关于勾股定理的几道道数学初二题......急啊
\u6570\u5b66\u521d\u4e8c\u52fe\u80a1\u5b9a\u7406\u9898\u89e3\uff1a\u2235S\u25b3ABC=20,\u6240\u4ee5AC*BC=40
\u7531\u9898\u610f\u53ef\u77e5
\u4ee5AC\u4e3a\u76f4\u5f84\u7684\u534a\u5706\u9762\u79ef\u4e3a\uff1a\uff08AC/2)^2*(\u03c0/2)
\u4ee5BC\u4e3a\u76f4\u5f84\u7684\u534a\u5706\u9762\u79ef\u4e3a\uff1a\uff08BC/2)^2*(\u03c0/2)
\u4ee5AB\u4e3a\u76f4\u5f84\u7684\u534a\u5706\u9762\u79ef\u4e3a\uff1a\uff08AB/2)^2*(\u03c0/2)
\u53c8\u2235S\u9634\u5f71=S\u534a\u5706AC+S\u534a\u5706CB-S\u534a\u5706AB+S\u25b3ABC
\u2234\u539f\u5f0f=\uff08AC/2)^2*(\u03c0/2)+\uff08BC/2)^2*(\u03c0/2)+\uff08AB/2)^2*(\u03c0/2)+20
=\uff08\uff08AC/2)^2+(BC/2)^2+\uff08AB/2)^2)*(\u03c0/2)+20
=((AC^2+BC^2-AB^2)/4)+20
\u7531\u52fe\u80a1\u5b9a\u7406\u5f97AC^2+BC^2=AB^2
\u5373AC^2+BC^2-AB^2=0,\u5e26\u5165\u539f\u5f0f\u5b50\u5f97
\u539f\u5f0f=\uff080/4)+20
=20
\u2234S\u9634\u5f71=20cm^2
\u7b54\uff1a\u2014\u2014\u2014\u2014\u2014\u2014
\u56fe\u662f\u5382\u95e8\u7684\u5c3a\u5bf8\uff0c\u4e0d\u662f\u706b\u8f66\uff0c\u706b\u8f66\u7684\u5bbd\u5ea6\u5c0f\u4e8e\u5382\u95e8\uff0c\u6a2a\u5411\u80af\u5b9a\u80fd\u8fc7\uff0c\u4f46\u662f\u8f66\u7684\u9ad8\u5ea6\u5927\u4e8e\u5382\u95e8\u957f\u65b9\u4f53\u7684\u90e8\u5206\uff0c\u6240\u4ee5\u8981\u4f60\u786e\u5b9a\u80fd\u4e0d\u80fd\u5728\u534a\u5706\u5f62\u7684\u7a7a\u95f4\u91cc\u901a\u8fc7\u3002
\u5982\u56fe\uff0c\u6211\u6dfb\u52a0\u7684\u90e8\u5206\u5c31\u662f\u8d27\u8f66\uff0cE\u662fAB\u4e2d\u70b9\uff0c\u8fd9\u9053\u9898\u76ee\u5c31\u662f\u8981\u7528\u52fe\u80a1\u5b9a\u7406\u5c31\u51faX\u7684\u957f\u5ea6\uff0c\u53ea\u8981X\u7684\u957f\u5ea6\u5c0f\u4e8eAB\u8fd9\u4e2a\u534a\u5706\u7684\u534a\u5f84\uff0c\u8d27\u8f66\u5c31\u80fd\u901a\u8fc7\uff08\u5e94\u8be5\u80fd\u7406\u89e3\u5427\uff09
AB\u534a\u5706\u7684\u534a\u5f84\u4e3a1
x=\u6839\u53f7\uff08y^2+z^2)=\u6839\u53f7\uff080.8^2+0.2^2)
x=\u6839\u53f70.68\u80af\u5b9a\u5c0f\u4e8e1
\u6240\u4ee5\u8d27\u8f66\u53ef\u4ee5\u901a\u8fc7
\u4e0d\u61c2\u7684\u5c31\u522b\u4e71\u56de\u7b54\uff0c\u6211\u770b\u7740\u5c31\u60f3\u7b11\uff0c\u8bef\u4eba\u5b50\u5f1f\u3002
∵M为AC中点
∴MC=MA
∵MD⊥BC，∠A=90°
∴△MCD、△MBD、△ABC为Rt△
∴CD²=MC²-MD²=MA²-MD²，MA²+AB²=MB²=BD²+MD²
∴MA²-MD²=BD²-AB²=CD²
∴BD²-CD²=AB²
M为AC中点,CM=AC/2
三角形ABC与三角形DMC相似
所以BC/CM=AC/CD
AC*CM=BC*CD
AC^2/2=CD*(CD+BD)
AC^2=2*CD^2+2CD*BD
又AC^2=BC^2-AB^2=(CD+BD)^2-AB^2=CD^2+2CD*BD+BD^2-AB^2
所以
2*CD^2+2CD*BD=CD^2+2CD*BD+BD^2-AB^2
得BD的平方-CD的平方=AB的平方
连接bm，由勾股定理，得，AB^2=BM^2-AM^2=（BD^2+DM^2）-AM^2,又因为AM=CM
所以AB^2=BD^2+DM^2-CM^2=BD^2+(CM^2-CD^2)-CM^2=BD^2-CD^2
即AB^2=BD^2-CD^2
对作线MD;
AB^2=BM^2-AM^2=BD^2+DM^2-AM^2
=BD^2+CM^2-CD^2-AM^2;
CM=AM;
因此AB^2=BD^2-CD^2;
连结BD
在ΔDBM中BD^2=BM^2-DM^2
在ΔMBA中BM^2=AB^2+AM^2
在ΔCDM中DM^2=CM^2-CD^2
将后两式代入BD^2=BM^2-DM^2
得BD^2=AB^2+AM^2-CM^2+CD^2
M为AC中点 AM=CM AM^2=CM^2 AM^2-CM^2=0
得BD^2=AB^2+CD^2
BD^2-CD^2 =AB^2
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绛旓細(ax^2-3x)(x^2-2x +1)=ax^4-2ax^3+ax^2-3x^3+6x^2-3x =ax^4+(-2a-3)x^3+(a+6)x^2-3x 鍥犱负灞曞紑寮忎腑涓嶅惈x^3 鎵浠(-2a-3)x^3涓殑-2a-3=0 a=-3/2
