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### Problem Summary
The number of diagonals \( D \) in a polygon is given by:
\[
D = Xn^2 + Yn
\]
where \( n \) is the number of sides.

We are asked to:
1. Use the number of diagonals in a triangle and a quadrilateral to show that the equations \( 3X + Y = 0 \) and \( 8X + 2Y = 1 \) hold.
2. Solve these simultaneous equations for \( X \) and \( Y \).
3. Use the values of \( X \) and \( Y \) to find the number of diagonals in an octagon.

### Solution

#### (a) Formulating the Equations

For a polygon with \( n \) sides:
- **Triangle** (\( n = 3 \)): A triangle has **0 diagonals**.
  \[
  D = X(3)^2 + Y(3) = 9X + 3Y
  \]
  Since \( D = 0 \), we get:
  \[
  9X + 3Y = 0
  \]
  Dividing by 3, we obtain:
  \[
  3X + Y = 0
  \]

- **Quadrilateral** (\( n = 4 \)): A quadrilateral has **2 diagonals**.
  \[
  D = X(4)^2 + Y(4) = 16X + 4Y
  \]
  Since \( D = 2 \), we get:
  \[
  16X + 4Y = 2
  \]
  Dividing by 2, we obtain:
  \[
  8X + 2Y = 1
  \]

These equations match the ones given in the problem:
\[
3X + Y = 0 \quad \text{and} \quad 8X + 2Y = 1
\]

#### (b) Solving for \( X \) and \( Y \)

We have the system:
1. \( 3X + Y = 0 \)
2. \( 8X + 2Y = 1 \)

Let's solve these equations simultaneously.

From the first equation:
\[
Y = -3X
\]

Substitute \( Y = -3X \) into the second equation:
\[
8X + 2(-3X) = 1
\]
\[
8X - 6X = 1
\]
\[
2X = 1
\]
\[
X = \frac{1}{2}
\]

Now, substitute \( X = \frac{1}{2} \) back into \( Y = -3X \):
\[
Y = -3\left(\frac{1}{2}\right) = -\frac{3}{2}
\]

So, the values are:
\[
X = \frac{1}{2}, \quad Y = -\frac{3}{2}
\]

#### (c) Finding the Number of Diagonals in an Octagon (\( n = 8 \))

For an octagon (\( n = 8 \)):
\[
D = Xn^2 + Yn
\]
Substitute \( X = \frac{1}{2} \), \( Y = -\frac{3}{2} \), and \( n = 8 \):
\[
D = \frac{1}{2}(8)^2 + \left(-\frac{3}{2}\right)(8)
\]
\[
D = \frac{1}{2}(64) - \frac{3}{2}(8)
\]
\[
D = 32 - 12 = 20
\]

So, the number of diagonals in an octagon is **20**.

### Final Answers
- **(a)** We verified the equations are correct.
- **(b)** \( X = \frac{1}{2} \) and \( Y = -\frac{3}{2} \).
- **(c)** The number of diagonals in an octagon is **20**.

Would you like further clarification on any step?

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### Follow-up Questions:
1. How can we derive the general formula for the number of diagonals in a polygon?
2. What would happen if we used different polygons to verify the equations?
3. How do these equations relate to the polygon diagonal formula \( D = \frac{n(n-3)}{2} \)?
4. How would the formula change if we considered 3D shapes instead of 2D polygons?
5. Can we generalize this method for polygons with an arbitrary number of diagonals?

#### Tip:
When solving simultaneous equations, always check if one equation can be simplified to express one variable in terms of the other. This simplifies substitution and reduces calculation steps.

The number of diagonals, D, that can be drawn in a polygon is given by the formula D = Xn^2 + Yn, where n is the number of sides. (a) By considering the number of diagonals in a triangle and a quadrilateral, show that 3X + Y = 0 and 8X + 2Y = 1. (b) Solve these simultaneous equations to find the values of X and Y. (c) Hence, find the number of diagonals in an octagon.

Learn how to calculate the number of diagonals in a polygon by using the formula D = Xn^2 + Yn. This problem involves setting up simultaneous equations to determine values of X and Y, and using these to find diagonals in an octagon. Suitable for students in Grades 10-12.

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