已知球面上有三点A、B、C，AB=6cm，BC=8cm，AC=10cm，且球心O到平面ABC的距离为12，则球的半径为（ ） 如图，已知⊙O上的三点A、B、C，且AB=AC=6cm，BC...
\u5df2\u77e5\u7403\u9762\u4e0a\u7684\u4e09\u70b9A\u3001B\u3001C\uff0cAB=6\uff0cBC=8\uff0cAC=10\uff0c\u7403\u7684\u534a\u5f84\u4e3a13\uff0c\u6c42\u7403\u5fc3\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u22356 2 +8 2 =10 2 \uff0c\u2234\u25b3ABC\u4e3aRt\u25b3\uff0e\u2235\u7403\u5fc3O\u5728\u5e73\u9762ABC\u5185\u7684\u5c04\u5f71M\u662f\u622a\u9762\u5706\u7684\u5706\u5fc3\uff0c\u2234M\u662fAC\u7684\u4e2d\u70b9\u4e14OM\u22a5AC\uff0e\u5728Rt\u25b3OAM\u4e2d\uff0cOM= O A 2 -A M 2 =12\uff0e\u2234\u7403\u5fc3\u5230\u5e73\u9762ABC\u7684\u8ddd\u79bb\u4e3a12\uff0e
\uff081\uff09\u8bc1\u660e\uff1a\u2235AB=AC\uff0c\u2234 AB = AC \uff0c\u2234\u2220AOB=\u2220AOC\uff1b\uff082\uff09\u8bbeOA\u4ea4BC\u4e8e\u70b9D\uff0c\u2235 AB = AC \uff0c\u2234OA\u22a5BC\uff0c\u2234BD= 1 2 BC= 1 2 \u00d710=5\uff08cm\uff09\uff0c\u2235AB=6cm\uff0c\u2234\u5728Rt\u25b3ABD\u4e2d\uff0cAD= A B 2 -B D 2 = 11 \uff08cm\uff09\uff0c\u2235OB=Rcm\uff0c\u5219OD=\uff08R- 11 \uff09cm\uff0c\u2235OB 2 =OD 2 +BD 2 \uff0c\u2234R 2 =\uff08R- 11 \uff09 2 +25\uff0c\u89e3\u5f97\uff1aR= 18 11 11 \uff08cm\uff09\uff1b\uff083\uff09\u22353\uff1c 11 \uff1c4\uff0c\u22344\uff1c 18 11 11 \uff1c6\u2234m=6\uff0cn=4\uff0e
如图所示：∵AB=6 cm，BC=8cm，CA=10cm，
∴∠CBA=90°
∴取AC的中点M，则球面上A、B、C三点所在的圆即为⊙M，连接OM，则OM即为球心到平面ABC的距离，
在Rt△OMA中，OM=13cm，MA=5cm，
∴OA=13cm，即球球的半径为13cm．
故选A．
绛旓細濡傚浘鎵绀猴細鈭礎B=6 cm锛孊C=8cm锛孋A=10cm锛屸埓鈭燙BA=90掳鈭村彇AC鐨勪腑鐐筂锛屽垯鐞冮潰涓夾銆丅銆丆涓夌偣鎵鍦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
绛旓細宸茬煡鐞冮潰涓婃湁A,B,C涓夌偣,宸茬煡AB=BC=AC=2鈭3,涓旂悆蹇冨埌骞抽潰ABC鐨勮窛绂讳负1,鍒欒鐞冪殑浣撶Н涓? 鍒嗕韩 寰俊鎵竴鎵 缃戠粶绻佸繖璇风◢鍚庨噸璇 鏂版氮寰崥 QQ绌洪棿 涓炬姤 娴忚19 娆 1涓洖绛 #鐑# 鍏ㄧ悆骞哥鎶ュ憡鍑虹倝鑺叞灞呮棣,瀵规浣犳庝箞鐪?bdwk灏忓彿 2013-09-14 bdwk灏忓彿 閲囩撼鏁:11 鑾疯禐鏁:28 LV4 鎿呴暱:鑰冪爺 ...
绛旓細瑙ｇ瓟锛氳В锛鐞冮潰涓婁笁鐐笰銆丅銆C锛骞抽潰ABC涓庣悆闈氦浜庝竴涓渾锛屼笁鐐笰銆丅銆丆鍦ㄨ繖涓渾涓娾埖AB=18锛孊C=24锛孉C=30锛孉C2=AB2+BC2锛屸埓AC涓鸿繖涓渾鐨勭洿寰勶紝AC涓偣M鍦嗗績鐞冨績O鍒板钩闈BC鐨勮窛绂诲嵆OM=鐞冨崐寰勭殑涓鍗=12R鈻砄MA涓紝鈭燨MA=90掳锛孫M=12R锛孉M=12AC=30脳12=15锛孫A=R鐢卞嬀鑲″畾鐞嗭紙12R锛2+152...
绛旓細瑕佸缓绔嬬┖闂村浘褰㈡蹇点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寮鏍瑰彿)銆
绛旓細AB²+BC²=AC²銆備笁瑙掑舰涓虹洿瑙掍笁瑙掑舰锛孉C涓衡柍ABC鎵鍦ㄥ渾鐨勭洿寰勩傚垯鐞冨崐寰勪负R²-锛0.5R锛²=AC²=900 R=20*3^0.5
绛旓細鎯虫兂鍐嶈锛屾亹鎬曞緱鍊熷姪绌洪棿鐩磋鍧愭爣绯汇 鏈変簡锛佲柍ABC鏄洿瑙掍笁瑙掑舰锛岀悆蹇冨埌骞抽潰ABC鐨勮窛绂诲氨鏄悆蹇冨埌AC杈逛腑鐐圭殑璺濈銆傝鐞冨崐寰勪负R锛岀敱鍕捐偂瀹氱悊鐭ワ紝RR-RR/4=225锛屽嵆3RR/4=225,RR/4=75锛孯R=300锛孯锛0锛屾晠R=10鈭3.
绛旓細鍋歄E鈯ラ潰ABC锛鍒橭E=r/2锛孫A=OB=OC=r锛岃涓夎褰BC鐨勫鍒囧渾涓衡姍O1锛屾樉鐒禘灏辨槸鈯橭1鐨勫渾蹇 鐢变綘缁欑殑鏉′欢鍙互绠楀嚭AB^2+BC^2=AC^2,鍗矨B鈯C锛屾墍浠C杩団姍O1鐨勫渾蹇冿紝鍗崇偣E鍦AC涓婏紝涓擜E=EC=AC/2=15锛孫E^2+AE^2=OA^2 (r/2)^2+15^2=r^2 r=10鈭3 鍗宠鐞冪殑鍗婂緞绛10鈭3 ...
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
绛旓細瑙ｇ瓟锛氳В锛氬鍥炬墍绀猴細鈭礎B=3cm锛孊C=4cm锛孉C=5cm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涓夌偣鎵鍦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OA=523cm锛孧A=2.5cm锛屸埓OM=52cm锛屽嵆鐞冨績鍒板钩闈BC鐨勮窛绂讳负52cm锛庢晠绛旀涓猴細52锛
