我们知道,从n边形的一个顶点出发,除了自己这个顶点和与自己相邻的两个顶点不能连成对角线这三条线段之外,从n边形一个顶点出发可以引(n-3)条对角线,因为n边形共有n个顶点,所以对角线总数为n(n-3),但其有一半是重复的,所以就再除以n(n-3)2得:,所以n边形的边数与对角线总数的2(和为n n-3)n(n-1)n+=.九年级“实际问题与一元二次方22 We know that starting from a vertex of an n-polygons, we can derive (n-3) from one vertex of the n-polygons except for the three vertices that can not be connected to each other, ) Diagonal because the n-sided polygons have n vertices, the total number of diagonals is n (n-3), but half of them are duplicates, so divide by n (n-3) 2: , So the number of sides of the n-polygons and the diagonal total number of 2 (and is n n-3) n (n-1) n + = 9th grade ”actual problem and the quadratic 22