△ABC的周长为30cm,把△ABC的边AC对折,使顶点C和A重合,折痕交BC与点D,交AC边于点E,连接AD,若AE=4cm
\u5982\u56fe\uff0c\u25b3ABC\u7684\u5468\u957f\u4e3a30\u5398\u7c73\uff0c\u628a\u4e09\u89d2\u5f62ABC\u7684\u8fb9AC\u5bf9\u6298\uff0c\u4f7f\u9876\u70b9C\u548c\u70b9A\u91cd\u5408\uff0c\u6298\u75d5\u4ea4BC\u4e8e\u70b9D\uff0c\u4ea4AC\u8fb9\u4e8e\u70b9E\uff0c\u8fde\u89e3\uff1a\u7531\u9898\u610f\u77e5\uff1a\u628a\u4e09\u89d2\u5f62ABC\u7684\u8fb9AC\u5bf9\u6298\uff0c\u4f7f\u9876\u70b9C\u548c\u70b9A\u91cd\u5408\uff0c\u6298\u75d5\u4ea4BC\u4e8e\u70b9D\uff0c
\u4ea4AC\u8fb9\u4e8e\u70b9E\uff0c\u8fde\u63a5AD\uff0c\u5219\u4e09\u89d2\u5f62ADE\u5168\u7b49\u4e8e\u4e09\u89d2\u5f62CDE\u3002
\u6240\u4ee5AD=CD\uff0cCE=AE=4\u5398\u7c73 \uff0cAC=4X2=8\u5398\u7c73\uff1b
\u56e0\u4e3a\u4e09\u89d2\u5f62ABC\u7684\u5468\u957f\u4e3a30\u5398\u7c73\uff0c
\u6240\u4ee5AB+BC=30-8=22\u5398\u7c73\uff1b
\u56e0\u4e3aAD=CD\uff0c
\u6240\u4ee5\u4e09\u89d2\u5f62ABD\u7684\u5468\u957f=AB+BD+AD
=AB+BD+CD
=AB+BC
=22\u5398\u7c73\u3002
\u89e3\uff1a
\u2235\u25b3CDE\u6cbfDE\u5bf9\u6298\u81f3\u25b3ADE
\u2234\u25b3ADE\u224c\u25b3CDE
\u2234CE\uff1dAE\uff0cCD\uff1dAD
\u2235AE\uff1d4
\u2234CE\uff1d4
\u2234AC\uff1dAE+CE\uff1d8
\u2235\u25b3ABC\u7684\u5468\u957f\u4e3a30
\u2234AB+BC+AC\uff1d30
\u2234AB+BC\uff1d30-AC\uff1d30-8\uff1d22
\u2235BC\uff1dBD+CD
\u2234BC\uff1dBD+AD
\u2234L\u25b3ABD\uff1dAB+BD+AD\uff1dAB+BC\uff1d22\uff08cm\uff09
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解:∵AE=4cm,
∴AC=8,
∵△ABC的周长为30cm,
∴AB+BC=22,
∵△ABD的周长=AB+AD+BD,AD=DC,
∴△ABD的周长=AB+AD+BD=AB+CD+BC-CD=AB+BC=22
故选择A.
30-2×4=22cm
∵折叠
∴AD=CD,AE=CE=4cm
∵C△ABD=AB+BD+AD,即=AB+BD+CD=C△ABC-AC=30-2*4=22cm
绛旓細浠庡唴鍒囧渾鐨勫渾蹇僌寮曚笁鏉$嚎鍒颁笁瑙掑舰鐨勪笁涓《鐐.澶栨帴涓夎褰㈣涓夋潯绾垮垎鎴愪簡涓変釜楂樼浉绛夌殑涓夎褰 闈㈢Н锛鈻矨BC=鈻砄AB+鈻砄BC+鈻砄AC 锛圓B+AC+BC锛*R*1/2=30
绛旓細9cm 30/2-6
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