasi faqat to'g'ri burchakli uchburchak (∠=90∘\angle = 90^\circ) uchungina o'rinlidir.

3. Ildiz chiqarish amalini noto'g'ri tushunish. ✗ a2+b2=a+b\sqrt{a^2 + b^2} = a + b deb yozish (masalan, 9+16=3+4=7\sqrt{9 + 16} = 3 + 4 = 7). ✓ Avval kvadratlar yig'indisi hisoblanadi, so'ngra ildiz olinadi: 9+16=25=5\sqrt{9 + 16} = \sqrt{25} = 5.

4. Burchak turini aniqlashda eng katta tomonni tanlamaslik. ✗ Ixtiyoriy tomonni cc deb olib, a2+b2a^2 + b^2 bilan solishtirish. ✓ Taqqoslashda cc sifatida doimo uchburchakning eng katta tomoni olinishi shart.


DTM uslubidagi test

Ushbu mavzuni mustahkamlash uchun berilgan 5 ta test savolini ishlab, javoblaringizni tekshiring.

Keyingi qadam

Pifagor teoremasi orqali to'g'ri burchakli uchburchak elementlarini hisoblashni o'rgandik. Keyingi bosqichda burchaklari teng, lekin o'lchamlari proporsional bo'lgan figuralar — o'xshash uchburchaklar mavzusiga o'tamiz.

Test

0 / 5 javob berildi

  1. 1.To'g'ri to'rtburchakning tomonlari 8 sm8\text{ sm} va 15 sm15\text{ sm}. Uning diagonali uzunligini toping.

  2. 2.To'g'ri burchakli uchburchakning gipotenuzasi 20 sm20\text{ sm}, bitta kateti 16 sm16\text{ sm}. Ikkinchi katetini toping.

  3. 3.Teng yonli uchburchakning yon tomoni 15 sm15\text{ sm}, asosi 18 sm18\text{ sm}. Asosga tushirilgan balandligini toping.

  4. 4.Tomonlari 5 sm,6 sm5\text{ sm}, 6\text{ sm} va 7 sm7\text{ sm} bo'lgan uchburchak burchaklariga ko'ra qanday uchburchak?

  5. 5.To'g'ri burchakli uchburchakda gipotenuzaga tushirilgan balandlik uni 3 sm3\text{ sm} va 12 sm12\text{ sm} li kesmalarga bo'ladi. Shu balandlikni toping.