试题分析:(1)连接AO、BO,过点A作AE⊥DC于点E,过点O作ON⊥DC于点N,ON交⊙O于点M,交AB于点F,则OF⊥AB.由OA =" OB" = 5m,AB = 8m,即可得到 ,∠AOB = 2∠AOF.在Rt△AOF中,根据∠AOF的正弦函数即可求得∠AOF 的度数,从而求得结果; (2)先根据勾股定理求的OF,即可得到FN,再根据等腰梯形的性质可得AE =" FN" = 3m,DC =" AB" + 2DE.解Rt△ADE即可得到DE = 2m,DC = 12m,根据 即可求得结果. (1)连接AO、BO,过点A作AE⊥DC于点E,过点O作ON⊥DC于点N,ON交⊙O于点M,交AB于点F,则OF⊥AB.
![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105023611-48747.png) ∵OA =" OB" = 5m,AB = 8m, ∴ ,∠AOB = 2∠AOF. 在Rt△AOF中,sin∠AOF = =" 0.8" = sin53°. ∴∠AOF = 53°,则∠AOB = 106°.即弧AB度数为106°; (2)∵ ,由题意得MN = 1m, ∴ . ∵四边形ABCD是等腰梯形,AE⊥DC,FN⊥AB, ∴AE =" FN" = 3m,DC =" AB" + 2DE. 在Rt△ADE中, , ∴DE = 2m,DC = 12m. ∴![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105023611-25422.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191105/20191105023613-16842.png) 答:U型槽的横截面积约为20m2. 点评:根据题意作出辅助线,构造出直角三角形及等腰梯形,再利用勾股定理进行求解是解此题的关键. |