It has 10 sides because using the formula 0.5*(102-30) = 35 diagonals
10 ... any polygon it is 2 less than the number of sides or vertices wince they are the same.
A polygon has as many angles as sides, so a polygon with 10 sides (decagon) would have 10 angles.
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
65. The number of diagonals in a polygon with n sides is given by: diagonals = n(n - 3) / 2 So for a triskaidecagon which has 13 sides n = 13 and: number_of_diagonals = 13(13 - 3) /2 = 13 x 10 / 2 = 65
38 diagonals
It has 10 sides because using the formula 0.5*(102-30) = 35 diagonals
A polygon with n sides has n*(n - 3)/2 diagonals. So n = 23 gives 23*20/10 = 230 diagonals
10 ... any polygon it is 2 less than the number of sides or vertices wince they are the same.
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
100 diagonals * * * * * No, it is 0.5*10*(10-3) = 35
Under normal scenarios, a polygon cannot have exactly 10 diagonals
A polygon has as many angles as sides, so a polygon with 10 sides (decagon) would have 10 angles.
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
65. The number of diagonals in a polygon with n sides is given by: diagonals = n(n - 3) / 2 So for a triskaidecagon which has 13 sides n = 13 and: number_of_diagonals = 13(13 - 3) /2 = 13 x 10 / 2 = 65
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!
No not normally. The formula of the number of diagonals is (n/2)(n - 3). (n/2)(n - 3) = 10 n(n - 3) = 20 n2 - 3n - 20 = 0 we cannot factor it as (n - r1)(n - r2), where r1 and r2 are natural numbers, so there is any polygon with 10 diagonals.