## Archive for January 2015

## PYTHAGOREAN THEOREM (Proof by Rearrangement: Part 1)

Here is another proof of the Pythagorean theorem.

Let us use a right triangle and name the shortest side as

Let us make three more of these so we have four congruent right triangles.

Now, let us arrange the four right triangles to form a square like this

In this figure, there are two squares formed. The first square is larger square, with side equal to

Let us focus on the inner square. The length of its side is equal to the hypotenuse of the four right triangles. It means that it has sides each measuring as

Let us take note of that the ares of the inner square is

Now, let us label the four right triangles as triangle 1, triangle 2, triangle 3 and triangle 4. This will make it easier for us to identify which triangle is moved later.

Let us rearrange the triangles. Let us move triangle 2 beside triangle 1, and triangle 4 beside triangle 3. In this case, each pair will form a rectangle.

Let us focus on the area being left by the two triangles and shade it with white.

If you notice, the white area can be divided into two like this

We formed two quadrilaterals but we are not yet sure if they are squares or not.

The smaller quadrilateral has a side that is equal to the shortest side of triangle 4. This means that this side measures

On the other hand, if we slide back triangle 2, we could see that the upper side of the small quadrilateral is also the shortest side of triangle 2.

This means that the upper side of the small quadrilateral measures

The area of the small square is

Now, let us look at the bigger quadrilateral shaded with white. One of its side (leftmost) has the same length as the longer side of triangle 2. It means that this side measures

On the other hand, if we slide back triangle 4, we could see that its longer side coincides with the lower side of the big white quadrilateral.

This means that the measure of the lower side of the big white quadrilateral is

The area of the big white square is

If we compare the two figures formed. Both of them has four (4) right triangles and the areas of these triangles are the same. It means that the area of the white inner square in the first figure is the same as the area of the two white squares in the second figure.

Therefore, for any right triangles

Let us use a right triangle and name the shortest side as

**a**, the longer side as**b**, and the hypotenuse as**c**.Let us make three more of these so we have four congruent right triangles.

Now, let us arrange the four right triangles to form a square like this

In this figure, there are two squares formed. The first square is larger square, with side equal to

**a+b**, while the second square is the inner square with side equal to**c**.Let us focus on the inner square. The length of its side is equal to the hypotenuse of the four right triangles. It means that it has sides each measuring as

**c.**Hence, the area of the square is**c^2.**Let us take note of that the ares of the inner square is

**c^2**.Now, let us label the four right triangles as triangle 1, triangle 2, triangle 3 and triangle 4. This will make it easier for us to identify which triangle is moved later.

Let us rearrange the triangles. Let us move triangle 2 beside triangle 1, and triangle 4 beside triangle 3. In this case, each pair will form a rectangle.

Let us focus on the area being left by the two triangles and shade it with white.

If you notice, the white area can be divided into two like this

We formed two quadrilaterals but we are not yet sure if they are squares or not.

The smaller quadrilateral has a side that is equal to the shortest side of triangle 4. This means that this side measures

**a**.On the other hand, if we slide back triangle 2, we could see that the upper side of the small quadrilateral is also the shortest side of triangle 2.

This means that the upper side of the small quadrilateral measures

**a**. Hence, the small quadrilateral is a**square**.The area of the small square is

Now, let us look at the bigger quadrilateral shaded with white. One of its side (leftmost) has the same length as the longer side of triangle 2. It means that this side measures

**b**.On the other hand, if we slide back triangle 4, we could see that its longer side coincides with the lower side of the big white quadrilateral.

This means that the measure of the lower side of the big white quadrilateral is

**b**. Hence, the big white quadrilateral is a**square**.The area of the big white square is

If we compare the two figures formed. Both of them has four (4) right triangles and the areas of these triangles are the same. It means that the area of the white inner square in the first figure is the same as the area of the two white squares in the second figure.

Therefore, for any right triangles

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Thank you and God bless!

Don't forget to like and share.... :)

Thank you and God bless!

Sunday, January 25, 2015