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 ![]() |
![阅读材料:已知,如图(1),在面积为S的△ABC中, BC=a,AC=](/uploads/image/z/16541192-56-2.jpg?t=%E9%98%85%E8%AF%BB%E6%9D%90%E6%96%99%EF%BC%9A%E5%B7%B2%E7%9F%A5%EF%BC%8C%E5%A6%82%E5%9B%BE%EF%BC%881%EF%BC%89%EF%BC%8C%E5%9C%A8%E9%9D%A2%E7%A7%AF%E4%B8%BAS%E7%9A%84%E2%96%B3ABC%E4%B8%AD%EF%BC%8C+BC%3Da%2CAC%3D%22b%2C%22+AB%3Dc%2C%E5%86%85%E5%88%87%E5%9C%86O%E7%9A%84%E5%8D%8A%E5%BE%84%E4%B8%BAr.%E8%BF%9E%E6%8E%A5OA%E3%80%81)
(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
![](http://img.wesiedu.com/upload/a/05/a0596265cbda53171d2ee5fc70fcc7d4.jpg)
![](http://img.wesiedu.com/upload/6/bf/6bfe68bce36967e250fd0bbfec466db0.jpg)
![](http://img.wesiedu.com/upload/7/91/791f618549f65b28691d9fd351ae21b9.jpg)
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
![](http://img.wesiedu.com/upload/e/74/e743a2f7548fafbeb7b97945dcf1d027.jpg)
(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
![](http://img.wesiedu.com/upload/2/2c/22cc5d42ee04cca23e74f51d5d932258.jpg)
![](http://img.wesiedu.com/upload/c/cb/ccb3e9d7869c4db058ad8f4b53588cda.jpg)
![](http://img.wesiedu.com/upload/6/9d/69da0e02095728c284d2aa017f9e32e5.jpg)
试题解析:(1)如图(2),连接OA、OB、OC、OD.···················································1分
∵
![](http://img.wesiedu.com/upload/f/cf/fcf4c818c9a6684869a771371e2762a3.jpg)
∴
![](http://img.wesiedu.com/upload/6/bf/6bfe68bce36967e250fd0bbfec466db0.jpg)
![](http://img.wesiedu.com/upload/3/b6/3b6a7f65202c5085385a7cfe65217f97.jpg)
(2)如图(3),过点D作DE⊥AB于点E,
则
![](http://img.wesiedu.com/upload/7/35/735f8b55bbc376bb855d36d23b60015b.jpg)
![](http://img.wesiedu.com/upload/5/90/5903aa11e8e3fce2f0ec1397345be8ad.jpg)
![](http://img.wesiedu.com/upload/d/21/d2188e43d534b96394b2408348759a58.jpg)
![](http://img.wesiedu.com/upload/b/02/b022a9184f8b8abf45ebf91df8aeb310.jpg)
∵AB∥DC,∴
![](http://img.wesiedu.com/upload/0/05/005ad9187f1c954fec7c6aecbf0e0215.jpg)
又∵
![](http://img.wesiedu.com/upload/6/21/62113205f8bb550ad1fbfe9156a7c783.jpg)
∴
![](http://img.wesiedu.com/upload/d/5c/d5c4a572e011db3b5d59cadec6f40391.jpg)
![](http://img.wesiedu.com/upload/7/91/791f618549f65b28691d9fd351ae21b9.jpg)
![](http://img.wesiedu.com/upload/a/05/a0596265cbda53171d2ee5fc70fcc7d4.jpg)
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