论文部分内容阅读
有一类数学题,在一定条件下可以先把问题推想到某种特殊情形,作为解题的突破口,这时问题往往能迎刃而解.例1如图1,在边长为a的正方形中,对角线AC与BD相交于O,P、Q分别为AB、BC上的点,且PO⊥QO,求四边形BPOQ的面积.分析因为P、Q分别为AB和BC边上的动点,且PO⊥QO,当P点与B点重合时,则Q点与C点重合于是四边形BPOQ就成了△BOC这种特殊情形,于是可猜想到四边形BPOQ的面积等于三角形BOC的面积,那么只要证明△BOP≌△COQ,即四边形BPOQ的面积等于1/4a~2
There is a class of mathematics problems, under certain conditions, the problem can be first pushed to a particular situation, as a breakthrough point for solving the problem, then the problem can often be solved. Example 1 In Figure 1, in the side of the square a, diagonal Lines AC and BD intersect at O, P, and Q are points on AB and BC, respectively, and PO QO, for the area of quadrilateral BPOQ. The analysis is performed because P and Q are the moving points on the sides of AB and BC, respectively. QO, when P coincides with point B, then the Q point coincides with point C. The quadrilateral BPOQ becomes a special case of △BOC. Therefore, we can guess that the area of the quadrilateral BPOQ is equal to the area of the triangle BOC. So just prove that BOP. ≌△COQ, the area of the quadrilateral BPOQ is equal to 1/4a~2