The number of diagonals in a polygon with n sides is given by n(n-3)/2 (see related question). So, 35 = n(n-3)/2 is what we have to solve. 70 = n(n-3) 70 = n2 - 3n n2 - 3n - 70 = 0 Factorizing the quadratic: (n-10)(n+7) = 0 This gives solutions n = 10 and n = -7. Clearly only the positive answer makes any sense (a shape cannot have minus seven sides!) so our shape has 10 sides: it is a decagon.
There are 35 diagonals in a 10 sided decagon
35
There are 560 diagonals by using the diagonal formula
It has 27.
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 0.5*(35^2 -3*35) = 560 diagonals
There are 35 diagonals in a 10 sided decagon
(35*(35-3))/2 (35*32)/2 1120/2 560 diagonals
It has 35 diagonals
35 diagonals
The number of diagonals is n(n-1)/2 - n substitute n=35 35(34)/2 - 35 = 560
octagon
35.
35
35
It has 10 sides because using the formula 0.5*(102-30) = 35 diagonals
There are 560 diagonals by using the diagonal formula