Which of the following cannot be number of diagonals of a polygon

Options

A. A)14
B. B)20
C. C)28
D. D)35

Answer

To determine which of the given options cannot be the number of diagonals of a polygon, we can use the formula for the number of diagonals \( D \) in a polygon with \( n \) sides:

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

Where \( n \) is the number of sides of the polygon. We need to check each option to see if it results in an integer value for \( n \).

### Step 1: Check Option A (14 diagonals)
Set \( D = 14 \):

\[ 14 = \frac{n(n-3)}{2} \]

Multiply both sides by 2:

\[ 28 = n(n-3) \]

Rearranging gives:

\[ n^2 - 3n - 28 = 0 \]

Now we can solve this quadratic equation using the quadratic formula:

\[ n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 1, b = -3, c = -28 \):

\[ n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-28)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ n = \frac{3 \pm \sqrt{9 + 112}}{2} = \frac{3 \pm \sqrt{121}}{2} = \frac{3 \pm 11}{2} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{14}{2} = 7 \)
2. \( n = \frac{-8}{2} = -4 \) (not valid)

Thus, \( n = 7 \) is valid, meaning a polygon with 7 sides can have 14 diagonals.

### Step 2: Check Option B (20 diagonals)
Set \( D = 20 \):

\[ 20 = \frac{n(n-3)}{2} \]

Multiply both sides by 2:

\[ 40 = n(n-3) \]

Rearranging gives:

\[ n^2 - 3n - 40 = 0 \]

Using the quadratic formula:

\[ n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-40)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ n = \frac{3 \pm \sqrt{9 + 160}}{2} = \frac{3 \pm \sqrt{169}}{2} = \frac{3 \pm 13}{2} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{16}{2} = 8 \)
2. \( n = \frac{-10}{2} = -5 \) (not valid)

Thus, \( n = 8 \) is valid, meaning a polygon with 8 sides can have 20 diagonals.

### Step 3: Check Option C (28 diagonals)
Set \( D = 28 \):

\[ 28 = \frac{n(n-3)}{2} \]

Multiply both sides by 2:

\[ 56 = n(n-3) \]

Rearranging gives:

\[ n^2 - 3n - 56 = 0 \]

Using the quadratic formula:

\[ n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-56)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ n = \frac{3 \pm \sqrt{9 + 224}}{2} = \frac{3 \pm \sqrt{233}}{2} \]

Since \( \sqrt{233} \) is not a perfect square, \( n \) will not be an integer. Thus, there is no polygon with 28 diagonals.

### Step 4: Check Option D (35 diagonals)
Set \( D = 35 \):

\[ 35 = \frac{n(n-3)}{2} \]

Multiply both sides by 2:

\[ 70 = n(n-3) \]

Rearranging gives:

\[ n^2 - 3n - 70 = 0 \]

Using the quadratic formula:

\[ n = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-70)}}{2 \cdot 1} \]

Calculating the discriminant:

\[ n = \frac{3 \pm \sqrt{9 + 280}}{2} = \frac{3 \pm \sqrt{289}}{2} = \frac{3 \pm 17}{2} \]

Calculating the two possible values for \( n \):

1. \( n = \frac{20}{2} = 10 \)
2. \( n = \frac{-14}{2} = -7 \) (not valid)

Thus, \( n = 10 \) is valid, meaning a polygon with 10 sides can have 35 diagonals.

### Conclusion
The only option that cannot represent the number of diagonals of any polygon is **Option C: 28**.