Q: How many sides would a polygon have if it contained 170 diagonals?

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A 4-sided polygon has 2 diagonals. A 5-sided polygon, 5. 6-sided, 9. By induction, an n-sided polygon has 'n*(n-3)/2' diagonals. "n" is how many sides a shape has. so, (7)*(7-3)/2 now we have, 7*4/2 so, 28/2 = 14 Therefore, a 7-sided polygon would have 14 diagonals.

A polygon with 27 sides would be called an icosakaiheptagon.

No, a cylinder is not a polygon. It is not a polygon because it has curved sides. In order to be considered a polygon, it would need to have flat sides. Also, a polygon is a 2 dimensional shape but a cylinder is a three dimensional shape.

A polygon with ten sides would be called a decagon.

If the only information that you have is the angle and the shape is a polygon, then you cannot determine how many sides the polygon consists of. The minimum number of sides is 3, but without seeing a picture, or knowing what type of polygon (e.g. right triangle or equalaterial triangle), then you won't be able to determine how many sides the polygon has.

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A polygon with n sides had n*(n-3)/2 diagonals. So a 20 sided polygon would have 20*17/3 = 170

If a polygon had 166 sides it would have 83 diagonals as each diagonal must join two vertices.

For a polygon with n sides, there would be n*(n-3)/2

It depends if the polygon is convex or concave but if it is a regular polygon it would have 560

5 diagonals * * * * * That is not correct since two of these would be lines joining the vertex to adjacent vertices (one on either side). These are sides of the polygon, not diagonals. The number of diagonals from any vertex of a polygon with n sides is n-3.

A regular octadecagon (an eighteen sided polygon) would have nine diagonals.

A 4-sided polygon has 2 diagonals. A 5-sided polygon, 5. 6-sided, 9. By induction, an n-sided polygon has 'n*(n-3)/2' diagonals. "n" is how many sides a shape has. so, (7)*(7-3)/2 now we have, 7*4/2 so, 28/2 = 14 Therefore, a 7-sided polygon would have 14 diagonals.

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.

number of diagonals in an n-sided polygon is n(n-1)/2 - n If the polygon has 12 sides, number of diagonals is 12(12-1)/2 -12 = (12)(11)/2 -12 = 132/2 - 12 = 66-12 = 54

47 sides. Take a vertex of an n-sided polygon. There are n-1 other vertices. It is already joined to its 2 neighbours, leaving n-3 other vertices not connected to it. Thus n-3 diagonals can be drawn in from each vertex. For n=50, n-3 = 50-3 = 47 diagonals can be drawn from each vertex. The total number of diagonals in an n-sided polygon would imply n-3 diagonals from each of the n vertices giving n(n-3). However, the diagonal from vertex A to C would be counted twice, once for vertex A and again for vertex C, thus there are half this number of diagonals, namely: number of diagonals in an n-sided polygon = n(n-3)/2.

Any regular polygon with 4n sides, where n is a positive integer.