Let's solve this step by step.

(a) Find the value of $x$.

To find the value of $x$, note that the formula for the number of diagonals in a convex polygon with $n$ sides is given by:

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

This formula calculates all the diagonals by connecting each vertex to every other vertex except itself and its adjacent vertices.

For $n = 6$: $d = \frac{6(6-3)}{2} = \frac{6 \cdot 3}{2} = 9$

So, $x = 9$.

(b) The number of diagonals is given by $d = pn^2 + qn$. Find $p$ and $q$.

We are given some values from the table:

• When $n = 3$, $d = 0$
• When $n = 4$, $d = 2$
• When $n = 5$, $d = 5$

Substitute these into $d = pn^2 + qn$ to create a system of equations:

1. For $n = 3, d = 0$: 0 = 9p + 3q \quad \Rightarrow \quad 3p + q = 0 \tag{1}

2. For $n = 4, d = 2$: 2 = 16p + 4q \quad \Rightarrow \quad 4p + q = 2 \tag{2}

3. For $n = 5, d = 5$: 5 = 25p + 5q \quad \Rightarrow \quad 5p + q = 5 \tag{3}

From equations (1) and (2), subtract (1) from (2): $(4p + q) - (3p + q) = 2 - 0$ $p = 2$

Substitute $p = 2$ into equation (1): $3(2) + q = 0 \quad \Rightarrow \quad q = -6$

So, $p = 2$ and $q = -6$.

The formula becomes: $d = 2n^2 - 6n$

(c) Can we construct a convex polygon such that the number of diagonals is 170?

We need to solve $d = 2n^2 - 6n$ for $d = 170$: $170 = 2n^2 - 6n$ $2n^2 - 6n - 170 = 0$ Divide through by 2: $n^2 - 3n - 85 = 0$

Solve this quadratic equation using the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here, $a = 1$, $b = -3$, $c = -85$. Substituting: $n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-85)}}{2(1)}$ $n = \frac{3 \pm \sqrt{9 + 340}}{2}$ $n = \frac{3 \pm \sqrt{349}}{2}$

Since $\sqrt{349} \approx 18.68$, we get: $n = \frac{3 + 18.68}{2} \quad \text{or} \quad n = \frac{3 - 18.68}{2}$ $n \approx 10.84 \quad \text{or} \quad n \approx -7.84$

Since $n$ must be a positive integer, we round $10.84$ to $11$.

Verification: For $n = 11$: $d = 2(11)^2 - 6(11)$ $d = 2(121) - 66 = 242 - 66 = 170$

Thus, a convex polygon with 11 sides can have 170 diagonals.

Final Answers:

(a) $x = 9$
(b) $p = 2$, $q = -6$
(c) Yes, a convex polygon with 11 sides has 170 diagonals.

Would you like further clarifications? Here are five related questions to expand on this:

1. What is the general formula for calculating diagonals in a polygon?
2. How do we derive the quadratic formula for solving problems like this?
3. Why is the formula for diagonals quadratic instead of linear?
4. Can this formula be used for polygons with fewer than 3 sides? Why or why not?
5. What other geometric properties can be modeled with quadratic equations?

Tip: When working with quadratics, always verify the reasonableness of your solution (e.g., whether it makes sense in the context of the problem).