The mumber of sides of two regular polygons are as 5 4 and the diereth

1.	The mumber of sides of two regular polygons are as 5 4 and the dieretheir angles is 9. Find the number of sides in the polygonsb
Answer» Let the no. of sides of the 1st polygon be 5n and that of the 2nd one be 4n. Then the sum of internal angles of the 1st polygon is (5n-2)x180 degrees and that of the 2nd is (4n-2)x180. Since they're regular polygons all internal angles are same. So internal angles of the 1st polygon are ((5n-2)x180)/5n degrees and that of the 2nd is ((4n-2)x180)/4n degrees. Then from question we can say : (((5n-2)x180)/5n) - (((4n-2)x180)/4n)=9 180 {((5n-2)/5n)-((4n-2) /4n) }=9 180 { 4(5n-2) - 5(4n-2) / 20n } = 9 solving we get n=2. So no of sides of the 1st polygons = 5n = 5 * 2 = 10 no of sides of the 1st polygons = 4n = 4 * 2 = 8

The mumber of sides of two regular polygons are as 5 4 and the dieretheir angles is 9. Find the number of sides in the polygonsb

Answer»

Let the no. of sides of the 1st polygon be 5n and that of the 2nd one be 4n.

Then the sum of internal angles of the 1st polygon is (5n-2)x180 degrees and that of the 2nd is (4n-2)x180.

Since they're regular polygons all internal angles are same.

So internal angles of the 1st polygon are ((5n-2)x180)/5n degrees and that of the 2nd is ((4n-2)x180)/4n degrees.

Then from question we can say

: (((5n-2)x180)/5n) - (((4n-2)x180)/4n)=9

180 {((5n-2)/5n)-((4n-2) /4n) }=9

180 { 4(5n-2) - 5(4n-2) / 20n } = 9

solving we get n=2.

So no of sides of the 1st polygons = 5n = 5 * 2 = 10

no of sides of the 1st polygons = 4n = 4 * 2 = 8