Question 1164701
the number of diagonals in a polygon is equal to n * (n-3) / 2.


if the number of sides of the larger polygon is 4 more than the number of sides of the smaller polygon, then:


the number of sides of the larger polygon is equal to n + 4, while the number of sides of the smaller polygon is equal to n.


the number of diagonals of the smaller polygon is equal to n * (n-1) / 2.


the number of diagonals of the larger polygon is equal to (n + 4) * (n + 4 - 3) / 2 which is equal to (n + 4) * (n + 1) / 2.


if 30 is the result when the number of diagonals of the smaller polygon is subtracted from the number of diagonals of the larger polygon, then your formula becomes:


(n + 4) * (n + 1) / 2 - n * (n - 3) / 2 = 30.


multiply both sides of this equation by 2 to get:


(n + 4) * (n + 1) - n * (n - 3) = 60.


simplify this to get n^2 + n + 4n + 4 - (n^2 - 3n) = 60


simplify this further to get n^2 + n + 4n + 4 - n^2 + 3n = 60


combine like terms to get:


8n + 4 = 60.


subtract 4 from both sides of the equation and simplify to get:


8n = 56


solve for n to get n = 56 / 8 = 7


that's the number of sides of the smaller polygon.


the number of the sides of the larger polygon is 7 + 4 = 11


to confirm, do the following:


when n = 11, the number of diagonals is equal to 11 * (11 - 3) / 2 = 11 * 8 / 2 = 11 * 4 = 44


when n = 7, the number of diagonals is equal to 7 * (7 - 3) / 2 = 7 * 4 / 2 = 7 * 2 = 14


the difference in number of diagonals is 44 minus 14 = 30


this confirms the solution is correct.


your solution is that the number of sides of the larger polygon is 11.