SOLUTION: what polygon has a number of sides that is five times the number of diagonals
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Question 652123: what polygon has a number of sides that is five times the number of diagonals
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
The way I understand geometry, the answer is .
I am betting there is a mix-up in the wording of the problem.
The way I understand geometry,
a diagonal of a polygon is a segment that connects two vertices of the polygon, but is not a side of the polygon.
If a polygon has sides, it will have n vertices.
Each vertex will be connected to other vertices by sides,
and can be connected to the remaining other vertices by a diagonal.
If we multiply the diagonals coming out of each vertex times , the number of vertices,
we get ,
but we are counting each diagonal twice.
The polygon with sides has diagonals.
To find what polygon has a number of sides that is five times the number of diagonals, we can set up an equation
or we can make a table comparing and .
For that table:
A triangle has sides and diagonals.
A quadrilateral has sides and diagonals.
From then on, the number of sides is equal to or less than the number of diagonals.
A pentagon has sides and diagonals.
A hexagon has sides and diagonals.
Either way (equation or table), the answer is .
If we wanted to know what polygon has a number of diagonals that is five times the number of sides, our equation would be
--> --> --> --> -->
Since does not make a polygon, we know is not .
--> --> or .
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