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uchburchaklardagi o'xshashlik

Raqamlar orasidagi o'xshashliklar

Ikki raqam bir xil shaklga ega bo'lsa, lekin bir xil o'lchamda bo'lishi shart bo'lmasa, ular o'xshash deb ataladi. Quyidagi rasmlarda o'xshash doiralar va o'xshash uchburchaklar ko'rsatilgan.

Uchburchaklarning o'xshashligi: Bir uchburchakning uchta burchagi mos ravishda boshqa uchburchakning uchta burchagiga teng bo'lgan va tegishli tomonlarning o'lchovlari orasidagi barcha nisbatlar teng bo'lgan ikkita uchburchak o'xshash deyiladi.

Ta'rifga ko'ra, ABC uchburchagi PQR uchburchagiga o'xshaydi, \(\triangle ABC \sim \triangle PQR\) agar:

1. \(\displaystyle \frac{a}{p} = \frac{b}{q} = \frac{c}{r}\)
2. \(\displaystyle \angle A = \angle P, \angle B = \angle Q, \angle C = \angle R\)

Shunday qilib, uchburchaklarning o'xshashligi ikkita narsani talab qiladi:

• Tegishli burchaklar teng bo'lishi kerak
• Tegishli tomonlar mutanosib bo'lishi kerak

UCHBURCHAKLARNING O'SHKARLIGI UCHUN UCHTA TEST

1. Burchak-burchak-burchak ( \(AAA\) ) O'XSHASH AKSIOMASI

Agar ikkita uchburchakning ikkita juft burchagi teng bo'lsa, ularning mos tomonlari proportsionaldir. ABC va \(DEF\) uchburchaklarida \(\displaystyle \angle A = \angle D, \angle B = \angle E, \angle C = \angle F\) , keyin \(\triangle ABC \sim \triangle DEF\) , ya'ni
\(\displaystyle \frac{BC}{EF} = \frac{AB}{DE}= \frac{AC}{DF}\)

2. Yon-burchak-yon ( \(SAS\) ) O'XSHASH AKSIOMASI

Agar ikkita uchburchakning mos keladigan burchaklari teng bo'lsa va tomonlari ular bilan mutanosib bo'lsa, uchburchaklar o'xshashdir.

Agar ABC uchburchakda va \(DEF\) , \(\angle A = \angle D\) va \(\frac{AB} { DE} = \frac{AC}{DF}\) keyin \(\triangle ABC \sim \triangle DEF\)

3. Side-Side-Side ( \(SSS\) ) O'XSHASH AKSIOMASI

Agar ikkita uchburchakning juft tomonlari mutanosib bo'lsa, uchburchaklar o'xshashdir. Agar ikkita uchburchak ABC va \(DEF\) bo'lsa, \(\displaystyle \frac{BC}{EF} = \frac{AB}{DE}= \frac{AC}{DF}\) u holda \(\triangle ABC \sim \triangle DEF\)

Teorema 1

Uchburchakning bir tomoniga parallel chizilgan to‘g‘ri chiziq qolgan ikki tomonini proporsional ravishda ajratadi. Aksincha, agar chiziq uchburchakning istalgan ikki tomonini proportsional ravishda ajratsa, chiziq uchinchi tomonga parallel bo'ladi.

\(\triangle ABC, \ DE \parallel BC\) da keyin

\(\displaystyle \frac{AD}{DB} = \frac{AE}{EC}\)

Teorema 2

O'xshash uchburchaklarning maydonlari tegishli tomonlardagi kvadratlarga proportsionaldir.

\(\displaystyle \frac{\textrm{maydoni }\triangle ABC}{\textrm{maydoni }\triangle DEF} = \) \(\displaystyle \frac{BC^2}{EF^2} = \frac{AB^2}{DE^2}= \frac{AC^2}{DF^2}\)

1-misol: \(\triangle ABC, PQ \parallel BC\) da. Agar AP/ \(PB\) = 1/2 va AQ = 2 sm bo'lsa. QC toping.

PQ BC ga parallel bo'lgani uchun
\(\displaystyle \frac{AP}{PB} = \frac{AQ}{QC}\)

\(\displaystyle\frac{1}{2} = \frac{2}{QC}\) ⇒ \(\displaystyle QC = 2 \times 2 = 4\)

2-misol: Uchburchak \(\triangle ABD, \triangle ACD \) o'xshash. BD = 2 sm va AB = 3 sm. Agar \(\triangle ABD \) uchburchakning maydoni 2 sm ² bo'lsa, \( \triangle ACD \) maydonini hisoblang.

\(\displaystyle \frac{\textrm{maydoni }\triangle ABD}{\textrm{maydoni }\triangle ADC} = \frac{4}{DC^2} = \frac{9}{AC^2}\)

\( \frac{2}{\textrm{ning maydoni }\triangle ADC} = \frac{4}{9}\)

\(\textrm{ning maydoni }\triangle ADC = \frac{2 \times 9}{4} = 4.5 cm^2\)