In a polygon, no three diagonals are concurrent. If the total number of points of intersection of diagonals interior to the polygon is 70, then the number of diagonals of the polygon is a. `20` b. `28` c. `8` d. none of these

Options

A. 20
B. 28
C. 8
D. None of these

Answer

To solve the problem, we need to find the number of diagonals in a polygon given that the number of points of intersection of the diagonals inside the polygon is 70.

### Step-by-Step Solution:

1. **Understanding the Intersection of Diagonals**:
The number of intersection points formed by the diagonals inside a polygon is given by the formula:
\[ \text{Number of intersection points} = \binom{n}{4} \]
where \( n \) is the number of vertices (or sides) of the polygon. This is because each intersection point is formed by the diagonals connecting four vertices.

2. **Setting Up the Equation**:
We know from the problem that the number of intersection points is 70. Therefore, we can set up the equation:
\[ \binom{n}{4} = 70 \]

3. **Expanding the Binomial Coefficient**:
The binomial coefficient can be expanded as:
\[ \binom{n}{4} = \frac{n(n-1)(n-2)(n-3)}{4!} = \frac{n(n-1)(n-2)(n-3)}{24} \]
Setting this equal to 70 gives us:
\[ \frac{n(n-1)(n-2)(n-3)}{24} = 70 \]

4. **Multiplying Both Sides by 24**:
To eliminate the fraction, multiply both sides by 24:
\[ n(n-1)(n-2)(n-3) = 70 \times 24 \]
Calculating the right side:
\[ 70 \times 24 = 1680 \]
So we have:
\[ n(n-1)(n-2)(n-3) = 1680 \]

5. **Finding \( n \)**:
Now we need to find \( n \) such that the product \( n(n-1)(n-2)(n-3) = 1680 \). We can test values for \( n \):
- For \( n = 8 \):
\[ 8 \times 7 \times 6 \times 5 = 1680 \]
This is correct.

6. **Calculating the Number of Diagonals**:
The number of diagonals \( D \) in a polygon with \( n \) sides is given by:
\[ D = \frac{n(n-3)}{2} \]
Substituting \( n = 8 \):
\[ D = \frac{8(8-3)}{2} = \frac{8 \times 5}{2} = \frac{40}{2} = 20 \]

### Final Answer:
The number of diagonals in the polygon is **20**.