球面上有三点A、B、C已知AB=18BC=24AC=30且球心到平面ABC的距离为球半径的1／2那 球面上有3点A B C 已知AB=18,BC=24,AC=3...
\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
\u8bbeO\u4e3a\u7403\u5fc3\uff0cO\u5728\u5e73\u9762ABC\u4e0a\u7684\u6295\u5f71\u4e3aO'\uff0c\u7403\u7684\u534a\u5f84\u4e3ar
\u56e0\u4e3aO\u5230A\uff0cB\uff0cC\u7684\u8ddd\u79bb\u76f8\u7b49\uff0cO'\u5230A\uff0cB\uff0cC\u7684\u8ddd\u79bb\u4e5f\u76f8\u7b49\uff0c\u6240\u4ee5O'\u4e3a\u4e09\u89d2\u5f62ABC\u7684\u5916\u5fc3\uff08\u4e2d\u5782\u7ebf\u4ea4\u70b9\uff09\u3002
\u6ce8\u610f\u5230AB^2+BC^2=AC^2\uff0c\u6240\u4ee5\uff0cO'\u5728AC\u4e0a\uff0c\u5e76\u4e14\u662fAC\u4e2d\u70b9\u3002\u6c42\u51faAO'=15
\u4e0b\u9762\u53ea\u770b\u4e09\u89d2\u5f62AOO'
AO=r\uff0cOO'=r/2\uff0cAO'=15\uff0c\u4e14\u89d2AO'O=90\u5ea6\uff08\u56e0\u4e3aO'\u662fO\u5728\u5e73\u9762ABC\u4e0a\u7684\u6295\u5f71\uff09
\u5f88\u5bb9\u6613\u6c42\u51far=10sqrt3
球心到ABC距离就是OD的长度，OA是球的半径，又知道球心到ABC距离为球半径的一半，可知三角形ODA是一个角为三十度的直角三角形。由AC=30可知DA=15，那么三角形ODA的斜边长就求出来是十倍根三了。
绛旓細鐢遍鎰忥紝鐭ヤ笁瑙掑舰ABC涓虹瓑鑵扮洿瑙掍笁瑙掑舰锛屼笖AC涓烘枩杈广傚垯鐞冨績O鍒癆C鐨勪腑鐐笻涓1锛屼笖OH鍨傜洿浜嶢C銆傝繛鎺A锛屽湪鐩磋涓夎褰HA涓紝鐢ㄥ嬀鑲″畾鐞嗗彲浠ユ眰鍑虹悆鐨勫崐寰凴銆備綋绉氨鍙互濂楀叕寮忎簡銆
绛旓細18^2+24^2=30^2 鈻矨BC涓虹洿瑙掍笁瑙掑舰 鈻矨BC澶栨帴鍦嗙殑鐩村緞涓30 鐞冨崐寰勪负R R^2-(R/2)^2=15^2 R^2*3/4=15^2 R^2=300 鐞冮潰绉疭 = 4蟺R^2 =4*3.14*300 =3768骞虫柟鍘樼背
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
绛旓細璁句笁瑙掑舰ABC鐨勫鎺ュ渾鍗婂緞涓簉,鐞冨崐寰勪负R,鐞冨績O鍒板钩闈BC鐨勮窛绂籬 鍥犱负AB^2+BC^2=3^2+4^2=5^2=AC^2 鎵浠BC鏄洿瑙掍笁瑙掑舰 鍒檙=AC/2=2.5 鍥燫^2=h^2+r^2 h^2=R^2-r^2=6.5^2-2.5^2=36 瑙ｅ緱h=6
绛旓細鑻ヨB=90 鍒欏搴斿姬AC=180搴 鍗犳嵁浜嗙悆鐨勪竴鍗 AC宀備笉鏄瓑浜庣洿寰?
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3=1.732, cos0.6=0.825, BC=5 鐒跺悗鐒跺悗鍚勪釜瑙掑害灏卞彲浠ヤ竴涓绠楀嚭鏉ワ紝璁惧嚭鍨傜嚎娈典氦鐐笶銆傛渶鍚庡彲浠ユ牴鎹粠涓涓偣鐨3涓悜閲忥紝寰楀埌琛ㄧずOE鐨勫悜閲忋傚嵆涓烘墍姹
绛旓細ABC涓虹瓑杈逛笁瑙掑舰 涓績鍒板悇涓《鐐硅窛绂讳负2涔樻牴鍙3 A鍜屼腑蹇冨拰鐞冨績缁勬垚涓涓洿瑙掍笁瑙掑舰 鎵浠ョ悆鐨勫崐寰勭殑骞虫柟涓2涔樻牴鍙3鐨勫钩鏂+2鐨勫钩鏂 鍗婂緞涓4 浣撶Н涓4/3蟺r^3=256/3蟺
绛旓細璁剧悆蹇冧负0 0B=0C=0A=R 涓夎褰BC涓虹洿瑙掍笁瑙掑舰 鐞冨績鍒板钩闈BC鐨勮窛绂荤瓑浜庣悆鍗婂緞鐨勪竴鍗 杩0鍋欰BC鐨勫瀭绾 浜や簬H H涓篈BC鐨勪腑鍨傜嚎浜ょ偣(鍥犱负0B=0C=0A,OH涓瀹 鎵浠A=HB=HC)鍙瘉H涓篈C涓偣 閭ｄ箞R^2=(R/2)^2+(5/2)^2 瑙ｅ緱R=5/鏍瑰彿3 鏈変粈涔堜笉鎳傚彲浠ラ棶鎴 鏂板勾蹇箰~!
绛旓細閫氳繃鐞冨績O鍋氬钩闈㈢殑鐨勫瀭绾匡紝鍐嶈繃鍨傝冻O鈥欒繛鎺ュ瀭瓒充笌A锛圔鎴朇锛屽摢涓偣鍧囧彲锛夛紝寤朵几鍒癇C涓庝箣鐩镐氦浜嶥鐐广宸茬煡锛骞抽潰ABC涓庣悆蹇僌鐨勮窛绂绘伆濂戒负鐞冨崐寰勭殑涓鍗婏紝閭ｄ箞AO鈥欏嵆涓虹悆鍗婂緞鐨勪簩鍒嗕箣鏇翠笁銆傚亣鎯充竴骞抽潰ABC锛骞抽潰涓庣悆鐨勭浉浜よ建杩逛负涓涓渾锛屼笖涓夌偣A銆丅銆丆鍧囧湪涓涓渾涓婏紝璇存槑锛岃鍦嗕负涓夎褰BC鐨勫...
