### Re: [數學]求圓半徑

G@ry 寫到:
........... 寫到:#ed_op#div#ed_cl##ed_op#font face="Verdana"#ed_cl#有一個直角三角形AOB，∠AOB=90°，以點O為圓心作一圓，該圓交#ed_op#br#ed_cl#OA於C，OB於D，並相切AB於T；已知AC=16，BD=135，求圓半徑。#ed_op#/font#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl#
#ed_op#br#ed_cl#設OT為x;#ed_op#br#ed_cl#圓切AB於一點T =&gt; OT(=半徑)垂直AB =&gt; OT*AB/2=三角形面積#ed_op#br#ed_cl#x = (16+x)*(135+x)/√((16+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+(135+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)/√(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481)#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl##ed_op#br#ed_cl#2x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+18481x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+27121x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+652320x+4665600#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#-8640x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-652320x-4665600=0#ed_op#br#ed_cl#待續...#ed_op#br#ed_cl#
#ed_op#br#ed_cl#終於完整解出：&nbsp; x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#-8640x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-652320x-4665600 =#ed_op#br#ed_cl#(x-120)*{x-2[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)-20]}*{x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+2[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)+40]x+8#ed_op#span style="color: rgb(255, 0, 127);"#ed_cl#y#ed_op#/span#ed_cl#}, #ed_op#br#ed_cl##ed_op#span style="color: rgb(255, 0, 127);"#ed_cl#y#ed_op#/span#ed_cl#=#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(4004225-625275√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(4004225+625275√41)+10[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)+24]#ed_op#br#ed_cl#若要再解#ed_op#span style="color: rgb(255, 0, 127);"#ed_cl#y#ed_op#/span#ed_cl#便要用上虛數來解(b#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#&lt;4ac)...太麻煩了~~~#ed_op#br#ed_cl#當中 120&gt;0;&nbsp; 2[#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970-630√41)+#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#√(3970+630√41)-20]~=2(-4+20-20)=-8&lt;0，#ed_op#br#ed_cl#故答案為120。#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#若有高人能再簡化這些數字，請務必賜教賜教，謝！... 小弟總覺得還是有辦法再行簡化的....只是小的道行太低.... : (#ed_op#br#ed_cl##ed_op#br#ed_cl#

### Re: [數學]求圓半徑

........... 寫到:#ed_op#div#ed_cl##ed_op#font face="Verdana"#ed_cl#有一個直角三角形AOB，∠AOB=90°，以點O為圓心作一圓，該圓交#ed_op#br#ed_cl#OA於C，OB於D，並相切AB於T；已知AC=16，BD=135，求圓半徑。#ed_op#/font#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl##ed_op#div#ed_cl##ed_op#/div#ed_cl#
#ed_op#br#ed_cl#設OT為x;#ed_op#br#ed_cl#圓切AB於一點T =&gt; OT(=半徑)垂直AB =&gt; OT*AB/2=三角形面積#ed_op#br#ed_cl#x = (16+x)*(135+x)/√((16+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+(135+x)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)/√(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481)#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#(2x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+302x+18481) = (x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+151x+2160)#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl##ed_op#br#ed_cl#2x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+18481x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl# = x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#+302x#ed_op#sup#ed_cl#3#ed_op#/sup#ed_cl#+27121x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#+652320x+4665600#ed_op#br#ed_cl#x#ed_op#sup#ed_cl#4#ed_op#/sup#ed_cl#-8640x#ed_op#sup#ed_cl#2#ed_op#/sup#ed_cl#-652320x-4665600=0#ed_op#br#ed_cl#待續...#ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl##ed_op#br#ed_cl#

### [數學]求圓半徑

#ed_op#DIV#ed_cl##ed_op#FONT face=Verdana#ed_cl#有一個直角三角形AOB，∠AOB=90°，以點O為圓心作一圓，該圓交#ed_op#BR#ed_cl#OA於C，OB於D，並相切AB於T；已知AC=16，BD=135，求圓半徑。#ed_op#/FONT#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl##ed_op#DIV#ed_cl##ed_op#/DIV#ed_cl#