Back to the topics list

kufanana katika pembetatu

Kufanana kati ya takwimu

Takwimu mbili zinasemekana kufanana ikiwa zina umbo sawa lakini si lazima ziwe na ukubwa sawa. Takwimu zifuatazo zinaonyesha miduara sawa na pembetatu sawa.

Kufanana kwa Pembetatu: Pembetatu mbili ambazo zina pembe tatu za pembetatu moja sawa kwa mtiririko huo na pembe tatu za pembetatu nyingine na uwiano wote kati ya kipimo cha pande zinazolingana sawa husemekana kuwa sawa.

Kulingana na ufafanuzi, pembetatu ABC ni sawa na pembetatu PQR , \(\triangle ABC \sim \triangle PQR\) ikiwa:

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

Kwa hivyo, kufanana kwa pembetatu kunahitaji vitu viwili:

• Pembe zinazolingana lazima ziwe sawa
• Pande zinazolingana lazima ziwe sawia

MAJARIBIO MATATU YA KUFANANA KWA PETRO

1. Angle-Angle-Angle ( \(AAA\) ) mhimili wa KUFANANA

Ikiwa pembetatu mbili zina jozi mbili za pembe sawa, pande zao zinazolingana ni sawia. Katika pembetatu ABC na \(DEF\) , \(\displaystyle \angle A = \angle D, \angle B = \angle E, \angle C = \angle F\) , kisha \(\triangle ABC \sim \triangle DEF\) , yaani
\(\displaystyle \frac{BC}{EF} = \frac{AB}{DE}= \frac{AC}{DF}\)

2. Upande-Angle-Upande ( \(SAS\) ) mhimili wa KUFANANA

Ikiwa pembetatu mbili zina jozi ya pembe zinazolingana sawa na pande zikiwemo sawia basi pembetatu zinafanana.

Ikiwa katika pembetatu ABC na \(DEF\) , \(\angle A = \angle D\) na \(\frac{AB} { DE} = \frac{AC}{DF}\) kisha \(\triangle ABC \sim \triangle DEF\)

3. Upande-Upande-Upande ( \(SSS\) ) mhimili wa KUFANANA

Ikiwa pembetatu mbili zina jozi zao za pande zinazolingana sawia basi pembetatu zinafanana. Ikiwa pembetatu mbili ABC na \(DEF\) , \(\displaystyle \frac{BC}{EF} = \frac{AB}{DE}= \frac{AC}{DF}\) basi \(\triangle ABC \sim \triangle DEF\)

Nadharia 1

Mstari wa moja kwa moja uliochorwa sambamba na upande mmoja wa pembetatu hugawanya pande nyingine mbili sawia. Kinyume chake, ikiwa mstari unagawanya pande zote mbili za pembetatu sawia basi mstari ni sambamba na upande wa tatu.

Katika \(\triangle ABC, \ DE \parallel BC\) basi

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

Nadharia 2

Maeneo ya pembetatu sawa ni sawia na mraba kwenye pande zinazofanana.

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

Mfano 1: Katika \(\triangle ABC, PQ \parallel BC\) . Ikiwa AP/ \(PB\) = 1/2 na AQ = 2 cm. Tafuta QC.

Kama PQ ni sambamba na BC, kwa hiyo
\(\displaystyle \frac{AP}{PB} = \frac{AQ}{QC}\)

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

Mfano wa 2: Pembetatu \(\triangle ABD, \triangle ACD \) zinafanana. BD = 2 cm na AB = 3 cm. Ikiwa eneo la pembetatu \(\triangle ABD \) ni 2 cm ² , hesabu eneo la \( \triangle ACD \) .

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

\( \frac{2}{\textrm{Eneo la }\triangle ADC} = \frac{4}{9}\)

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