阅读材料:已知,如图(1),在面积为S的△ABC中, BC=a,AC="b," AB=c,内切圆O的半径为r.连接OA、
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阅读材料:已知,如图(1),在面积为S的△ABC中, BC=a,AC="b," AB=c,内切圆O的半径为r.连接OA、OB、OC�
阅读材料: 已知,如图(1),在面积为S的△ABC中, BC=a,AC="b," AB=c,内切圆O的半径为r.连接OA、OB、OC,△ABC被划分为三个小三角形. ∵ . ∴ . (1)类比推理:若面积为S的四边形ABCD存在内切圆(与各边都相切的圆),如图(2),各边长分别为AB=a,BC=b,CD=c,AD=d,求四边形的内切圆半径r; (2)理解应用:如图(3),在等腰梯形ABCD中,AB∥DC,AB=21,CD=11,AD=13,⊙O 1 与⊙O 2 分别为△ABD与△BCD的内切圆,设它们的半径分别为r 1 和r 2 ,求 的值. |
(1) (2) .
试题分析:(1)如图,连接OA、OB、OC、OD,则△AOB、△BOC、△COD和△DOA都是以点O为顶点、高都是r的三角形,根据 即可求得四边形的内切圆半径r.
(2)过点D作DE⊥AB于点E,分别求得AE的长,进而BE 的长,然后利用勾股定理求得BD的长;然后根据 , ,两式相除,即可得到 的值.
试题解析:(1)如图(2),连接OA、OB、OC、OD.···················································1分
∵ ·3分
∴ ························································································4分
(2)如图(3),过点D作DE⊥AB于点E,
则
·························································6分
∵AB∥DC,∴ .
又∵ ,
∴ .即 .···········································································9分
试题分析:(1)如图,连接OA、OB、OC、OD,则△AOB、△BOC、△COD和△DOA都是以点O为顶点、高都是r的三角形,根据 即可求得四边形的内切圆半径r.
(2)过点D作DE⊥AB于点E,分别求得AE的长,进而BE 的长,然后利用勾股定理求得BD的长;然后根据 , ,两式相除,即可得到 的值.
试题解析:(1)如图(2),连接OA、OB、OC、OD.···················································1分
∵ ·3分
∴ ························································································4分
(2)如图(3),过点D作DE⊥AB于点E,
则
·························································6分
∵AB∥DC,∴ .
又∵ ,
∴ .即 .···········································································9分
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