Question

已知△ABC相似△A1B1C1，相似比为k（k＞1），且△ABC的三边长分别为a、b、c（a＞b＞c），△A1B1C1的三边

长分别为a1、b1、c1

Answer 1

解:
(1)证:Q△ ABC ∽△ A1 B1C1 ,且相似比为 k ( k > 1), a = k, a = ka1. ∴ a1 又Q c = a1, a = kc.

(2)解:取 a = 8,b = 6,c = 4,同时取a1 = 4,b1 = 3,c1 = 2.
此时 a b c = = = 2,△ ABC ∽△ A1 B1C1 且 c = a1.
∴ a1 b1 c1

(3)解:不存在这样的△ ABC 和△ A1 B1C1 .
理由如下:若 k = 2, a = 2a1,b = 2b1,c = 2c1. 则 又Q b = a1, = b1 , c ∴ a = 2a1 = 2b = 4b1 = 4c, ∴ b = 2c.
∴ b + c = 2c + c < 4c = a ,而 b + c > a, -1- 故不存在这样的△ ABC 和△ A1 B1C1 ,使得 k = 2.

回答完毕

Answer 2

因为a:a1=K所以a1=a/K(先交叉相乘得:a1xk=a再两边同除以k)同理:b1=a/k,c1=c/k