求初二勾股定理的经典题目 初二数学勾股定理难一点的应用题，要有答案。谢谢。
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AC===4\uff08m\uff09\uff0c
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\u2234BC=3+2=5km\uff0c
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若△ABC的三边a,b,c满足a^2+b^2+c^2+50=6a+8b+10c，求△ABC的面积。
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
如图，C为线段BD上一动点，分别过点B,D作AB⊥BD，ED⊥BD，连结AC,EC。已知AB=5,DE=1,BD=8,设CD=X.
(1)用含X的代数式表示AC+CE的最小值。
（2）试求AC+CE的最小值。
（3）根据（2）中的规律和结论，请构图求出代数式（√x²+4）+（√（12-x)²+9）的最小值。
（1）AC＋CE＝√(AB²＋BC²)＋√(CD²＋DE²
＝√[5²＋(8－x)²]＋√(x²＋1²)
＝√(x²＋1)＋√[(8－x)²＋25]
（2）设点A关于BD的对称点为A'，连接A'E交BD于点C
由三角形两边之和大于第三边可知，A'C＋CE最小
又由对称性可知AC＝A'C，故AC＋EC最小
由勾股定理可知，A'E＝√[8²＋(5＋1)²]＝10
故AC＋CE的最小值为10
（3）如图，AB⊥BD，ED⊥BD，AB＝2，DE＝3，BD＝12，C在BD上，设BC＝x
则AC＋EC＝√(x²＋4)＋√[(12－x)²＋9]
设点A关于BD的对称点为A'，连接A'E交BD于点C
由（2）可知AC＋EC最小
在Rt△A'EF中，A'F＝BD＝12，EF＝DE＋DF＝DE＋A'B＝DE＋AB＝3＋2＝5
故A'E＝√(A'F²＋EF²)＝√12²＋5²)＝13
即√(x²＋4)＋√[(12－x)²＋9]的最小值为13
绛旓細瑙ｏ細濡傚浘鎵绀 锛1锛夊湪鐩磋涓夎褰CD涓紝BD=鈭(AB²-AD²)=鈭(150²-90²)=120km 120/20=6(灏忔椂)鍙伴涓績缁忚繃6灏忔椂浠嶣鐐圭Щ鍒癉鐐广傦紙2锛60/6=5(灏忔椂)锛屽嵆娓镐汉闇瑕5灏忔椂鎵嶈兘鎾ょ鍒板畨鍏ㄥ湴鐐癸紝鍥犳锛屽湪鎺ュ埌鍙伴璀︽姤鍚庣殑1灏忔椂鍐呭繀椤绘挙绂伙紝鏈濂介夋嫨娌緽C鏂瑰悜鎾ょ ...
绛旓細鍥炵瓟锛氱鍗侀 a²+b²=c² a+b=14 c=10 鍒檃=6 b=8 闈㈢Н鏄 24
绛旓細璁炬按娣盭灏恒傝嵎鑺遍珮涓0.5+X(鍗虫枩杈)銆備竴鐩磋杈逛负2灏(钀借姳璺濇牴浜屽昂鏁)锛屽彟涓鐩磋杈逛负姘存繁涓篨銆傛嵁鍕捐偂瀹氱悊鍙煡锛歑^2+2^2=(0.5+X)^2 瑙ｅ緱X=3.75灏猴紝鍗虫按娣变负3.75灏
绛旓細鍕捐偂瀹氱悊棰樼洰濡備笅锛1銆佺洿瑙掍笁瑙掑舰涓鐩磋杈逛负12cm锛屾枩杈归暱涓13cm锛屽垯瀹冪殑闈㈢Н鏄灏戯紵2銆佹湁鍏牴缁嗘湪妫掞紝瀹冧滑鐨勯暱鍒嗗埆鏄2锛4锛6锛8锛10锛12锛堝崟浣嶏細cm锛夛紝棣栧熬杩炵粨鑳芥惌鎴愮洿瑙掍笁瑙掑舰鐨勪笁鏍圭粏鏈ㄦ鍒嗗埆鏄灏戯紵3銆佸凡鐭ワ紝鍦ㄢ柍ABC涓紝AB=15锛孉C=13锛岄珮AD=12锛屽垯鈻矨BC鐨勫懆闀夸负澶氬皯锛4銆佸湪Rt鈻矨BC涓...
绛旓細璁続D=x锛孊D=鏍瑰彿(AB^2-AD^2)=鏍瑰彿(25-x^2)CD=鏍瑰彿(AC^2-AD^2)=鏍瑰彿(49-x^2)BD+CD=BC=6 鏍瑰彿(25-x^2)+鏍瑰彿(49-x^2)=6 瑙ｅ緱 x=2鏍瑰彿6 鎵浠D=2鏍瑰彿6 涓夎褰BC鐨勯潰绉=1/2*BC*AD=1/2*6*2鏍瑰彿6=6鏍瑰彿6 2銆佽杩欎釜涓夎褰㈢殑鑵伴暱涓簒锛屽垯搴曡竟=2*鏍瑰彿(x^2-4^2)=2...
绛旓細4銆佲柍ABC涓,AB=AC,鈭燘AC=120掳,AB=12cm,鍒橞C杈逛笂鐨勯珮AD = 锛縞m.5銆佸凡鐭ワ細鈻矨BC涓紝鈭燗CB=90掳锛孋D鈯B浜嶥锛孊C= ,DB=2cm ,鍒橞C锛濓伎 cm, AB= 锛縞m, AC= 锛縞m.6銆佸鍥撅紝鏌愪汉娆叉í娓′竴鏉℃渤锛岀敱浜庢按娴佺殑褰卞搷锛屽疄闄呬笂宀稿湴鐐笴鍋忕娆插埌杈剧偣B200m锛岀粨鏋滀粬鍦ㄦ按涓疄闄呮父浜520m锛屾眰璇ユ渤娴佺殑...
绛旓細1.璁句笁涓繛缁嚜鐒舵暟涓簒,x+1,x+2,鏍规嵁鍕捐偂瀹氱悊寰,x^2+(x+1)^2=(x+2)^2瑙ｅ緱x=3,x=-1,涓夎褰㈢殑杈归暱涓嶈兘涓鸿礋鏁. 绛旀娌℃湁浜 2.杩炴帴AC鍙緱鐩磋涓夎褰BC寰楁枩杈笰C涓5,寰楃洿瑙掍笁瑙掑舰DAC,瑙扗AC涓90搴,涓や釜鐩磋涓夎褰㈤潰绉浉鍔犲嵆鍙. 绛旀锛36 3,鍏堢敾鍑哄浘褰,寤堕暱AD,BC浜や簬鐐笶寰楄AEB30...
绛旓細灏卞彲浠ユ眰鍑烘瀛愭粦鍔ㄤ簡8绫炽傝В棰樿繃绋嬪涓 瑙ｏ細鐢遍鍙煡锛岃繖鏃舵瀛愪笌鍦伴潰鍜屽褰㈡垚浜嗕竴涓洿瑙掍笁瑙掑舰 鍦≧t涓夎褰腑锛孉B^2+BC^2=AC^2 AB^2+7^2=25^2 姹傚嚭 AB=24 鍚庣粡婊戝姩锛屾瀛愪笌鍦伴潰鍜屽褰㈡垚鐨勪粛涓虹洿瑙掍笁瑙掑舰 鍚岀悊姹傚嚭BC鎾囦负15 鍒欐瀛愭粦鍔ㄤ簡15-7=8锛堢背锛夌瓟锛氭瀛愭粦鍔ㄤ簡8绫炽
绛旓細鐢鍕捐偂瀹氱悊鍙煡锛孉'E锛濃垰[8²锛(5锛1)²]锛10 鏁匒C锛婥E鐨勬渶灏忓间负10 锛3锛夊鍥撅紝AB鈯D锛孍D鈯D锛孉B锛2锛孌E锛3锛孊D锛12锛孋鍦˙D涓婏紝璁綛C锛漻 鍒橝C锛婨C锛濃垰(x²锛4)锛嬧垰[(12锛峹)²锛9]璁剧偣A鍏充簬BD鐨勫绉扮偣涓篈'锛岃繛鎺'E浜D浜庣偣C 鐢憋紙2锛夊彲鐭C锛婨C...
绛旓細濡傚浘7鎵绀猴紝AB鏄竴鏉′笢瑗胯蛋鍚戠殑椹矾锛屽湪A鐐圭殑涓滃崡鏂瑰悜1000鈭2 鐨勫湴鏂规湁涓鎵涓C锛岀幇鏈変竴鎷栨媺鏈鸿嚜瑗垮悜涓滆椹讹紝鎷栨媺鏈哄彂鍑虹殑鍣煶800 鑼冨洿鍐呭潎鏈夊奖鍝嶏紝璇ユ嫋鎷夋満鍦ㄨ椹惰繃绋嬩腑瀵逛腑瀛鏈夊奖鍝嶅悧锛熶负浠涔堬紵杩嘋鐐逛綔AD鈯B锛岃鍨傝冻涓篋 鈭碈D鏄偣C鍒扮洿绾緼B鏈鐭殑璺濈 鈭佃BAC=45掳 鈭磋DCA=45掳 鈭碅D=...
