O'xshashlik alomatlari

Bu nima va nega kerak

Smartfonda fotosuratni kattalashtirganda (zoom) rasm o'lchami o'zgaradi, lekin shakli buzilmaydi. Burchaklar saqlanadi, barcha masofalar esa bir xil nisbatda kattalashadi yoki kichrayadi.

Geometriyada bunday shakllar o'xshash shakllar deyiladi.

Ikkita uchburchak o'xshashligini aniqlash uchun ta'rifdagi 6 ta shartni (3 ta burchak tengligi va 3 ta tomon proporsionalligi) tekshirish shart emas. Buni osonlashtiruvchi o'xshashlik alomatlari mavjud.

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Asosiy g'oya va ta'riflar

Ta'rif. Agar $\triangle ABC$ va $\triangle A_1B_1C_1$ ning mos burchaklari teng: $\angle A = \angle A_1, \quad \angle B = \angle B_1, \quad \angle C = \angle C_1$ va mos tomonlari proporsional bo'lsa: $\frac{AB}{A_1B_1} = \frac{BC}{B_1C_1} = \frac{AC}{A_1C_1} = k,$ bu uchburchaklar o'xshash uchburchaklar deyiladi ($\triangle ABC \sim \triangle A_1B_1C_1$). $k$ — o'xshashlik koeffitsiyenti.

       B                         B1
      / \                       /  \
     /   \                     /    \
    /     \                   /      \
   A-------C                 A1--------C1

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Uchburchaklar o'xshashligining 3 ta alomati

1-alomat: Ikkita burchak bo'yicha (BB)

[!IMPORTANT] 1-alomat (Burchak-Burchak). Agar bir uchburchakning ikkita burchagi ikkinchi uchburchakning ikkita burchagiga mos ravishda teng bo'lsa, bunday uchburchaklar o'xshashdir.

$\angle A = \angle A_1, \quad \angle B = \angle B_1 \implies \triangle ABC \sim \triangle A_1B_1C_1.$

Ichki burchaklar yig'indisi $180^\circ$ bo'lgani uchun uchinchi burchaklar ham avtomatik teng bo'ladi.

2-alomat: Ikki tomon proporsionalligi va burchak bo'yicha (TBT)

[!IMPORTANT] 2-alomat (Tomon-Burchak-Tomon). Agar bir uchburchakning ikki tomoni ikkinchi uchburchakning ikki tomoniga proporsional bo'lib, bu tomonlar orasidagi burchaklar teng bo'lsa, bunday uchburchaklar o'xshashdir.

$\frac{AB}{A_1B_1} = \frac{AC}{A_1C_1} \quad \text{va} \quad \angle A = \angle A_1 \implies \triangle ABC \sim \triangle A_1B_1C_1.$

3-alomat: Uchta tomon proporsionalligi bo'yicha (TTT)

[!IMPORTANT] 3-alomat (Tomon-Tomon-Tomon). Agar bir uchburchakning uchta tomoni ikkinchi uchburchakning mos uchta tomoniga proporsional bo'lsa, bunday uchburchaklar o'xshashdir.

$\frac{AB}{A_1B_1} = \frac{BC}{B_1C_1} = \frac{AC}{A_1C_1} = k \implies \triangle ABC \sim \triangle A_1B_1C_1.$

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Muhim xossalar

Agar $\triangle ABC \sim \triangle A_1B_1C_1$ va o'xshashlik koeffitsiyenti $k$ bo'lsa:

1. Perimetrlar va chiziqli elementlar nisbati: $\frac{P}{P_1} = \frac{h}{h_1} = \frac{m}{m_1} = \frac{l}{l_1} = k.$
2. Yuzalar nisbati: O'xshash uchburchaklar yuzalari nisbati koeffitsiyent kvadratiga teng: $\frac{S}{S_1} = k^2.$

To'g'ri burchakli uchburchaklarda to'g'ri burchak doimo tengligi sababli bitta o'tkir burchak tengligi yoki katetlar proporsionalligi yetarlidir.

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Yechilgan misollar

1-misol. 1-alomat (BB) bo'yicha o'xshashlik

$\triangle ABC$ da $\angle A = 50^\circ, \angle B = 70^\circ$. $\triangle DEF$ da $\angle E = 70^\circ, \angle F = 60^\circ$. Uchburchaklar o'xshashmi?

• $\triangle ABC$ da $\angle C = 180^\circ - (50^\circ + 70^\circ) = 60^\circ$.
• Ikkala uchburchakda $\angle B = \angle E = 70^\circ$ va $\angle C = \angle F = 60^\circ$.
• BB alomatiga ko'ra $\triangle ABC \sim \triangle DEF$.

Javob: O'xshash, BB alomatiga ko'ra.

2-misol. 2-alomat (TBT) bo'yicha tomonni topish

$\triangle ABC$ da $AB = 6\text{ sm}, AC = 8\text{ sm}, \angle A = 40^\circ$. $\triangle A_1B_1C_1$ da $A_1B_1 = 9\text{ sm}, A_1C_1 = 12\text{ sm}, \angle A_1 = 40^\circ$. $BC = 10\text{ sm}$ bo'lsa, $B_1C_1$ tomonni toping.

• $\angle A = \angle A_1 = 40^\circ$.
• $\dfrac{A_1B_1}{AB} = \dfrac{9}{6} = 1{,}5$ va $\dfrac{A_1C_1}{AC} = \dfrac{12}{8} = 1{,}5$.
• TBT alomatiga binoan $\triangle ABC \sim \triangle A_1B_1C_1$, $k = 1{,}5$.
• $B_1C_1 = 1{,}5 \cdot 10 = 15\text{ sm}$.

Javob: $15\text{ sm}$.

3-misol. Trapetsiyada diagonallar kesishishi

$ABCD$ trapetsiyada asoslar $BC = 8\text{ sm}, AD = 24\text{ sm}$. Diagonallar $O$ nuqtada kesishadi. $AC = 20\text{ sm}$ bo'lsa, $AO$ va $OC$ kesmalarni toping.

• $BC \parallel AD \implies \angle CBO = \angle ADO, \angle BCO = \angle DAO$.
• BB alomatiga binoan $\triangle BOC \sim \triangle DOA$.
• $\dfrac{AO}{OC} = \dfrac{AD}{BC} = \dfrac{24}{8} = 3 \implies AO = 3 \cdot OC$.
• $4 \cdot OC = 20 \implies OC = 5\text{ sm}, AO = 15\text{ sm}$.

Javob: $AO = 15\text{ sm}, OC = 5\text{ sm}$.

4-misol. Yuzalar va perimetrlar nisbati

Ikki o'xshash uchburchakning perimetrlari $24\text{ sm}$ va $36\text{ sm}$. Kichik uchburchakning yuzi $32\text{ sm}^2$ bo'lsa, katta uchburchak yuzini toping.

• $k = \dfrac{P_2}{P_1} = \dfrac{36}{24} = \dfrac{3}{2}$.
• $\dfrac{S_2}{S_1} = k^2 = \left(\dfrac{3}{2}\right)^2 = \dfrac{9}{4}$.
• $S_2 = 32 \cdot \dfrac{9}{4} = 72\text{ sm}^2$.

Javob: $72\text{ sm}^2$.

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Mashq

1. $\triangle ABC$ da $\angle A = 45^\circ, \angle B = 75^\circ$. $\triangle MNP$ da $\angle M = 45^\circ, \angle P = 60^\circ$. Bu uchburchaklar o'xshashmi?
2. Tomonlari $6\text{ sm}, 8\text{ sm}, 12\text{ sm}$ bo'lgan uchburchakka o'xshash uchburchakning kichik tomoni $15\text{ sm}$. Qolgan ikki tomonini toping.
3. $\triangle ABC$ va $\triangle A_1B_1C_1$ da $\angle B = \angle B_1 = 55^\circ$, $AB = 8\text{ sm}, BC = 12\text{ sm}$, $A_1B_1 = 12\text{ sm}, B_1C_1 = 18\text{ sm}$. Uchburchaklar o'xshashmi? $k$ ni toping.
4. To'g'ri burchakli $\triangle ABC$ da to'g'ri burchak uchidan gipotenuzaga $CD$ balandlik tushirilgan. $AD = 4\text{ sm}, DB = 9\text{ sm}$ bo'lsa, $CD$ balandlikni toping.
5. $ABCD$ trapetsiyada asoslar $BC = 6\text{ sm}, AD = 15\text{ sm}$, diagonallari $O$ nuqtada kesishadi. $BD = 28\text{ sm}$ bo'lsa, $BO$ va $OD$ kesmalarni toping.
6. Ikki o'xshash uchburchakning mos tomonlari $4\text{ sm}$ va $10\text{ sm}$. Kichik uchburchak perimetri $28\text{ sm}$ bo'lsa, kattasining perimetrini toping.
7. Ikki o'xshash uchburchakning yuzalari $18\text{ sm}^2$ va $50\text{ sm}^2$. Kichik uchburchakning balandligi $6\text{ sm}$ bo'lsa, katta uchburchakning mos balandligini toping.
8. $\triangle ABC$ da $D \in AB, E \in AC$ va $\angle ADE = \angle ACB$. $AD = 3\text{ sm}, AE = 4\text{ sm}, AC = 9\text{ sm}$ bo'lsa, $AB$ tomonni toping.

Javoblar va yechim yo'llari

1. $\triangle ABC$ da $\angle C = 180^\circ - (45^\circ + 75^\circ) = 60^\circ$. Ikkala uchburchakda $45^\circ$ va $60^\circ$ teng, demak BB alomatiga ko'ra o'xshash. Javob: O'xshash, BB alomati.
2. $k = 15 : 6 = 2{,}5$. Qolgan tomonlar: $8 \cdot 2{,}5 = 20\text{ sm}$, $12 \cdot 2{,}5 = 30\text{ sm}$. Javob: 20 sm va 30 sm.
3. $\angle B = \angle B_1$, $\dfrac{12}{8} = \dfrac{18}{12} = 1{,}5$. TBT alomatiga ko'ra o'xshash, $k = 1{,}5$. Javob: O'xshash, k = 1,5.
4. $\triangle ACD \sim \triangle CBD \implies \dfrac{CD}{DB} = \dfrac{AD}{CD} \implies CD^2 = 36 \implies CD = 6\text{ sm}$. Javob: 6 sm.
5. $\triangle BOC \sim \triangle DOA \implies \dfrac{BO}{OD} = \dfrac{6}{15} = \dfrac{2}{5}$. $7x = 28 \implies x = 4$. $BO = 8\text{ sm}, OD = 20\text{ sm}$. Javob: BO = 8 sm, OD = 20 sm.
6. $k = 10 : 4 = 2{,}5$. $P_2 = 28 \cdot 2{,}5 = 70\text{ sm}$. Javob: 70 sm.
7. $\dfrac{S_2}{S_1} = \dfrac{50}{18} = \dfrac{25}{9} \implies k = \dfrac{5}{3}$. $h_2 = 6 \cdot \dfrac{5}{3} = 10\text{ sm}$. Javob: 10 sm.
8. $\angle A$ umumiy, $\angle ADE = \angle C \implies \triangle ADE \sim \triangle ACB$. $\dfrac{AD}{AC} = \dfrac{AE}{AB} \implies \dfrac{3}{9} = \dfrac{4}{AB} \implies AB = 12\text{ sm}$. Javob: 12 sm.

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Keng tarqalgan xatolar

1. O'xshashlikda mos tomonlarni chalkashtirish. ✗ $\triangle ADE \sim \triangle ACB$ bo'lganda $AD$ ga $AB$ tomonni mos deb xato olish. ✓ Mos tomonlar teng burchaklar qarshisida yotadi: $\angle ADE = \angle C$ bo'lsa, $AE$ tomon $AB$ ga, $AD$ tomon esa $AC$ ga mos keladi.

2. 2-alomatda burchakning joylashuvini e'tibordan chetda qoldirish. ✗ Ikkita tomon proporsional bo'lib, ixtiyoriy burchak teng bo'lsa ham o'xshash deb o'ylash. ✓ Burchak albatta proporsional tomonlar orasidagi burchak bo'lishi shart!

3. Yuzalar nisbatini adashtirish. ✗ O'xshash uchburchaklar yuzalari nisbati $k$ ga teng deb hisoblash. ✓ Yuzalar nisbati o'xshashlik koeffitsiyentining kvadratiga teng: $\dfrac{S_1}{S_2} = k^2$.

4. O'xshashlikni tenglik bilan aralashtirish. ✗ O'xshash uchburchaklarning tomonlari ham teng bo'ladi deb hisoblash. ✓ O'xshash shakllarning faqat burchaklari teng, tomonlari esa proporsional bo'ladi ($k=1$ bo'lgandagina tomonlar teng bo'ladi).

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DTM uslubidagi test

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

Keyingi qadam

Uchburchaklar o'xshashlik alomatlarini to'liq o'zlashtirdik. Ushbu alomatlar 8–9-sinflardagi planimetriya, aylanada burchaklar va kesmalar hamda DTM imtihonlaridagi murakkab masalalarni yechishda asosiy vosita hisoblanadi.