It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
54 Diagonals. * * * * * A polygon with n sides has 1/2*n*(n-3) diagonals. So a decagon would have 1/2*10*7 = 35 diagonals. How the Community answer got 54 is anybody's guess!
A diagonal line of a polygon is a line that joins any two vertices not already joined by a side. A polygon with n sides has n(n-3)/2 diagonals → a decagon has 10(10-3)/2 = 10 × 7 ÷ 2 = 35 diagonals
A decagon has 10 sides and 35 diagonals
An octagon has 8 sides and 20 diagonals so the question doesn't make sense
It will have ten sides
(35*(35-3))/2 (35*32)/2 1120/2 560 diagonals
The formula to calculate the number of diagonals in a polygon is n(n-3)/2, where n represents the number of vertices. Setting this formula equal to 560 and solving for n, we get n(n-3)/2 = 560. By solving this quadratic equation, we find that the polygon has 20 vertices.
It depends if the polygon is convex or concave but if it is a regular polygon it would have 560
In a polygon with n sides, the number of diagonals that can be drawn from one vertex is given by the formula (n-3). Therefore, in a 35-sided polygon, you can draw (35-3) = 32 diagonals from one vertex.
It has 35 diagonals
There are 560 diagonals by using the diagonal formula
54 Diagonals. * * * * * A polygon with n sides has 1/2*n*(n-3) diagonals. So a decagon would have 1/2*10*7 = 35 diagonals. How the Community answer got 54 is anybody's guess!
1oo diagonals, because 1 side has 10 diagonals so 10 sides (10x10) = 100 diagonals! This is the correct answer..... a decagon has 35 diagonals. The equation for the this is D=n(n-3)/2 meaning diagonals = number of sides (N) multiplied by number of sides minus 3. and that number over 2. D=10(10-3)/2
The number of diagonals is n(n-1)/2 - n substitute n=35 35(34)/2 - 35 = 560
100 diagonals * * * * * No, it is 0.5*10*(10-3) = 35
35 diagonals