O'xshash uchburchaklar

Bu nima va nega kerak

Xaritalar, ekran masshtabi yoki bino maketlarini ko'rganmiz. Ular asliga nisbatan kichikroq, lekin shakli mutlaqo bir xil saqlanadi: burchaklar o'zgarmaydi, masofalar esa ma'lum nisbatda o'zgaradi.

Geometriyada shakli bir xil, lekin o'lchamlari nechtadir marta kattalashgan yoki kichraygan figuralar o'xshash figuralar deyiladi. 7-sinfda o'rganilgan teng uchburchaklar — shakli ham, o'lchami ham bir xil bo'lgan figuralardir. O'xshash uchburchaklarda esa burchaklar teng saqlanadi, tomonlar esa proporsional ravishda o'zgaradi.

Qadimgi yunon olimi Fales Misr piramidalarining balandligini aynan o'xshash uchburchaklar yordamida — quyosh nuri tushirgan soyalarni solishtirib aniqlagan.

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Asosiy g'oya va ta'riflar

O'xshash uchburchaklar ta'rifi

[!IMPORTANT] Ta'rif. Agar ikkita uchburchakning mos burchaklari teng va mos tomonlari proporsional bo'lsa, bunday uchburchaklar o'xshash uchburchaklar deyiladi.

Belgilanishi: $\triangle ABC \sim \triangle A_1B_1C_1$.

        B                      B1
       / \                    /  \
      /   \                  /    \
     /_____\                /______\
    A       C              A1       C1

Shartlar:

1. Mos burchaklar teng: $\angle A = \angle A_1, \quad \angle B = \angle B_1, \quad \angle C = \angle C_1$
2. Mos tomonlar proporsional: $\frac{AB}{A_1B_1} = \frac{BC}{B_1C_1} = \frac{AC}{A_1C_1} = k$

Bu yerda $k$ — o'xshashlik koeffitsiyenti. Agar $k = 1$ bo'lsa, uchburchaklar teng bo'ladi ($\triangle ABC \cong \triangle A_1B_1C_1$).

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Asosiy teoremalar va xossalar

1. Perimetrlar nisbati

[!IMPORTANT] Teorema. O'xshash uchburchaklar perimetrlarining nisbati o'xshashlik koeffitsiyentiga teng: $\frac{P(\triangle ABC)}{P(\triangle A_1B_1C_1)} = k$

2. Mos chiziqli elementlar nisbati

O'xshash uchburchaklarda mos balandliklar, medianalar, bissektrisalar va aylanalar radiuslari nisbati ham $k$ ga teng: $\frac{h_a}{h_{a1}} = \frac{m_a}{m_{a1}} = \frac{l_a}{l_{a1}} = \frac{r}{r_1} = \frac{R}{R_1} = k$

3. Yuzalar nisbati teoremasi

[!IMPORTANT] Teorema. O'xshash uchburchaklar yuzalarining nisbati o'xshashlik koeffitsiyentining kvadratiga teng: $\frac{S}{S_1} = k^2$

Tomon $k$ marta, unga tushirilgan balandlik ham $k$ marta kattalashgani uchun yuza $k \cdot k = k^2$ marta ortadi.

4. Asosiy lemma

[!IMPORTANT] Lemma. Uchburchakning ikki tomonini kesib o'tuvchi va uchinchi tomoniga parallel to'g'ri chiziq berilgan uchburchakka o'xshash uchburchak ajratadi. Agar $MN \parallel AC$ bo'lsa, $\triangle MBN \sim \triangle ABC$.

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Yechilgan misollar

1-misol. Tomonlar proporsionalligi bo'yicha hisoblash

$\triangle ABC \sim \triangle A_1B_1C_1$. Tomonlar: $AB = 6\text{ sm}, BC = 8\text{ sm}, AC = 10\text{ sm}$. Agar $\triangle A_1B_1C_1$ ning eng kichik tomoni $9\text{ sm}$ bo'lsa, qolgan tomonlarini va perimetrini toping.

• Eng kichik tomon $AB = 6\text{ sm}$.
• Koeffitsiyent: $k = \frac{9}{6} = 1{,}5$.
• Qolgan tomonlar: $B_1C_1 = 1{,}5 \cdot 8 = 12\text{ sm}$, $A_1C_1 = 1{,}5 \cdot 10 = 15\text{ sm}$.
• Perimetr: $P = 9 + 12 + 15 = 36\text{ sm}$.

Javob: $12\text{ sm}, 15\text{ sm}$, perimetri $36\text{ sm}$.

2-misol. Perimetrlar nisbati orqali tomonni topish

Ikkita o'xshash uchburchakning perimetrlari $24\text{ sm}$ va $36\text{ sm}$. Birinchi uchburchakning bir tomoni $8\text{ sm}$ bo'lsa, ikkinchi uchburchakning unga mos tomonini toping.

• Koeffitsiyent: $k = \frac{P_1}{P_2} = \frac{24}{36} = \frac{2}{3}$.
• Tomon: $\frac{8}{a_2} = \frac{2}{3} \implies 2 a_2 = 24 \implies a_2 = 12\text{ sm}$.

Javob: $12\text{ sm}$.

3-misol. Yuzalar nisbati va noma'lum tomon

Ikkita o'xshash uchburchakning yuzalari $18\text{ sm}^2$ va $50\text{ sm}^2$. Katta uchburchakning bir tomoni $15\text{ sm}$ bo'lsa, kichik uchburchakning mos tomonini toping.

• Yuzalar nisbati: $\frac{S_1}{S_2} = \frac{18}{50} = \frac{9}{25} = k^2 \implies k = \frac{3}{5}$.
• Mos tomon: $a_1 = 15 \cdot \frac{3}{5} = 9\text{ sm}$.

Javob: $9\text{ sm}$.

4-misol. Soya yordamida balandlikni aniqlash

Bo'yi $1{,}8\text{ m}$ bo'lgan kishining soyasi $1{,}2\text{ m}$. Daraxtning soyasi esa $8\text{ m}$. Daraxtning balandligini toping.

• Uchburchaklar o'xshashligidan: $\frac{H}{1.8} = \frac{8}{1.2} \implies H = \frac{1.8 \cdot 8}{1.2} = 12\text{ m}$

Javob: $12\text{ m}$.

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Mashq

1. $\triangle ABC \sim \triangle A_1B_1C_1$. Koeffitsiyent $k = 3$. Kichik uchburchakning perimetri $16\text{ sm}$ bo'lsa, katta uchburchak perimetrini toping.
2. O'xshash uchburchaklar mos tomonlarining nisbati $4 : 5$. Katta uchburchak yuzi $75\text{ sm}^2$ bo'lsa, kichik uchburchak yuzini hisoblang.
3. Tomonlari $5\text{ sm}, 7\text{ sm}, 9\text{ sm}$ bo'lgan uchburchakka o'xshash uchburchakning perimetri $42\text{ sm}$. Uning tomonlarini toping.
4. O'xshash uchburchaklarning mos balandliklari $6\text{ sm}$ va $10\text{ sm}$. Ularning yuzalari nisbatini toping.
5. Vertikal tayoqning bo'yi $2\text{ m}$, soyasi $1{,}5\text{ m}$. Minora soyasi $30\text{ m}$ bo'lsa, minora balandligini hisoblang.
6. $\triangle ABC$ da $AC$ asosga parallel o'rta chiziq o'tkazilgan. Ajralgan kichik uchburchak yuzi $\triangle ABC$ yuzining qanday qismini tashkil qiladi?
7. Ikkita o'xshash uchburchakning yuzalari $32\text{ sm}^2$ va $50\text{ sm}^2$. Ularning perimetrlari nisbatini toping.
8. $\triangle ABC \sim \triangle DEF$. $AB = 8\text{ sm}, BC = 10\text{ sm}, AC = 12\text{ sm}$. Agar $DE = 12\text{ sm}$ bo'lsa, $EF$ va $DF$ tomonlarni toping.
9. Uchburchak asosi $18\text{ sm}$, balandligi $12\text{ sm}$. Asosga parallel kesma balandlikni uchidan boshlab $1:2$ nisbatda bo'lsa, hosil bo'lgan kichik uchburchak yuzi butun uchburchak yuzining qanday qismini tashkil etadi?
10. O'xshash uchburchaklarning perimetrlari $18\text{ sm}$ va $27\text{ sm}$, yuzalari yig'indisi $65\text{ sm}^2$. Har bir uchburchakning yuzini toping.

Javoblar va yechim yo'llari

1. $P_2 = 3 \cdot 16 = 48\text{ sm}$. Javob: 48 sm.
2. $\dfrac{S_1}{S_2} = \left(\dfrac{4}{5}\right)^2 = \dfrac{16}{25} \implies S_1 = 75 \cdot \dfrac{16}{25} = 48\text{ sm}^2$. Javob: 48 sm².
3. $P_1 = 5 + 7 + 9 = 21\text{ sm}$. $k = 42 : 21 = 2$. Tomonlar: $10\text{ sm}, 14\text{ sm}, 18\text{ sm}$. Javob: 10 sm, 14 sm, 18 sm.
4. $\left(\dfrac{6}{10}\right)^2 = \left(\dfrac{3}{5}\right)^2 = \dfrac{9}{25}$. Javob: 9 : 25.
5. $\dfrac{H}{2} = \dfrac{30}{1{,}5} = 20 \implies H = 40\text{ m}$. Javob: 40 m.
6. O'rta chiziq uchun $k = \dfrac{1}{2}$. Yuza $k^2 = \left(\dfrac{1}{2}\right)^2 = \dfrac{1}{4}$ qismini tashkil qiladi. Javob: 1/4 qismini.
7. $\dfrac{S_1}{S_2} = \dfrac{32}{50} = \dfrac{16}{25} \implies k = \dfrac{4}{5}$. Perimetrlar nisbati ham $4:5$. Javob: 4 : 5.
8. $k = 12 : 8 = 1{,}5$. $EF = 10 \cdot 1{,}5 = 15\text{ sm}$, $DF = 12 \cdot 1{,}5 = 18\text{ sm}$. Javob: EF = 15 sm, DF = 18 sm.
9. Balandlik qismlari $1x$ va $2x$, to'liq balandlik $3x$. $k = \dfrac{1}{3} \implies k^2 = \dfrac{1}{9}$. Javob: 1/9 qismini.
10. $k = \dfrac{18}{27} = \dfrac{2}{3} \implies \dfrac{S_1}{S_2} = \dfrac{4}{9}$. $4x + 9x = 65 \implies 13x = 65 \implies x = 5$. $S_1 = 20\text{ sm}^2, S_2 = 45\text{ sm}^2$. Javob: 20 sm² va 45 sm².

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Keng tarqalgan xatolar

1. Yuzalar nisbatini $k$ ga teng deb olish. ✗ Tomonlari 2 marta katta uchburchakning yuzi ham 2 marta katta deb hisoblash. ✓ Yuza $k^2$ marta ortadi: $\dfrac{S_1}{S_2} = k^2 = 2^2 = 4$.

2. Mos tomonlarni noto'g'ri taqqoslash. ✗ Eng katta tomonni boshqa uchburchakning ixtiyoriy tomoniga bo'lish. ✓ Proporsiya faqat teng burchaklar qarshisidagi mos tomonlar orasida tuziladi.

3. Perimetrlar nisbatini kvadratga ko'tarish. ✗ $\dfrac{P_1}{P_2} = k^2$ deb xato qilish. ✓ Perimetr chiziqli kattalik, shuning uchun $\dfrac{P_1}{P_2} = k$.

4. Harflar ketma-ketligini buzish. ✗ $\triangle ABC \sim \triangle A_1B_1C_1$ da mos burchaklar tartibini aralashtirish. ✓ O'xshashlik belgilanishida mos uchlar aynan bir xil o'rinda yozilishi shart.

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DTM uslubidagi test

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

Keyingi qadam

O'xshash uchburchaklar xossalarini o'rgandik. Keyingi darsda proporsional kesmalar haqidagi klassik qonun — Fales teoremasi bilan tanishamiz.