Let's reanalyze the problem step by step, directly referencing the image to clarify each part.

(a) Find the value of $x$

From the given table:

• The number of diagonals $d$ in a polygon is determined by the formula: $d = \frac{n(n-3)}{2}$

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

Thus, the value of $x$ is $9$.

(b) Find the values of $p$ and $q$

The number of diagonals is also expressed as: $d = pn^2 + qn$

Using the given table data, we form equations by substituting $n$ and $d$ into the formula.

From $n = 3$, $d = 0$:

0 = p(3^2) + q(3) \implies 0 = 9p + 3q \implies 3p + q = 0 \tag{1}

From $n = 4$, $d = 2$:

2 = p(4^2) + q(4) \implies 2 = 16p + 4q \tag{2}

Solve the system of equations $(1)$ and $(2)$:

1. From $(1)$: Solve for $q$ in terms of $p$: $q = -3p$

2. Substitute $q = -3p$ into $(2)$: $2 = 16p + 4(-3p) \implies 2 = 16p - 12p \implies 2 = 4p \implies p = \frac{1}{2}$

3. Substitute $p = \frac{1}{2}$ into $q = -3p$: $q = -3 \left(\frac{1}{2}\right) \implies q = -\frac{3}{2}$

Thus: $p = \frac{1}{2}, \quad q = -\frac{3}{2}$

(c) Can we construct a convex polygon such that $d = 170$?

The number of diagonals $d$ is given by: $d = \frac{n(n-3)}{2}$

Rewriting for $d = 170$: $\frac{n(n-3)}{2} = 170 \implies n(n-3) = 340$

This simplifies to the quadratic equation: $n^2 - 3n - 340 = 0$

Solve using the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}, \quad a = 1, \, b = -3, \, c = -340$ $n = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-340)}}{2(1)} \implies n = \frac{3 \pm \sqrt{9 + 1360}}{2} \implies n = \frac{3 \pm \sqrt{1369}}{2}$ $n = \frac{3 \pm 37}{2}$

Calculate the two roots: $n = \frac{3 + 37}{2} = 20 \quad \text{and} \quad n = \frac{3 - 37}{2} = -17$

Since $n$ must be a positive integer, $n = 20$.

Thus, a convex polygon with 20 sides has 170 diagonals, so the answer is yes.

Final Answers

(a) $x = 9$
(b) $p = \frac{1}{2}, \, q = -\frac{3}{2}$
(c) Yes, a convex polygon with 20 sides has 170 diagonals.

Would you like additional explanations or details on any of the steps? Here are 5 related questions to explore further:

1. What is the logic behind the formula $d = \frac{n(n-3)}{2}$ for the number of diagonals?
2. Can we derive $d = pn^2 + qn$ directly from the original formula for diagonals?
3. Why do we reject $n = -17$ in part (c)?
4. How can this formula for diagonals be used in real-world geometry problems?
5. How does solving quadratic equations help in interpreting polygon properties?

Tip: Always verify your results for quadratic equations by substituting the solution back into the original equation!