第37页 - 苏科版数学补充习题九年级上下册答案

$解：∵△ABC∽△A'B'C'$

$∴\frac {AB}{A'B'}=\frac {AC}{A'C'}=\frac {BC}{B'C'}=\frac {AB+BC+AC}{A'C'+B'C'+A'C'}=\frac {60}{72}$

$又∵AB=15\ \mathrm {cm}，B'C'=24\ \mathrm {cm}$

$∴A'B'=18\ \mathrm {cm}，BC=20\ \mathrm {cm}$

$解：∵DE//BC$

$∴∠DEB=∠EBC，∠EDC=∠DCB$

$∴△DOE∽△COB$

$由△DOE与△COB的面积比为4：9$

$可得△DOE与△COB的相似比为2：3$

$∴\frac {DE}{BC}=\frac 23$

$∵DE//BC$

$∴\frac {AE}{AC}=\frac {DE}{BC}=\frac 23$

$解：(1)△ABC的周长=20×\frac 3{3+2}=12(\mathrm {cm})$

$△A'B'C'的周长=20×\frac 2{3+2}=8(\mathrm {cm})$

$(2)∵△ABC与△A'B'C'的相似比为\frac 32$

$∴面积比为\frac 94$

$∴S_{△ABC}=5÷\frac {9-4}9=9(\mathrm {cm^2})，S_{△A'B'C'}=5÷\frac {9-4}4=4(\mathrm {cm^2})$