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Math problems in the first semester
1. The number of diagonals in a polygon with 2n sides is 6 times that of a polygon with n sides. Find the number of sides of these two polygons.

C(2n,2)-2n=6[C(n,2)-n]

2n(2n- 1)/2-2n = 6n(n- 1)/2-6n

n=6

So it is a hexagon and a dodecagon.

2 The sum of the degrees of the inner and outer angles of a polygon is 1350, and the number of sides of the polygon is found.

(n-2)* 180+x= 1350

1350- 180 & lt; (n-2)* 180 & lt; 1350

8.5 & ltn & lt9.5

n=9

3. A pair of internal angles of quadrilateral are complementary, and the degree ratio of three adjacent internal angles is 2∶3∶7. What are the four inner corners?

Let the internal angles be 2x 3x 7x y respectively.

2x+3x+7x+y=360

12x+y=360

2x+7x= 180 (two pairs of internal angles are complementary)

x=20 y= 120

So the four internal angles are: 40 60140120.