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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 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
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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Don't forget to like and share.... :)
Thank you and God bless!
Sunday, January 25, 2015
PYTHAGOREAN THEOREM EXPLORATION 2 (CUT-OUTS)
Pythagorean Theorem Exploration 2 Cutouts |
You may download this template for personal use and for your math class activity. The length of the sides on the second page measures 6.5 inches. This is also the measure of the length of the hypotenuse of the right triangle on the first page.
Your comments and suggestions are welcome here. Write them in the comment box below.
Don't forget to like and share.... :)
Thank you and God bless!
Don't forget to like and share.... :)
Thank you and God bless!
Friday, November 21, 2014
PYTHAGOREAN THEOREM (Exploration 1)
The Pythagoras' Theorem |
The theorem states that the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the remaining sides. That is c^2 = a^2 + b^2, where c is the hypotenuse, a and b are the lengths of the remaining sides.
Here is one of the theorem. Take note that the theorem only applies to right triangles.
Start with a cutout of any right triangle.
Let us name the shortest side as a, the longer side as b and the hypotenuse as c.
Let us make three (3) more of the right triangle.
Our next step is to form a square out of the four (4) triangles, making sure that the combined length of a and b will be the length of the side of the square that we are forming.
Let us take a look at the different parts of the square we formed. If we separate the whole square, we can get its area in terms of a and b.
The middle part is also a square with sides equal to the hypotenuse of the right triangle. The area can also be obtained.
The area of the four right triangles outlining the sides of the whole square is as follows
We take note that the area of the whole square is equal to the area of the small square in the middle added to the area of the four right triangles. That is
Thus, c^2 = a^2 + b^2.
Hope this will help you understand the derivation of the Pythagoras' Theorem.
Your comments and suggestions are welcome here. You may write them in the comment box below. Thank you!
Saturday, May 24, 2014
Sunday, March 23, 2014
EXTERIOR ANGLE THEOREM (Part 1: Exploration)
On the other hand, how many are you are familiar with the exterior angle theorem? What is the relationship of the exterior angles with the interior angles? The answer for these questions can be summarized by the "Exterior Angle Theorem".
In this post, let us discover where this theorem come from by using some illustrative examples.
Materials needed: ruler, protractor, calculator and pencil
Procedure:
1) On a piece of paper, draw one (1)
acute triangle, one (1) right triangle, and one (1) obtuse triangle. Name each
triangle as triangle ABC, with AC as the base.
2) Extend the base of AC to a point
D. Points A, C and D should be on the same line and C is between A and D.
- angle BCD is the exterior angle of each triangle
- angles A and B are the remote (non-adjacent) interior angles of each triangle
4) Look at the rightmost part of the table. Compare the sum of the measures of angles A and B and the exterior angle.
5) We can now make a conclusion/generalization about the relationship of the measurement of the exterior angle and the remote interior angles.
Here is the pdf copy of the activity, if to be conducted in a classroom setting. You are free to download it and use for your class. Hope it will be beneficial for you.
Your comments and suggestions are accepted here. Just write them in the comment box below. Thank you!
Thursday, March 20, 2014
THE SUM OF THE ANGLES OF A TRIANGLE (Part 2: Exploration)

This is the second part of the proof that the sum of the measures of the angles of a triangle is 180 degrees. This is actually the second method. The first one is more on using manipulatives or visual representations. This time, let us use basic mathematical concepts in proving.
This method is applicable to any type of triangle.
Let us use only one of the triangles. The process will be the same for the other triangles. To start with, let us name the triangle as ABC.
Now, let us draw a line parallel to base AC passing through B. Let us name this line as line BD or line BE or line ED, in any way you want it.
Since the focus of this proof is on the angles, let us rename each angle using numbers. It will be easier for us to determine the angles using the numbers instead of using three letters.
In this case,
Since we have drawn a line parallel to line AC, then we could say that side AB and side side BC are transversals of the parallel lines BD and line AC. Let us recall the concept of alternating interior angles for parallel lines.
Since we know that these alternate interior angles are always equal, then in the figure that we have formed, angle 1 = angle 4 and angle 3 = angle 5.
If you notice, angles 2, 4 and 5 form a straight line. Let us recall
That means the sum of angles 4, 2 and 5 is 180 degrees, because they form a straight line.
From the illustrations above, let us recall that angle 1 = angle 4 and angle 3 = angle 5.
Further, it means that the sum of angles 1, 2 and 3 is also 180 degrees.
Therefore, we can conclude that
You can also use the other sides of the triangle for the proof. The same process will be used in each of the cases.
Here is the summary of the proof in pdf form. You may download and print for academic use. Hope it will become useful to you.
You comments ad suggestions are welcome here. Write them down in the comment box below. Thank you!