球面上有三个点A、B、C,已知AB=18,BC=24,AC=30,且球心到平面ABC的距离半径的二分之一，求这个球的半径。
\u7403O\u7403\u9762\u4e0a\u6709\u4e09\u70b9A\u3001B\u3001C\uff0c\u5df2\u77e5AB=18\uff0cBC=24\uff0cAC=30\uff0c\u4e14\u7403\u534a\u5f84\u662f\u7403\u5fc3O\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u76842\u500d\uff0c\u6c42\u7403O\u7684\u8868\u9762\u89e3\u7b54\uff1a\u89e3\uff1a\u7403\u9762\u4e0a\u4e09\u70b9A\u3001B\u3001C\uff0c\u5e73\u9762ABC\u4e0e\u7403\u9762\u4ea4\u4e8e\u4e00\u4e2a\u5706\uff0c\u4e09\u70b9A\u3001B\u3001C\u5728\u8fd9\u4e2a\u5706\u4e0a\u2235AB=18\uff0cBC=24\uff0cAC=30\uff0cAC2=AB2+BC2\uff0c\u2234AC\u4e3a\u8fd9\u4e2a\u5706\u7684\u76f4\u5f84\uff0cAC\u4e2d\u70b9M\u5706\u5fc3\u7403\u5fc3O\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u5373OM=\u7403\u534a\u5f84\u7684\u4e00\u534a=12R\u25b3OMA\u4e2d\uff0c\u2220OMA=90\u00b0\uff0cOM=12R\uff0cAM=12AC=30\u00d712=15\uff0cOA=R\u7531\u52fe\u80a1\u5b9a\u7406\uff0812R\uff092+152=R2\uff0c34R2=225\u89e3\u5f97R=103\u7403\u7684\u8868\u9762\u79efS=4\u03c0R2=1200\u03c0\uff08\u9762\u79ef\u5355\u4f4d\uff09
\u60f3\u60f3\u518d\u8bf4\uff0c\u6050\u6015\u5f97\u501f\u52a9\u7a7a\u95f4\u76f4\u89d2\u5750\u6807\u7cfb\u3002 \u6709\u4e86\uff01\u25b3ABC\u662f\u76f4\u89d2\u4e09\u89d2\u5f62\uff0c\u7403\u5fc3\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u5c31\u662f\u7403\u5fc3\u5230AC\u8fb9\u4e2d\u70b9\u7684\u8ddd\u79bb\u3002\u8bbe\u7403\u534a\u5f84\u4e3aR\uff0c\u7531\u52fe\u80a1\u5b9a\u7406\u77e5\uff0cRR-RR/4=225\uff0c\u53733RR/4=225,RR/4=75\uff0cRR=300\uff0cR\uff1e0\uff0c\u6545R=10\u221a3.
若角B=90则对应弧AC=180度
占据了球的一半
AC岂不是等于直径?
绛旓細鈭碘柍ABC鏄竴涓竟闀夸负l鐨勭瓑杈逛笁瑙掑舰锛屸埓鈻矨BC鐨勯珮AD=32锛庤鈻矨BC鐨勫鎺ュ渾鍦嗗績璁句负O'锛屽緱鍒癆O'=23AD=33鍐嶈鐞冨績涓篛锛屽洜涓虹悆蹇僌鍒板钩闈BC鐨勮窛绂荤瓑浜庣悆鍗婂緞鐨13锛屾墍浠O'=13OA锛Rt鈻砄O'A涓紝O'A2+OO'2=OA2锛屽嵆锛33锛2+19OA2=OA2鈭89OA2=13锛屾晠OA=64锛屽嵆鐞冨崐寰勬槸64鏁呯瓟妗堜负64 ...
绛旓細濡傚浘鎵绀猴細鈭礎B=6 cm锛孊C=8cm锛孋A=10cm锛屸埓鈭CBA=90掳鈭村彇AC鐨勪腑鐐筂锛屽垯鐞冮潰涓夾銆丅銆C涓鐐规墍鍦ㄧ殑鍦嗗嵆涓衡姍M锛岃繛鎺M锛屽垯OM鍗充负鐞冨績鍒板钩闈BC鐨勮窛绂伙紝鍦≧t鈻砄MA涓紝OM=13cm锛孧A=5cm锛屸埓OA=13cm锛屽嵆鐞冪悆鐨勫崐寰勪负13cm锛庢晠閫堿锛
绛旓細杩欐槸涓涓┖闂村嚑浣曢锛岃寤虹珛绌洪棿鍥惧舰姒傚康銆傜悆蹇冿紙鍋囪涓篛鐐癸級锛屼笁瑙掑舰ABC鎵鍦ㄥ钩闈㈢殑澶栨帴鍦嗙殑鍦嗗績涓篗锛 鏍规嵁棰樻剰锛屼笁瑙掑舰OMA涓虹洿瑙掍笁瑙掑舰锛屼笖OM涓烘墍姹傜殑鐞冨績鍒板钩闈BC鐨勮窛绂汇傜敱浜庝笁瑙掑舰ABC涓虹洿瑙掍笁瑙掑舰锛屾墍浠ュ鎺ュ渾鐨勫渾蹇冨湪杈AC涓婏紝涓斾负鍏朵腑鐐癸紝鎵浠M=5锛屾墍浠ワ細OA^2=OM^2+AM^2,OM=10*(2...
绛旓細瑙ｉ杩囩▼濡備笅锛氳A锛坅1,a2,a3)B(b1,b2,b3)C(c1,c2,c3)鍗婂緞R 鐞冨績O(x,y,z)|AO|^2=(x-a1)^2+(y-a2)^2+(z-a3)^2 鍚屾牱寰楀埌|BO|锛寍CO| AO=BO=CO=R 瑙ｄ笁鍏冧竴娆℃柟绋嬬粍鍗冲緱銆
绛旓細鍦ㄤ笁瑙掑舰ABC涓紝鏍规嵁浣欏鸡瀹氱悊锛宑osA=(18^2+30^2-24^2)/(2*18*30)=3/5,sinA=4/5,璁句笁瑙掑舰澶栨帴鍦嗗崐寰剅,2r=a/sinA=24/(4/5),r=15,璁剧悆鍗婂緞涓篟锛屽鸡蹇冭窛涓篸,d=R/2,鏍规嵁鍕捐偂瀹氱悊锛孯^2=(R/2)^2+r^2,R=10鈭3锛岀悆闈绉疭=4蟺R^2=1200蟺....
绛旓細鈭6 2 +8 2 =10 2 锛屸埓鈻ABC涓篟t鈻筹紟鈭电悆蹇僌鍦ㄥ钩闈BC鍐呯殑灏勫奖M鏄埅闈㈠渾鐨勫渾蹇冿紝鈭碝鏄AC鐨勪腑鐐逛笖OM鈯C锛庡湪Rt鈻砄AM涓紝OM= O A 2 -A M 2 =12锛庘埓鐞冨績鍒板钩闈BC鐨勮窛绂讳负12锛
绛旓細骞抽潰ABC涓鐞冮潰鐨勫垏闈㈡槸涓涓渾銆侫BC涓夌偣鍧囧湪杩欎釜鍦嗗懆涓娿傚洜涓轰笁瑙掑舰ABC鏄竴涓洿瑙掍笁瑙掑舰锛屾墍浠C鏄鍦嗙殑鐩村緞銆傜悆蹇僶鍒板钩闈BC鐨勫瀭瓒宠偗瀹氬湪BC鐨勪腑鐐广傛墍浠ワ紝鐞冪殑鍗婂緞涓烘牴鍙(11+5^2)=鏍瑰彿36=6 鐞冪殑浣撶Н涓4PI*6^3/3=288PI
绛旓細鐢遍鎰AB=5锛孊C=12锛孉C=13锛屽彲鐭モ垹ABC=90掳锛岄潰ABC涓庣悆蹇冭窛绂讳负3R2锛屾濂芥槸鐞冨績鍒癆C鐨勪腑鐐圭殑璺濈锛屾墍浠ョ悆鐨勫崐寰勬槸锛歊2=132鎵浠=13锛庢晠绛旀涓猴細13锛
绛旓細AB=BC=CA=2 鈭碅BC涓虹瓑杈逛笁瑙掑舰 鍏跺鎺ュ渾鍗婂緞r = ( AB/2) /cos30掳 = 1/(鏍瑰彿3/2) = 2/鏍瑰彿3 A銆丅銆丆鐨勭晫闈㈠拰鐞冨績鐨勮窛绂荤瓑浜庣悆鍗婂緞R鐨勪竴鍗 鈭碦^2 = (R/2)^2 + r^2 鈭3R^2/4 = r^2 = 4/3 R^2 = (4/3)^2 R=4/3 鐞冭〃闈㈢Н = 4蟺R^2 = 4蟺*(4/3)...
绛旓細棣栧厛姹傚嚭鎴潰ABC鐨勫鎺ュ渾鐨勫崐寰勩侫C=BC=6锛孉B=4锛屽欢闀緼O浜C浜嶥锛屼笖AD鍨傜洿BC锛鏍规嵁鐩磋鈻矨BD姹傚嚭AD=4鈭2銆傛牴鎹洿瑙掆柍OBD锛岀粨鍚堝嬀鑲″畾鐞嗗緱锛歄B*OB=OD*OD+BD+BD,鍏蜂綋浠ｅ叆锛歊*R=锛4鈭2-R锛夛紙4鈭2-R锛+2*2銆傛暣鐞嗗緱R=2鈭2銆傝繃鐞冮潰涓婁笁鐐笰 ,B,C鐨勬埅闈㈠埌鐞冨績鐨勮窛绂籨绛変簬鐞冨崐寰剅鐨...
