如图，△ABC的周长为30厘米，把三角形ABC的边AC对折，使顶点C和点A重合，折痕交BC于点D，交AC边于点E，连接
\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\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
\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
∵EF是折痕
∴EF垂直平分AC
∴DA=DC
∵AE=4
∴AC=8
∵△ABC的周长为30
∴AB+BC=30-8=22
∴△ABD的周长=AB+AD+BD=AB+CD+BD=AB+BC=22cm
绛旓細鍥犱负涓夎褰ABC鐨鍛ㄩ暱涓30鍘樼背锛屾墍浠B+BC=30-8=22鍘樼背锛涘洜涓篈D=CD锛屾墍浠ヤ笁瑙掑舰ABD鐨勫懆闀=AB+BD+AD =AB+BD+CD =AB+BC =22鍘樼背銆
绛旓細鐢遍鎰忓彲鐭ワ細DE涓篈C鐨勪腑鍨傜嚎 AE=CE=AC/2锛孉D=CD 鈻矨BD鐨鍛ㄩ暱=AB+BD+AD=AB+BD+CD=AB+BC=30-AC=30-4脳2=22
绛旓細s=(AB+AC+BC)*h/2=30*6/2=90 cm^3
绛旓細瑙ｏ細鈭礒F鏄姌鐥 鈭碋F鍨傜洿骞冲垎AC 鈭碊A=DC 鈭礎E=4 鈭碅C=8 鈭碘柍ABC鐨鍛ㄩ暱涓30 鈭碅B+BC=30-8=22 鈭粹柍ABD鐨勫懆闀=AB+AD+BD=AB+CD+BD=AB+BC=22cm
绛旓細S鈻ABC=S鈻矨PB+S鈻矨PC+S鈻矪PC =AB脳BP梅2+AC脳PE梅2+BC脳PD梅2 =1/2PD脳锛圓B+AC+BC锛=1/2脳3脳30 =45锛堝钩鏂瑰帢绫筹級绛旓細涓夎褰BC鐨勯潰绉槸45骞虫柟鍘樼背銆備笁瑙掑舰鎬ц川锛1 銆佸湪骞抽潰涓婁笁瑙掑舰鐨勫唴瑙掑拰绛変簬180掳锛堝唴瑙掑拰瀹氱悊锛夈2 銆佸湪骞抽潰涓婁笁瑙掑舰鐨勫瑙掑拰绛変簬360掳 (澶栬鍜屽畾鐞)銆3銆 ...
绛旓細A 鈭礎E=4cm锛鈭碅C=8锛屸埖鈻矨BC鐨勫懆闀夸负30cm锛屸埓AB+BC=22锛屸埖鈻矨BD鐨勫懆闀=AB+AD+BD锛孉D=DC锛屸埓鈻矨BD鐨勫懆闀=AB+AD+BD=AB+CD+BC-CD=AB+BC=22鏁呴堿
绛旓細杩欐牱瀵规姌涔嬪悗锛屽彲寰桝D=DC锛孉E=EC锛屾墍浠C=8cm锛孉B+BC=22cm 鍙堝洜涓鈻ABD鐨勫懆闀=AB+BD+AD=AB+BD+DC=AB+BC=22cm
绛旓細濡傚浘鎵绀,涓夎褰ABC鐨勫懆闀夸负30cm,鐐筆,Q鍚屾椂浠嶢鐐瑰嚭鍙戞部鐫涓夎褰㈢殑涓鏉¤竟杩愬姩,鐐筆鐨勯熷害鏄1cm/绉,鐐筈鐨勯熷害鏄2cm/绉,鍑犵鍚庝袱鐐归娆￠噸鍚?... 濡傚浘鎵绀,涓夎褰BC鐨勫懆闀夸负30cm,鐐筆,Q鍚屾椂浠嶢鐐瑰嚭鍙戞部鐫涓夎褰㈢殑涓鏉¤竟杩愬姩,鐐筆鐨勯熷害鏄1cm/绉,鐐筈鐨勯熷害鏄2cm/绉,鍑犵鍚庝袱鐐归娆￠噸鍚? 灞曞紑 鎴戞潵...
绛旓細30cm梅锛1+2锛cm/s=10s
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
