梯形被分成了四份，C的面积是20平方厘米，D的面积是50平方厘米，梯形的面积是多少？
\u5df2\u77e5\u68af\u5f62\u7684\u9762\u79ef\u662f20\u5e73\u65b9\u5398\u7c73\uff0c\u4e0a\u5e95\u662f4\u5398\u7c73\uff0c\u4e0b\u5e95\u662f6\u5398\u7c73\u3002\u68af\u5f62\u9762\u79ef\u7b49\u4e8e\u4e0a\u5e95\u52a0\u4e0b\u5e95\u7684\u548c\u4e58\u9ad8\u9664\u4ee52\uff0c\u8bbe\u9ad8\u4e3a\u672a\u77e5\u6570\uff0c\u5219\u5f97\u51fa\u65b9\u7a0b\uff1a
\uff084+6\uff09\u00d7\u672a\u77e5\u6570\u00f72=20
\u672a\u77e5\u6570=4
\u9a8c\u7b97\uff1a\uff084+6\uff09\u00d74\u00f72=20
\u6240\u4ee5\uff0c\u9ad8\u4e3a4
\u81f3\u4e8e\u9634\u5f71\u90e8\u5206\u9762\u79ef\uff0c\u6211\u770b\u4e0d\u61c2\u4f60\u753b\u7684\u56fe\uff0csorry \u2026\u2026
\u68af\u5f62\u7684\u9762\u79ef=\uff08\u4e0a\u5e95+\u4e0b\u5e95\uff09*20
\u4e0b\u5e95=\uff1f\uff1f\uff1f
\u8ba1\u7b97\u4e0d\u51fa\u7684...\u6c57..
\u4f60\u9898\u51fa\u9519\u4e86..
因为ABCD均为直角三角形 且D为等腰直角三角形 则D的边长为10 则C的变长为4 B的面积就是4*4/2=8
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绛旓細濡傚浘銆傚洜涓猴細AD鈥朆 C锛鎵浠ワ細鈻矨BC鍜屸柍DBC鏄悓搴曠瓑楂樼殑涓や釜涓夎褰 鎵浠ワ細杩欎袱涓笁瑙掑舰鐨勯潰绉兘鏄18骞虫柟鍘樼背 鎵浠ワ細鈻矨OB鐨勯潰绉=18-12=6锛堝钩鏂瑰帢绫筹級鐢变簬锛氣柍BCO鍜屸柍DCO鐨勯珮鐩哥瓑锛屾墍浠ワ細OD/OB=6/12=1/2 鑰屸柍AOD鈭解柍COB 鎵浠ワ細S鈻矨OD/12=1/4 (鐩镐技涓夎褰㈢殑闈㈢Н姣旂瓑浜庣浉浼兼瘮鐨勫钩鏂癸級鎵...
绛旓細瑙ｇ瓟锛氳В锛氣埖姊舰ABCD鐨勫洓涓《鐐圭殑鍧愭爣鍒嗗垾涓篈锛-1锛0锛夛紝B锛5锛0锛锛孋锛2锛2锛夛紝D锛0锛2锛夛紝鈭姊舰鐨勯潰绉负锛(6+2)脳22=8锛屸埖鐩寸嚎y=kx+2灏姊舰鍒嗘垚闈㈢Н鐩哥瓑鐨勪袱閮ㄥ垎锛屸埓鐩寸嚎y=kx+2涓嶢D銆丄B鍥存垚鐨勪笁瑙掑舰鐨勯潰绉负4锛岃鐩寸嚎涓巟杞翠氦浜庣偣锛坸锛0锛夛紝鈭12锛坸+1锛壝2=4锛屸埓x=3锛...
绛旓細璁剧瓑鑵姊舰閫嗘椂閽堟爣鍙凤紝ABCD 杩嘊銆丆鍒嗗埆浣淏E銆丆F鍒嗗埆鍨傜洿AD浜嶦銆丗 鍥犱负瑕佹眰4涓闈㈢Н褰㈢姸鐩稿悓鐨勬褰 鎵浠ヨAD=y锛孊E鍚戝彸骞崇Щx涓崟浣嶉暱搴︼紝寰桞'E'锛孋F鍚戝乏鍚岀悊寰桟'F'鍒欓渶瑕丼鍥涜竟褰'E'C'F'=浜屽峉姊舰AE'BB'鎵浠(2x+y/4)=h(y/2-2x)h涓洪珮 寰楀埌y=16x 鎵浠ュ钩绉粂/16涓崟浣嶉暱搴 F'...
绛旓細姊舰闈㈢Н=锛堜笂搴+涓嬪簳锛墄楂樏2锛鐢讳竴涓珮2鍘樼背锛屼笂搴0.5鍘樼背锛屼笅搴1.5鍘樼背鐨勭洿瑙掓褰㈠氨鍙互浜嗐
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绛旓細璁惧瑙掔嚎浜ょ偣涓篛 S涓=S涔 S鐢诧細S涔=4:6=2:3 鍒橭A锛歄D=2:3 鐢蹭笌涓欑浉浼 鎵浠ワ紝S鐢诧細S涓=4:9 鍗充笝=9 姊舰闈㈢Н=4+6+6+9=25
绛旓細鍏堕珮鐨勫叧绯绘槸杩欐牱鐨勶細骞抽潰鍥句腑鐨勯珮OA鏄洿瑙傚浘涓璒A'闀垮害鐨2鍊嶏紝濡傜洿瑙傚浘锛孫A'鐨勯暱搴︽槸鐩磋鍥句腑姊舰鐨勯珮鐨2鍊嶏紝鐢辨骞抽潰鍥句腑姊舰鐨勯珮OA鐨勯暱搴︽槸鐩磋鍥句腑姊舰楂樼殑2脳2=22鍊嶏紝鏁呭叾闈㈢Н鏄褰A鈥睟鈥睠鈥茬殑闈㈢Н22鍊锛屾褰OA鈥睟鈥C鈥鐨勯潰绉负2锛鎵浠ュ師姊舰鐨勯潰绉槸4锛庢晠搴旈塂锛
绛旓細C鍜孌鐢诲弽浜哹?瑙ｏ細璁続B = a 姊舰鐨楂樹负h锛屸柍AOB杈笰B涓婄殑楂樹负h1锛屸柍DOC杈笴D涓婄殑楂樹负h2锛h1 = 2S锛圓OB锛壝 AB = 50/a h = 2S锛圓BC锛壝 AB = 120/a h2 = h - h2 = 120/a-50/a = 70/a S(DOC) 锛歋(AOB) = h2² 锛歨1²S(DOC) = 25 脳 (h2/h1)...
绛旓細鎴戞妸宸﹁竟鐨C鎹㈡垚B 瑙ｏ細鍥犱负AD鈭C 鎵浠モ柍AOD鈭解柍COB 鍥犱负闈㈢Н姣旂瓑浜庣浉浼兼瘮鐨勫钩鏂 鎵浠O:CO=DO:BO=1锛2 鈻矨OD 銆佲柍COD鍒嗗埆浠O銆丆O涓哄簳锛岄珮鐩稿悓锛屾墍浠鈻矯OD 锛歋鈻矨OD =CO:AO=2:1 鎵浠鈻矯OD=10 鍚岀悊S鈻矨OB=1/2S鈻矯OB=10 姊舰ABCD鐨勯潰绉=5+20+10+10=45 銆愭暟瀛﹁緟瀵煎洟銆戝洟闃熶负鎮...
