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116问答网 > 已知△abc中，ab=7,bc:av=4:3,试求这个角的周长取值范围

已知△abc中，ab=7,bc:av=4:3,试求这个角的周长取值范围

2025-03-29 15:17:58

推荐回答（2个）

回答1：

BC:AC=4:3
设BC=4X，AC=3X

BC+AC>AB
7X>7
X>1

AB+AC>BC
7+3X>4X
X<7

所以:17+7*1<7+7X<7+7*7
14<△ABC周长<56

回答2：

因为BC:AC=4:3，所以BC=4AC/3 △ABC周长为7+BC+AC=7+（7AC/3）
由于三角形的任意两边之和都大于第三边，所以：BC+AC＞AB（7）
即（7AC/3）＞7 所以 △ABC周长：7+BC+AC=7+（7AC/3）＞14

ab不是最大边时 ab+ac>bc
即7+3k>4k
7>k
当7=k时
周长=7+49=56
所以周长<56
14<周长<56

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