初二勾股定理数学题
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\u5219\u6709X2-Y2=169
\u5f97X=85,Y=84
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30\u6d77\u91cc
2\u3001\u662f12\u7c73
2、由折叠可得AF=AB ∠ADC=∠AFE △ABE全等于△AFE并可推出△ABF相似于△FCE 同时可得AB∥FC=BF∥CE
AF=AD=BC=10 由勾股定理可得 BF=6 CF=BC—BF=10-6=4 则由AB∥FC=BF∥CE得
8/4=6/CE CE=3
3、由给出条件可得 AB=10 △ADC’∽△ABC BC=BC'
则AC'/AC=C'D/BC AC'=AB-BC'=AB-BC=10-6=4
4/8=C‘D/6 C’D=3
△ADC’=C'D*AC'/2=3*4/2=6
1、√(25*25-20*20)-√(25*25-24*24)=8
2、由折叠可得AF=AB ∠ADC=∠AFE △ABE全等于△AFE并可推出△ABF相似于△FCE 同时可得AB∥FC=BF∥CE
AF=AD=BC=10 由勾股定理可得 BF=6 CF=BC—BF=10-6=4 则由AB∥FC=BF∥CE得
8/4=6/CE CE=3
3、由给出条件可得 AB=10 △ADC’∽△ABC BC=BC'
则AC'/AC=C'D/BC AC'=AB-BC'=AB-BC=10-6=4
4/8=C‘D/6 C’D=3
△ADC’=C'D*AC'/2=3*4/2=6
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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
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绛旓細瑙ｏ細璁続D=x锛屽垯CD=12-x FC²=DF²+CD²=16+(12-x)²BC²=36 濡傛灉FC涓烘枩杈癸紝鏈夛細16+(12-x)²=36+x²x=124/24=31/6 濡傛灉浠D涓烘枩杈癸紝鏈夛細x²=16+(12-x)²+36 x=49/6 濡傛灉浠C涓烘枩杈癸紝鏈夛細36=16+(12-x)²+x²...
绛旓細AC^2-CD^2=AD^2=AB^2-BD^2 鎵浠,7^2-CD^2=13^2-(12-CD)^2 49-CD^2=169-144+24CD-CD^2 CD=1
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绛旓細1銆佺敱浜20^2=12^2+16^2,鎵浠モ垹CDB=90锛岃AD=x,鍒欙細(12+x)^2=x^2+16^2.姹傚嚭x=14/3,鈻矨BC鐨勫懆闀=20+(12+14/3)*2=160/3.2銆佷粠绗2姝ュ紑濮嬪嚭閿欍-鍙峰啓鎴+鍙蜂簡锛岃屼笖浠庣2姝ュ埌绗3姝ヤ篃鏄敊鐨勩傛渶缁堢瓟妗堟槸姝ｇ‘鐨刢 ²=a ²+b ²銆
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绛旓細璁綜D=X, 鍦ㄧ洿瑙掍笁瑙掑舰ACD涓庣洿瑙掍笁瑙掑舰ABD涓侫D^2=AC^2-X^2 AD^2=AB^2-(BC+X)^2 鎵浠 AC^2-X^2=AB^2-(BC+X)^2 10^2-X^2=17^2-(9+X)^2 100-X^2=289-(81+18X+X^2)18X=108 X=6 鎵浠 AD^2=10^2-6^2 =64 AD=8 ...
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