To solve this, we use the formula for the number of diagonals of a regular polygon, which is given by:

$D = \frac{n(n-3)}{2}$

Where:

• $D$ is the number of diagonals
• $n$ is the number of sides of the polygon

According to the problem, the number of diagonals subtracted from the number of sides results in zero:

$n - D = 0$

Substitute $D$ from the diagonal formula into the equation:

$n - \frac{n(n-3)}{2} = 0$

Now, solve for $n$. Start by multiplying the entire equation by 2 to eliminate the fraction:

$2n - n(n-3) = 0$

Simplify the expression:

$2n - n^2 + 3n = 0$

$5n - n^2 = 0$

Factor the equation:

$n(5 - n) = 0$

Thus, $n = 0$ or $n = 5$.

Since $n = 0$ is not a valid number of sides for a polygon, the only solution is $n = 5$.

Therefore, the polygon has 5 sides, meaning it is a regular pentagon.

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Would you like more details, or do you have any other questions?

Here are 5 related questions for further practice:

1. What is the number of diagonals of a polygon with 7 sides?
2. Derive the formula for the number of diagonals of a polygon.
3. How many sides does a polygon have if it has 35 diagonals?
4. What is the sum of the interior angles of a regular pentagon?
5. How many diagonals does a hexagon have?

Tip: A polygon with $n$ sides will always have fewer diagonals than sides, except in a triangle (where both are zero).