Uchburchak yuzi

Bu nima va nega kerak

Yer maydonlarini o'lchash, tom qoplamalari va qurilish konstruksiyalarini hisoblashda eng ko'p uchraydigan shakl bu uchburchakdir. Ko'pburchaklarni diagonallar orqali uchburchaklarga ajratish mumkin bo'lgani uchun, uchburchak yuzini hisoblash butun geometriya uchun tayanch hisoblanadi.

Parallelogramm yuzi $S = a \cdot h$ ekanini bilamiz. Agar parallelogramm diagonali bo'ylab kesilsa, ikkita teng uchburchak hosil bo'ladi. Shu sababli uchburchak yuzi unga mos parallelogramm yuzining aynan yarmiga tengdir: $S = \frac{1}{2} a h$.

Ushbu darsda biz uchburchak yuzini hisoblashning asosiy usullarini — balandlik orqali, to'g'ri burchakli uchburchak uchun, burchak sinusi yordamida hamda Geron formulasini o'rganamiz.

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Asosiy g'oya va ta'riflar

Uchburchak yuzi uning tekislikda egallagan sohasining o'lchovidir.

          B
         /|\
        / | \
       /  |  \
      /   |h  \
     /    |    \
    /_____|_____\
   A      H      C
      (asos, a)

Chizmada: $AC = a$ — asos, $BH = h_a$ — shu asosga tushirilgan balandlik.

1. Asosiy formula (asos va balandlik)

[!IMPORTANT] Teorema. Uchburchakning yuzi asosi bilan shu asosga tushirilgan balandligi ko'paytmasining yarmiga teng: $S = \frac{1}{2} a \cdot h_a = \frac{1}{2} b \cdot h_b = \frac{1}{2} c \cdot h_c$

Isbot g'oyasi: $ABC$ uchburchakni to'ldirib, asosi $a$ va balandligi $h$ bo'lgan parallelogramm hosil qilinadi. Parallelogramm yuzi $a \cdot h$ bo'lib, $BC$ diagonali uni ikkita teng uchburchakka bo'ladi. Demak, $S = \frac{1}{2} a h$.

2. To'g'ri burchakli uchburchak yuzi

To'g'ri burchakli uchburchakda katetlar bir-biriga perpendikulyar bo'lgani uchun bittasi asos, ikkinchisi balandlik bo'ladi:

[!IMPORTANT] $S = \frac{1}{2} a \cdot b$ Bu yerda $a, b$ — katetlar. Gipotenuzaga tushirilgan balandlik $h_c$ uchun: $S = \frac{1}{2} c \cdot h_c \implies h_c = \frac{a \cdot b}{c}$.

3. Ikki tomon va ular orasidagi burchak orqali

[!IMPORTANT] $S = \frac{1}{2} a b \sin \gamma$ Uchburchak yuzi ikki tomoni va ular orasidagi burchak sinusi ko'paytmasining yarmiga teng.

Chunki balandlik $h_a = b \sin \gamma$ ga teng.

4. Geron formulasi (uch tomon ma'lum bo'lganda)

[!IMPORTANT] $S = \sqrt{p(p-a)(p-b)(p-c)}$ Bu yerda $p = \dfrac{a + b + c}{2}$ — yarim perimetr.

5. Teng tomonli uchburchak yuzi

Barcha tomonlari $a$ va burchaklari $60^\circ$ bo'lgani uchun: $S = \frac{a^2 \sqrt{3}}{4}$

6. Mediananing yuzaga doir xossasi

[!NOTE] Yuzalari teng shakllar tengdosh deyiladi. Uchburchakning medianasi uni ikkita tengdosh uchburchakka ajratadi ($S_1 = S_2$). Uchala mediana kesishganda uchburchak 6 ta tengdosh bo'lakka bo'linadi.

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Yechilgan misollar

1-misol. Asos va balandlik orqali yuzani hisoblash

Uchburchakning asosi $18\text{ sm}$, unga tushirilgan balandligi esa $12\text{ sm}$. Uchburchakning yuzini toping.

• Asosiy formula: $S = \frac{1}{2} a h$.
• Hisoblash: $S = \frac{1}{2} \cdot 18 \cdot 12 = 9 \cdot 12 = 108\text{ sm}^2$.

Javob: $108\text{ sm}^2$.

2-misol. To'g'ri burchakli uchburchak balandligi

Katetlari $9\text{ sm}$ va $12\text{ sm}$ bo'lgan to'g'ri burchakli uchburchakning gipotenuzasiga tushirilgan balandligini toping.

• Gipotenuza: $c = \sqrt{9^2 + 12^2} = \sqrt{81 + 144} = \sqrt{225} = 15\text{ sm}$.
• Yuza katetlar orqali: $S = \frac{1}{2} \cdot 9 \cdot 12 = 54\text{ sm}^2$.
• Balandlik: $h_c = \frac{2S}{c} = \frac{108}{15} = 7{,}2\text{ sm}$.

Javob: $7{,}2\text{ sm}$.

3-misol. Geron formulasi bilan hisoblash

Tomonlari $13\text{ sm}, 14\text{ sm}$ va $15\text{ sm}$ bo'lgan uchburchakning yuzini toping.

• Yarim perimetr: $p = \frac{13 + 14 + 15}{2} = \frac{42}{2} = 21\text{ sm}$.
• Ayirmalar: $p - a = 8$, $p - b = 7$, $p - c = 6$.
• Geron formulasi: $S = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{(7 \cdot 3) \cdot 8 \cdot 7 \cdot (2 \cdot 3)} = \sqrt{7^2 \cdot 3^2 \cdot 16} = 7 \cdot 3 \cdot 4 = 84\text{ sm}^2$

Javob: $84\text{ sm}^2$.

4-misol. Burchak sinusi orqali yuzani topish

Teng yonli uchburchakning yon tomoni $10\text{ sm}$, uchidagi burchagi $120^\circ$. Yuzini toping.

• Formula: $S = \frac{1}{2} a b \sin \gamma$.
• Hisoblash: $\sin 120^\circ = \frac{\sqrt{3}}{2}$. $S = \frac{1}{2} \cdot 10 \cdot 10 \cdot \frac{\sqrt{3}}{2} = 25\sqrt{3}\text{ sm}^2$

Javob: $25\sqrt{3}\text{ sm}^2$.

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Mashq

1. Asosi $16\text{ sm}$, unga tushirilgan balandligi $7\text{ sm}$ bo'lgan uchburchak yuzini hisoblang.
2. To'g'ri burchakli uchburchakning katetlari $7\text{ sm}$ va $24\text{ sm}$. Uning yuzini toping.
3. Teng tomonli uchburchakning tomoni $8\text{ sm}$. Uning yuzini hisoblang.
4. Uchburchakning yuzi $45\text{ sm}^2$, bitta tomoni $10\text{ sm}$. Shu tomonga tushirilgan balandlikni toping.
5. Tomonlari $5\text{ sm}, 5\text{ sm}$ va $6\text{ sm}$ bo'lgan teng yonli uchburchakning yuzini toping.
6. Uchburchakning ikki tomoni $8\text{ sm}$ va $12\text{ sm}$, ular orasidagi burchak $150^\circ$. Yuzini toping.
7. Tomonlari $10\text{ sm}, 17\text{ sm}$ va $21\text{ sm}$ bo'lgan uchburchak yuzini Geron formulasi bilan toping.
8. To'g'ri burchakli uchburchakning gipotenuzasi $26\text{ sm}$, bitta kateti $10\text{ sm}$. Uning yuzini toping.
9. $ABC$ uchburchak yuzi $48\text{ sm}^2$. $AD$ mediana o'tkazilgan. $\triangle ABD$ yuzini toping.
10. Rombning diagonallari $14\text{ sm}$ va $18\text{ sm}$. Diagonallar hosil qilgan 4 ta uchburchakdan birining yuzini toping.

Javoblar va yechim yo'llari

1. $S = \dfrac{1}{2} \cdot 16 \cdot 7 = 56\text{ sm}^2$. Javob: 56 sm².
2. $S = \dfrac{1}{2} \cdot 7 \cdot 24 = 84\text{ sm}^2$. Javob: 84 sm².
3. $S = \dfrac{a^2 \sqrt{3}}{4} = \dfrac{64\sqrt{3}}{4} = 16\sqrt{3}\text{ sm}^2$. Javob: 16√3 sm².
4. $45 = \dfrac{1}{2} \cdot 10 \cdot h \implies 5h = 45 \implies h = 9\text{ sm}$. Javob: 9 sm.
5. Asosga balandlik tushirilsa, u asosni $3\text{ sm}$ dan bo'ladi. $h = \sqrt{5^2 - 3^2} = 4\text{ sm}$. $S = \dfrac{1}{2} \cdot 6 \cdot 4 = 12\text{ sm}^2$. Javob: 12 sm².
6. $\sin 150^\circ = \sin 30^\circ = 0{,}5$. $S = \dfrac{1}{2} \cdot 8 \cdot 12 \cdot 0{,}5 = 24\text{ sm}^2$. Javob: 24 sm².
7. $p = (10+17+21):2 = 24$. $S = \sqrt{24 \cdot 14 \cdot 7 \cdot 3} = \sqrt{7056} = 84\text{ sm}^2$. Javob: 84 sm².
8. Ikkinchi katet: $\sqrt{26^2 - 10^2} = 24\text{ sm}$. $S = \dfrac{1}{2} \cdot 10 \cdot 24 = 120\text{ sm}^2$. Javob: 120 sm².
9. Mediana uchburchakni teng ikkiga ajratadi: $S_{ABD} = 48 : 2 = 24\text{ sm}^2$. Javob: 24 sm².
10. Diagonallar rombni 4 ta teng to'g'ri burchakli uchburchakka bo'ladi: $S_1 = \dfrac{1}{2} \cdot 7 \cdot 9 = 31{,}5\text{ sm}^2$. Javob: 31.5 sm².

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Keng tarqalgan xatolar

1. Balandlikni o'ziga tegishli bo'lmagan tomonga ko'paytirish. ✗ $a$ tomonni $b$ tomonga tushirilgan balandlik $h_b$ ga ko'paytirish. ✓ Balandlik faqat o'zi tushirilgan asosga ko'paytiriladi: $S = \dfrac{1}{2} a \cdot h_a = \dfrac{1}{2} b \cdot h_b$.

2. $\frac{1}{2}$ koeffitsiyentini yozishni unutish. ✗ Formulani $S = a \cdot h$ deb yozish. ✓ Uchburchak parallelogrammning yarmi bo'lgani uchun doimo 2 ga bo'linadi: $S = \dfrac{1}{2} a \cdot h$.

3. Geron formulasida to'liq perimetrni ishlatish. ✗ Formuladagi $p$ o'rniga perimetr $P = a + b + c$ ni qo'yish. ✓ Formuladagi $p$ — bu yarim perimetr, ya'ni $p = \dfrac{a + b + c}{2}$.

4. Burchak sinusini e'tibordan chetda qoldirish. ✗ Boshqa burchakli uchburchaklar uchun ham $S = \dfrac{1}{2} a \cdot b$ deb hisoblash. ✓ Bu faqat to'g'ri burchakli uchburchak uchun xos. Boshqa hollarda $S = \dfrac{1}{2} a b \sin \gamma$ formulasi ishlatiladi.

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DTM uslubidagi test

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

Keyingi qadam

Uchburchak yuzasi va to'g'ri burchakli uchburchak xossalarini mustahkamlab oldik. Endi to'g'ri burchakli uchburchak tomonlari orasidagi bog'liqlikni ifodalovchi eng asosiy qonun — Pifagor teoremasiga o'tamiz.