北師大版初中九年級數(shù)學(xué)上冊第四章圖形的相似1成比例線段第2課時比例的其他性質(zhì)課件

第四章圖形的相似1成比例線段第2課時比例的其他性質(zhì)基礎(chǔ)過關(guān)全練知識點1比例的合比、分比性質(zhì)1.(一題多解)(2023四川甘孜州中考)若

=2,則

=

.1解析解法一(利用分式的性質(zhì)):∵

=2,∴

=

-1=2-1=1.解法二(利用比例的合比、分比性質(zhì)):∵

=2=

,∴

=

=1.2.(2024上海嘉定期末)如果5a=3b(a、b都不等于零),那么

=

.-

解析∵5a=3b,∴

=

,∴

=

=-

.3.(一題多解)如果

=

,那么

的值為

.解析解法一:∵

=

,∴3(a-b)=a,∴a=

b,∴

=

=

.解法二:由

=

,得1-

=

,∴

=

,∴

=

,∴

=

+1=

+1=

.解法三:∵

=

,∴

=-

,∴

=

,即

=

,∴

=

,∴

=

=

.知識點2比例的等比性質(zhì)4.(2024貴州畢節(jié)威寧期末)已知

=

=

=

,若b+d+f=9,則a+c+e=

(

)A.12

B.15

C.16

D.18A解析∵

=

=

=

,∴

=

,∵b+d+f=9,∴a+c+e=12.故選A.5.已知

=

=

=m(a+b+c≠0),求m的值.解析∵

=

=

=m(a+b+c≠0),∴

=m,∴

=m,∴m=3.6.已知a,b,c是△ABC的三邊長,且

=

=

≠0.(1)求

的值.(2)若△ABC的周長為60,求各邊的長.解析

(1)設(shè)

=

=

=k,則a=5k,b=4k,c=6k,∴

=

=

=

.(2)解法一(參數(shù)法):∵△ABC的周長為60,∴a+b+c=60,∴5k+

4k+6k=60,∴k=4,∴a=5k=20,b=4k=16,c=6k=24,∴三角形各邊的長分別

為20,16,24.解法二(等比性質(zhì)法):∵△ABC的周長為60,∴a+b+c=60,∴

=

=

=

=

=4,∴a=20,b=16,c=24.∴三角形各邊的長分別為20,16,24.能力提升全練7.(2022上海嘉定期末,5,★★☆)若實數(shù)a,b,c,d滿足

=

,則下列四個選項中正確的是

(

)A.

=

B.

=

C.

=

D.

=

A解析

A項,由比例的合比性質(zhì)可知

=

,等式成立;B項,當(dāng)a+b=c+d=0時,等式不成立;C項,當(dāng)b+d=0時,等式不成立;

D項,

=

不一定成立.8.(2024河南許昌二中月考,13,★★☆)在△ABC和△DEF中,

若

=

=

=

,且△ABC與△DEF的周長之差為10cm,則△ABC的周長為

cm.25解析∵

=

=

=

,∴

=

,∴

=

,∴設(shè)△ABC的周長為5xcm,△DEF的周長為3xcm,∵△ABC與△DEF的周長之差為10cm,∴5x-3x=10,∴x=5,∴△ABC的周長=25cm.9.(2024湖南師大附中期末,14,★★☆)已知

=

=

=

=m,則m的值為

.

3或-1解析分兩種情況:當(dāng)a+b+c+d≠0時,根據(jù)等比性質(zhì)可得m=

=

=3;當(dāng)a+b+c+d=0時,a+b+c=-d,∴m=

=

=-1.綜上所述,m的值為3或-1.10.(一題多解)(2024陜西西安鐵一中學(xué)月考,20,★★☆)已知a、b、c是△ABC的三邊長,且滿足

=

=

,a+b+c=30,試判斷△ABC的形狀,并說明理由.解析解法一(等比性質(zhì)法):△ABC是直角三角形.理由:∵

=

=

,a+b+c=30,∴

=

=

=2,∴

=2,

=2,

=2,∴a=5,b=12,c=13,則b2+a2=c2,∴△ABC是直角三角形.解法二(參數(shù)法):△ABC是直角三角形.理由:設(shè)

=

=

=k,則a=3k-1,b=4k+4,c=8k-3,∵a+b+c=30,∴3k-1+4k+4+8k-3=30,∴k=2,∴a=5,b=12,c=13,則b2+a2=c2,∴△ABC是直角三角形.素養(yǎng)探究全練11.(運算能力)已知

=

=

,若a,b,c均為非零的實數(shù),求

的值.解析①當(dāng)a+b+c≠0時,由等比性質(zhì),得

=

=

=

=1,∴a+b-c=c,a-b+c=b,-a+b+c=a.∴a+