北师大版初中九年级数学上册第四章图形的相似1成比例线段第2课时比例的其他性质课件

第四章图形的相似1成比例线段第2课时比例的其他性质基础过关全练知识点1比例的合比、分比性质1.(一题多解)(2023四川甘孜州中考)若

=2,则

=

.1解析解法一(利用分式的性质):∵

=2,∴

=

-1=2-1=1.解法二(利用比例的合比、分比性质):∵

=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.(一题多解)如果

=

,那么

的值为

.解析解法一:∵

=

,∴3(a-b)=a,∴a=

b,∴

=

=

.解法二:由

=

,得1-

=

,∴

=

,∴

=

,∴

=

+1=

+1=

.解法三:∵

=

,∴

=-

,∴

=

,即

=

,∴

=

,∴

=

=

.知识点2比例的等比性质4.(2024贵州毕节威宁期末)已知

=

=

=

,若b+d+f=9,则a+c+e=

(

)A.12

B.15

C.16

D.18A解析∵

=

=

=

,∴

=

,∵b+d+f=9,∴a+c+e=12.故选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的三边长,且

=

=

≠0.(1)求

的值.(2)若△ABC的周长为60,求各边的长.解析

(1)设

=

=

=k,则a=5k,b=4k,c=6k,∴

=

=

=

.(2)解法一(参数法):∵△ABC的周长为60,∴a+b+c=60,∴5k+

4k+6k=60,∴k=4,∴a=5k=20,b=4k=16,c=6k=24,∴三角形各边的长分别

为20,16,24.解法二(等比性质法):∵△ABC的周长为60,∴a+b+c=60,∴

=

=

=

=

=4,∴a=20,b=16,c=24.∴三角形各边的长分别为20,16,24.能力提升全练7.(2022上海嘉定期末,5,★★☆)若实数a,b,c,d满足

=

,则下列四个选项中正确的是

(

)A.

=

B.

=

C.

=

D.

=

A解析

A项,由比例的合比性质可知

=

,等式成立;B项,当a+b=c+d=0时,等式不成立;C项,当b+d=0时,等式不成立;

D项,

=

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

若

=

=

=

,且△ABC与△DEF的周长之差为10cm,则△ABC的周长为

cm.25解析∵

=

=

=

,∴

=

,∴

=

,∴设△ABC的周长为5xcm,△DEF的周长为3xcm,∵△ABC与△DEF的周长之差为10cm,∴5x-3x=10,∴x=5,∴△ABC的周长=25cm.9.(2024湖南师大附中期末,14,★★☆)已知

=

=

=

=m,则m的值为

.

3或-1解析分两种情况:当a+b+c+d≠0时,根据等比性质可得m=

=

=3;当a+b+c+d=0时,a+b+c=-d,∴m=

=

=-1.综上所述,m的值为3或-1.10.(一题多解)(2024陕西西安铁一中学月考,20,★★☆)已知a、b、c是△ABC的三边长,且满足

=

=

,a+b+c=30,试判断△ABC的形状,并说明理由.解析解法一(等比性质法):△ABC是直角三角形.理由:∵

=

=

,a+b+c=30,∴

=

=

=2,∴

=2,

=2,

=2,∴a=5,b=12,c=13,则b2+a2=c2,∴△ABC是直角三角形.解法二(参数法):△ABC是直角三角形.理由:设

=

=

=k,则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,则b2+a2=c2,∴△ABC是直角三角形.素养探究全练11.(运算能力)已知

=

=

,若a,b,c均为非零的实数,求

的值.解析①当a+b+c≠0时,由等比性质,得

=

=

=

=1,∴a+b-c=c,a-b+c=b,-a+b+c=a.∴a+