Finding the Number of Sides of a Polygon Given Its Diagonals

The question asks us to determine the number of sides of a polygon that has exactly 44 diagonals. To solve this, we need to use the formula relating the number of sides of a polygon to the number of its diagonals.

Formula for Polygon Diagonals

The number of diagonals, \(D\), in a polygon with \(n\) sides is given by the formula:

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

This formula comes from the fact that from each vertex of an \(n\)-sided polygon, we can draw diagonals to \(n-3\) other vertices (we cannot draw a diagonal to the vertex itself or to its two adjacent vertices). Since each diagonal connects two vertices, we divide the total count \(n(n-3)\) by 2 to avoid counting each diagonal twice.

Setting up the Equation

We are given that the polygon has 44 diagonals. Using the formula, we can set up the equation:

\( 44 = \frac{n(n-3)}{2} \)

Solving for the Number of Sides (n)

Now, we need to solve this equation for \(n\), the number of sides.

Multiply both sides by 2:

\( 44 \times 2 = n(n-3) \)

\( 88 = n^2 - 3n \)

Rearrange the equation to form a quadratic equation:

\( n^2 - 3n - 88 = 0 \)

We can solve this quadratic equation by factoring, completing the square, or using the quadratic formula. Factoring is often the quickest method if possible. We look for two numbers that multiply to -88 and add up to -3. These numbers are 8 and -11.

So, we can factor the quadratic equation as:

\( (n+8)(n-11) = 0 \)

This equation gives two possible solutions for \(n\):

• \( n+8 = 0 \implies n = -8 \)
• \( n-11 = 0 \implies n = 11 \)

The number of sides of a polygon must be a positive integer, and it must be at least 3 (as a polygon requires a minimum of 3 sides). Therefore, the solution \(n = -8\) is not valid in the context of polygon sides.

The valid solution is \(n = 11\).

Verification

Let's check if a polygon with 11 sides has 44 diagonals using the formula \( D = \frac{n(n-3)}{2} \):

\( D = \frac{11(11-3)}{2} = \frac{11 \times 8}{2} = \frac{88}{2} = 44 \)

This matches the given number of diagonals.

Therefore, the number of sides of the polygon is 11.

Revision Table: Polygon Diagonals

Additional Information: Deriving the Diagonal Formula

Let's understand how the formula \( D = \frac{n(n-3)}{2} \) is derived.

• Consider a polygon with \(n\) vertices.
• From each vertex, you can draw a line segment to every other vertex. There are \(n-1\) other vertices from any given vertex.
• So, from one vertex, you can draw \(n-1\) lines to other vertices.
• If we do this for all \(n\) vertices, we get a total of \(n \times (n-1)\) lines.
• These lines include both the sides of the polygon and the diagonals.
• An \(n\)-sided polygon has exactly \(n\) sides.
• The number of diagonals is the total number of lines minus the number of sides: \( n(n-1) - n \).
• \( n(n-1) - n = n^2 - n - n = n^2 - 2n = n(n-2) \).
• However, this count \(n(n-2)\) counts each diagonal twice (once from each endpoint). For example, the diagonal from vertex A to vertex B is counted when starting from A and again when starting from B.
• Therefore, we must divide by 2 to get the unique number of diagonals.
• The number of unique diagonals is \( \frac{n(n-2)}{2} \). Wait, something is wrong here. The correct formula is \( \frac{n(n-3)}{2} \). Let's re-evaluate the step "from each vertex... draw diagonals to \(n-3\) other vertices".

Correct Derivation:

• From any single vertex in an \(n\)-sided polygon, we can draw line segments to \(n-1\) other vertices.
• Of these \(n-1\) vertices, two are adjacent (connected by a side) and one is the vertex itself.
• We cannot draw a diagonal to the vertex itself or to the two adjacent vertices because these form the sides of the polygon, not diagonals.
• So, from each vertex, the number of diagonals we can draw is \(n-1 - 2 = n-3\).
• There are \(n\) vertices, so the total number of possible diagonal segments starting from all vertices is \(n \times (n-3)\).
• Since each diagonal connects two vertices, the line segment from vertex A to vertex B is the same diagonal as the line segment from vertex B to vertex A. Our count \(n(n-3)\) counts each diagonal twice.
• To get the actual number of unique diagonals, we divide by 2.
• Thus, the number of diagonals \( D = \frac{n(n-3)}{2} \). This confirms the formula used in the solution.