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Тема: “Подобни триъгълници- упражнение върху признаците за подобност”. Изготвил: Верка Тодорова Додникова, старши учител по математика в ЕГ”Бертолт Брехт”-гр.Пазарджик. - PowerPoint PPT Presentation
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: - : , -.
: ABC PQR , :
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( V ) , .ABCPQR..
1: , =12 cm. .D, 1:2, .. .D , , .. DE, - 6 .1.3.2.4.!!!!
ABC AB=10 cm, AC=6 cm. M, N P AB, BC AC , AMNP . 14 cm. . 2.1.2.3.4.!!!!
3. ABC AB O , H M AB AC. , AHC ~ OMC. R=4 cm CH =6 cm, BC=?1.2.3.4.!!!!
4. M N AB AC ABC, AC=24 cm CM=18 cm. P CM , CP=16 cm. , PN AC. ABC .1.2.3.4.!!!!
5. =15 m. =20 m. M N , AM=12 m. AN=9 m. , ~ NM BCNM .ABCMN15cm12cm20cm9cm
6.!3.2.! ABCD B=9cm. D=4cm., 6cm. , 2,5cm. !!4.1.D9cm4cm6cm
7. DE ABCD 2:3, . , AB=28cm. AD=20cm.1.3.2.4.ABCDEMN!!!!
1: D:DC=1:2 =12 .AD= 4 cm.; DC= 8 cm. DE= x, x>0BC= x+6 AED :1. 2. AED = ABC ( , DE ) ~ AED ( )
ABC PNC PNAB 1. CAB = CPN 2. ACB = PCN ABC PNC (I )
PC = 6 - AP P = 14 2(PN+AP) = 14 . AP = 4,5 cm PN = 2,5 cm 2:
3: AHC OMC 1. OMC = AHC = 90 2. ACH AHC ~ OMC ( I )AC=2.MC, .O .O .
.M AC OM OMC= 90; .H AB OH AHC= 90
4: AMC PNC 1. 2. NCP 1 2. II- AMC PNC AMC PNC=90 PN AC PN AC PN AC .P PN CM .P PC=R R=16 cm.
5: NMABCMN15cm12cm20cm9cm ~ NM ( )2.) A- ~ NM ABC=ANM , ANM + CNM =180 ( ) ABC + CNM =180 BCNM .
6: AD = x, x >0 BC = x + 2,5 CDD9cm4cm6cm ~ CD ( )2.) BAC= ACD () +2,5
7:ABCDEMNEM ~ CDM ( ) ? ABCD BN ~ DCN ( ) ?, AN=AD+DN