# What Is The Formula For Area Of A Right Triangle Pascal’s Triangle and Square Numbers

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## Pascal’s Triangle and Square Numbers

Below are the first few lines of Pascal’s triangle:

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 21 15 6 1

The numbers in bold are the third diagonal when Pascal’s triangle is drawn in the middle. These are triangle numbers made from the sums of consecutive whole numbers (eg 15 = 1 + 2 + 3 + 4 + 5) and we can form square numbers from them. We only need to add the consecutive numbers from them and we get square numbers. To get the first square number, we need to add 0 to the list:

0, 1, 3, 6, 10, 15, 21…

0 + 1 = 1 = 1^2

1 + 3 = 4 = 2^2

3+ 6 = 9 = 3^2

6 + 10 = 16 = 4^2

10 + 15 = 25 = 5^2

15 +21 = 36 = 6^2

By the way, you can also get square numbers if you take the differences of the numbers on the 4th diagonal of Pascal’s triangle two places apart. The fourth diagonal is 1, 4, 10, 20 35… and the differences are 1-0 = 1, 4-0 = 4, 10-1 = 9, 20-4 = 16, 35-10 = 25, and so on.

You can use different methods to understand why the addition of consecutive triangle numbers produces square numbers. First, if you know that the formula for the nth triangle number is (n^2 + n)/2, then the previous triangle number is n less than that because it is the same sum of numbers but (n-1) and not n as the last one the number you add. If we add these two numbers together, we get

(n^2 + n)/2 + (n^2 + n)/2 – n

= (1/2)n^2 + n/2 + (1/2)n^2 + n/2 – n

= n^2 + n – n

= n^2

If you did not like this method, we can also show this result graphically. Triangle numbers get their name from the fact that you can create them by adding up the number of points that make up a triangle of different sizes, and square numbers from the number of points that make up squares of different sizes. So all you need to do is make a square out of the two triangles of points. If you try this with coins or counters or on paper and make right triangles, you should find that you can make a square out of two triangles, but one has to be one counter smaller on each side. Okay, so it wasn’t a rigorous method to prove it, but it was a lot easier than doing a lot of algebra, wasn’t it?

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