Question Number 12

Today, we will explore how to count the number of diagonals in an n-sided polygon. Can anyone tell me how many sides a triangle has?

Three sides!

Correct! Now, how many diagonals does a triangle have?

Zero, because you can't draw a diagonal in a triangle.

Exactly! So, our formula for the number of diagonals is $\frac{n(n-3)}{2}$. Let's break this down. Who can explain what n represents?

N represents the number of sides in the polygon, right?

Great job! Now let's calculate the number of diagonals in a polygon with four sides, or a square.

Using the formula, it would be $\frac{4(4-3)}{2} = 2$ diagonals.

That's right! We see that a square has two diagonals. This gives us a foundational understanding of our topic.

Next, let's discuss how to prove this formula using mathematical induction. What is our base case when n equals three?

The base case shows that for a triangle, we have no diagonals.

That’s correct. Now let's assume this holds for a polygon with k sides. What do we need to show for k + 1?

We need to show that it holds true that $D(k + 1) = \frac{(k + 1)((k + 1)-3)}{2}$.

Excellent! Now, how would we approach counting the new diagonals when transitioning from k to k + 1?

We consider the diagonals from the added vertex to the existing vertices minus the two adjacent sides.

Exactly, and how many new diagonals does it add?

k - 2 new diagonals!

Brilliant! You all grasped this very well. Can someone summarize the inductive reasoning?

If it holds for k, we proved it holds for k + 1. Thus, it’s true for all n greater than or equal to three!