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68\u3001\u5df2\u77e5\uff1a\u5982\u56fe, \u25b3ABC\u4e2d, \u2220C=90\u00b0, AC=BC, AD\u5e73\u5206\u2220CAB\u4ea4BC\u4e8eD,DE\u22a5AB\u4e8eE, \u4e14AB=6\uff0e\u6c42\uff1aEB\u7684\u957f\uff0e 69\u3001 \u5982\u56fe\uff0c\u5c71\u5761AB\u7684\u5761\u5ea6\u4e3a5\u223612\uff0c\u4e00\u8f86\u6c7d\u8f66\u4ece\u5c71\u811a\u4e0bA\u5904\u51fa\u53d1\uff0c\u628a\u8d27\u7269\u8fd0\u9001\u5230\u8ddd\u5c71\u811a500 m\u9ad8\u7684B\u5904\uff0c\u6c42\u6c7d\u8f66\u4eceA\u5230B\u6240\u884c\u9a76\u7684\u8def\u7a0b\uff0e
方法一：a²+b²+c²-6a-8b-10c+50=0(a²-6a+9)+(b²-8b+16)+(c²-10c+25)=0
(a-3)²+(b-4)²+(c-5)²=0
所以a-3=0，b-4=0，c-5=0
a=3，b=4，c=5
因为3²+4²=5²
即a²+b²=c²
由勾股定理的逆定理得
以a,b,c为三边的三角形是直角三角形，a,b是直角边，c是斜边
面积为3*4/2=6
方法二：△ABC是直角三角形
a²+b²+c²-6a-8b-10c+50=0
(a²-6a+9)+(b²-8b+16)+(c²-10c+25)=0
(a-3)²+(b-4)²+(c-5)²=0
所以(a-3)²=0,(b-4)²=0,(c-5)²=0
a-3=0，b-4=0，c-5=0
a=3，b=4，c=5
a²+b²=c²
所以△ABC是直角三角形
面积为3*4/2=6
祝你学习天天向上，加油！！！！！！！！！！！
（a-3)^2+(b-4)^2+(c-5)^2=0
a=3 b=4 c=5
为直角三角形S=(1/2)*3*4=6
△ABC是直角三角形
a²+b²+c²-6a-8b-10c+50=0
(a²-6a+9)+(b²-8b+16)+(c²-10c+25)=0
(a-3)²+(b-4)²+(c-5)²=0
所以(a-3)²=0,(b-4)²=0,(c-5)²=0
a-3=0，b-4=0，c-5=0
a=3，b=4，c=5
a²+b²=c²
所以△ABC是直角三角形
解：∵a²＋b²＋c²－6a－8b－10c+50=0
∴(a²－6a+9)+(b²－8b+16)+(c²－10c+25)=0
∴(a－3)²+(b－4)²+(c－5)²=0
又∵(a－3)²≥0,(b－4)²≥0,(c－5)²≥0
∴{a=3 ,b=4,c=5}
∵a²＋b²=3²+4²=25
c²=5²=25
∴a²＋b²=c²
∴△ABC是直角三角形
∴S△ABC=½×a×b
=½×3×4
=6
答：S△ABC=6.
绛旓細鈭碦T鈻矪PN涓紙鈭燩BN=90掳)鍕捐偂瀹氱悊锛歅N骞虫柟=BP骞虫柟+BN骞虫柟 鈭碝N骞虫柟=AM骞虫柟+BN骞虫柟
绛旓細鐢变簬:AD鏄枩杈笲C涓婄殑楂 鎵浠ワ細AD²=BD*DC 鍗筹細DC=AD²/BD=9/3(鈭3)=鈭3 鎵浠ワ細BC=BD+DC=4鈭3 鐢鍕捐偂瀹氱悊姹傚緱AC=2鈭3 2銆佽В锛氬鍥 鍦≧T鈻矨BC涓紝鏍规嵁宸茬煡鏉′欢姹傚緱锛欱C=2,AC=2鈭3 杩嘋浣淐E鈯B浜嶦锛屽垯锛氣柍CED鏄瓑鑵扮洿瑙掍笁瑙掑舰,鏈塂E=CE=鈭3 鐢卞嬀鑲″畾鐞嗘眰寰楋細CE=鈭...
绛旓細4銆佸湪鈻矨BD涓 鈭碘垹ABC=45掳鈭燗DB=90掳 鈭粹垹BAD=45掳锛堜笁瑙掑舰鍐呰鍜岋級鈭碘垹BAD=鈭燗BC=45掳 鈭碅D=BD=6锛堜竴涓笁瑙掑舰涓紝绛夎瀵圭瓑杈癸級5銆佸湪绛夎叞鐩磋鈻矨BE涓 鈭礏E=2鈭2 鈭碅B=AE=鈭氾紙2鈭2锛²=2 浠嶥鐐逛綔DF鈯C浜C浜嶧 鍦ㄧ瓑鑵扮洿瑙掆柍涓 鈭碘垹DEF=45掳 鈭碋F=DF=鈭氾紙鈭2锛²...
绛旓細绛旀鏄疌锛48鎴36鎴30锛汣 杩囩▼濡備笅锛1銆佽В锛氱洿瑙掕竟涓3:4锛岄偅涔堣鐩磋杈瑰垎鍒负3k鍜4k,鍒欐牴鎹鍕捐偂瀹氱悊锛屾枩杈归暱涓5k=25锛岃В寰梜=5 璁句笁瑙掑舰鏂滆竟涓婄殑楂樹负h 鏍规嵁涓夎褰㈤潰绉叕寮忥紝杩欎釜涓夎褰㈤潰绉负3k路4k/2=5k路h/2 h=2.4k=2.4脳5=12 閫夋嫨C绛旀銆2銆12鍒嗚В璐ㄥ洜鏁12=2脳2脳3 閭ｄ箞鏃㈢劧涓夎竟...
绛旓細瑙C=x锛屽垯BC=20-x 鍒欑敱鍕捐偂瀹氱悊 AC^2+AD^2=DC^2 BC^2+BE^2=CE^2 DC绛変簬CE 鍗矨C^2+AD^2=BC^2+BE^2 鍗硏²+8²=锛20-x锛²+12²瑙ｅ緱x=12
绛旓細鈻矪DP鈮屸柍CDF锛孊P=CF锛屸垹C=鈭燚BP D鏄疨F涓偣锛孌E鈯F锛屾墍浠ヤ笁瑙掑舰EPF鏄瓑鑵颁笁瑙掑舰锛孍P=EF 鍦ㄢ柍BEP涓紝鈭燛BD+鈭燚BP=鈭燛BD+鈭燙=90掳 涓擯E=EF锛孊P=CE锛孭E²=BP²+BE²鎵浠 EF²=CF²+BE²,EF=13 绗簩绉嶆柟娉 瑙ｏ細鈭燙=鈭燚AE=45掳锛屸垹ADE+鈭燛DB=90掳...
绛旓細1銆佽В锛氣埖鈭燙=90 锛孉C=6 锛孋B=8 鈭寸敱鍕捐偂瀹氱悊寰楋細AB=鈭氾箼AC²+BC²锕=鈭氾箼6²锕8²锕=10锕欍帩锕氬張鈭碘柍ACD鈮屸柍AED锕欐部鎶樼棔杞村绉帮級鈭碈D=ED锛孉E=AC=6銕濓紝BE=10-6=4銕 鍙堚埖 鈭燙=鈭燗ED=鈭燚EB=90 鈭燘=鈭燘 鈭粹柍ACB鈭解柍DEB 鈭碊E锛廇C=EB锛廋B 鈭碊E锛6=...
绛旓細瑙ｏ細鐢盇B=BC+2寰桝B-BC=2 鍐嶇敱鍕捐偂瀹氱悊寰楋細AB^2-BC^2=AC^2=36 鍒嗚В鍥犲紡寰楋細(AB-BC)(AB+BC)=36 鎵浠B+BC=26/(AB-BC)=36/2=18銆傦紙璇存槑锛氣淾2鈥濊〃绀哄钩鏂癸級
绛旓細灏嗕竴鏍归暱24cm鐨勬湪妫掞紝缃簬搴曢潰鐩村緞涓5cm锛岄珮涓12cm鐨勫渾鏌卞舰姘存澂涓紝璁炬湪妫掗湶鍦ㄨ瀛愬闈㈢殑闀夸负Xcm锛屽垯X鐨勫彇鍊艰寖鍥存槸 11cm锛淴锛12cm 瑙ｆ瀽锛氬綋鏈ㄦ鏂滅潃鏀剧疆鏃讹紝鏉瓙閲岄暱鏄垰锛5²+12²锛夛紳13cm锛岄湶鍑虹殑闀垮害鏈鐭紝鏄24-13锛11cm;褰撴湪妫掔珫鐩存斁缃椂锛屾澂瀛愰噷鐨勯暱鏄12cm锛岄湶鍑虹殑闀垮害鏈闀匡紝鏄...
绛旓細鏈鐩稿綋浜庝粠鍙充笅瑙掑埌涓婅竟鐨勪腑鐐癸紝鍒╃敤鍕捐偂瀹氱悊锛岀洿瑙掕竟鍒嗗埆涓4m鍜屽簳闈㈠懆闀跨殑涓鍗婏紝涔熷氨鏄簳闈㈠渾鍛ㄧ殑涓鍗娿傝繖涓洿瑙掍笁瑙掑舰鐨勫簳灏辨槸鍦嗙殑鍛ㄩ暱锛屽嵆4鍏=12.56m 楂樻槸4m 鏂滆竟鏄牴鍙凤紙涓ゅ彧鑴氳竟鐨勫钩鏂瑰拰锛夊嵆 鏍瑰彿(12.56²+4²)鈮13.18 濡傚浘锛孉鏄铏庯紝B鏄铏紝AB鏄蛋鐨勮矾绾縹...
