Dear Secondary Math students, now that you have a better visualization of the concept of Pythagoras’ Theorem from ** Math Lobby**’s earlier Secondary Math article on

**, you might be curious whether how do we know that it holds true to the definition?**

__Pythagoras Theorem__

__Visual proving of the Pythagoras’ Theorem__

__Visual proving of the Pythagoras’ Theorem__

Hence, we will now take a look at how we can derive this relationship between the sides of the right-angled triangle with the hypotenuse of the triangle. Let’s say we are given a right-angled triangle of sides **a** & **b**, with a hypotenuse of **c**.

We will now have 4 identical right-angled triangles, pieced together to make the hypotenuse of the triangles form a smaller square out of the bigger square (See Fig. 1).

After some rearranging of the identical triangles (See Fig. 2), we are able to see that despite the number of identical triangles remained the same, the area of the smaller square (**c²**) in Fig. 1 is actually equivalent to the area of the two squares combined (**a² + b²**) (See Fig. 3).

Therefore, we can conclude that **a² + b² = c²** to be true.

__Algebraic proving of the Pythagoras’ Theorem__

__Algebraic proving of the Pythagoras’ Theorem__

Given the same right-angled triangle with sides **a** & **b**, with a hypotenuse of **c**, we can try to prove this algebraically as well. With reference back to Fig. 1, since we want to prove that **a2 + b2 = c2** is true, let’s equate the sum of the areas of the identical triangles **plus** the area of the smaller square, with the area of the big square. Let’s see how it’s done:

Formula for area of triangle: ½ x base x height, and since we have 4 identical triangles:

Sum of the areas of the identical triangles + Area of the smaller square = Area of the big square

4 (½ x **a** x **b**) + **c²** = (**a** + **b**)²

2**ab **+ **c²** = **a²** + 2**ab** +** b²**

After rearranging and cancelling out terms, we will end up with: **a²+ b² = c²**

And therefore, this is how you prove Pythagoras’ Theorem algebraically!

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