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Visual and Geometric Proof of Algebraic Identities

noreply@blogger.com (Mathrealm) — Mon, 28 Sep 2020 15:06:00 +0000

There are a lot of algebraic identities. One of the most commonly used identity is the identity

(a+b)^2 = a^2 + 2ab + b^2.

Did you ever wondered where did it came from?

There are various ways on how to prove this algebraic identity. One of which is using the visual and geometric way. It used basic concepts such as areas of squares and rectangles to derive the identity. In the following video, the prerequisite concepts are also included to support the derivation process. The visual proof or geometric representation is shown in detailed and in step-by-step manner. This is for you to easily understand the concept. You may watch the following video:

The same algebraic identity can also be proven using basic algebraic processes. Some of the prerequisite concepts included are multiplication law of indices and distributive property of multiplication. These prerequisites help in the derivation process of the algebraic identity (a+b)^2 = a^2 + 2ab + b^2. You may watch the complete details in this video:

Further, the algebraic identity can also be used as a guide in expanding the square of any binomials. An acronym S-2P-S is introduced in the following video for you to easily remember the process of expanding the square of binomials the fastest way. Here is the complete discussion of the acronym with various examples:

Hope you will learn from these videos about algebraic identities.

Your comments and suggestions are welcome here. Write them in the comment box below. Thank you and God bless!

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PYTHAGOREAN THEOREM (Proof by Rearrangement: Part 1)

noreply@blogger.com (Mathrealm) — Sun, 25 Jan 2015 05:00:00 +0000

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

Your comments and suggestions are welcome here. Write them in the comment box below.
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PYTHAGOREAN THEOREM EXPLORATION 2 (CUT-OUTS)

noreply@blogger.com (Mathrealm) — Fri, 21 Nov 2014 10:34:00 +0000

Pythagorean Theorem Exploration 2 Cutouts

Here is the second part of the previous post (Pythagorean Exploration 1 cut-outs). The first one uses the other two sides (a and b) of the right triangle to form the sides of the square pattern. This time, the hypotenuse (c) will be used for the sides of the square. You may use the discussion in Pythagorean Exploration 2 as a guide.

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!

PYTHAGOREAN THEOREM EXPLORATION 1 (CUT-OUTS)

noreply@blogger.com (Mathrealm) — Mon, 17 Nov 2014 14:53:00 +0000

Pythagoras Theorem Cutouts

In the previous posts, two different explorations were presented for the derivation of the Pythagoras Theorem. The first exploration can be found here and the second here.

The Pythagoras Theorem states that:
"The square of the length of the hypotenuse (denoted by c) is equal to the sum of the squares of the lengths of the other two sides (denoted by a and b)"

Pythagoras Theorem

This post is for the downloadable cutouts for each of the explorations. You may download them and use them for your class or for your kids to learn.

Here is the cutout for the first exploration of the Pythagoras Theorem. The file consist of two pages. The first page consists of four congruent right triangles. The second page is a square.

Cut along the broken lines on the right triangles and use them to fill in the square on the second page using the EXPLORATION 1 as a guide.

You comments and suggestions are welcome here. You may write them in the comment box below.

PYTHAGOREAN THEOREM (Exploration 2)

noreply@blogger.com (Mathrealm) — Sun, 17 Aug 2014 13:51:00 +0000

The Pythagorean Theorem

Here is another proof for the Pythagorean Theorem. You can see the first part here.

Start with a cutout of a right triangle.

Cut three more cutouts.

Let us name the sides of the right triangle with c as the hypotenuse.

Now, let us form a square using the four (4) triangles. This time, use the hypotenuse as the sides of the square.

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 c.

The middle part is also a square with sides equal to the difference of sides b and a. The area can be obtained as

On the other hand, the area of the four triangles outlining the sides of the whole square is as follows.

Combining their areas, we obtain

Thus, c^2 = a^2 + b^2.

Your questions, comments and suggestions are welcome here. Kindly write them in the comment box below.

PYTHAGOREAN THEOREM (Exploration 1)

noreply@blogger.com (Mathrealm) — Sat, 24 May 2014 05:29:00 +0000

The Pythagoras' Theorem

The Pythagorean theorem is also known as the Pythagoras' theorem. It is named after a famous Greek mathematician, Pythagoras, who is credited for its proof.

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!

EXTERIOR ANGLE THEOREM (Part 2: Proof)

noreply@blogger.com (Mathrealm) — Sun, 23 Mar 2014 12:01:00 +0000

Here is the two-column proof for the EXTERIOR ANGLE THEOREM. You may download and print it for your perusal.

Your comments and suggestions are welcome here. You may write them in the comment box below.

EXTERIOR ANGLE THEOREM (Part 1: Exploration)

noreply@blogger.com (Mathrealm) — Thu, 20 Mar 2014 15:42:00 +0000

Many are already familiar with the angles of a triangle. Many can easily point out which are the interior angles and even draw the exterior angles.

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

3) Measure each of the angles and complete the table below:

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!

THE SUM OF THE ANGLES OF A TRIANGLE (Part 2: Exploration)

noreply@blogger.com (Mathrealm) — Fri, 28 Jun 2013 06:55:00 +0000

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,

Let us take note that the angles now of the triangle are angle 1, angle 2 and angle 3.

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!

THE SUM OF THE ANGLES OF A TRIANGLE (Part 1: Exploration)

noreply@blogger.com (Mathrealm) — Mon, 24 Jun 2013 11:22:00 +0000

Everybody knows that the sum of the measures of the angles of any triangle is 180 degrees. If I may ask each one of you why, one of the reasons that I may probably hear is that "Our math teacher told us!" - which should not be the case. You should know how to show that the sum of the angles of any triangle is really 180 degrees. Where do 180 degrees come from? Why 180 degrees? Why not 360 degrees?

The purpose of this post is to show you how to prove that the sum of the angles of any triangle is really 180 degrees. This is the first method, which is the elementary way - the easiest way, to prove it. The other methods will also be posted here.

PROVE: The sum of the measures of the angles of any triangle is 180 degrees.

PRE-REQUISITE: The sum of the measures of the angles that form a straight line is 180 degrees.

MATERIALS NEEDED: colored papers or cardboards, pair of scissors, ruler, marker

PROCEDURES:

1. Cut three different types of triangles, classified according to angles. Use the colored papers or cardboards for the triangles. One should be a right triangle, another should be an acute triangle and the last should be an obtuse triangle.

2. For us to easily identify the angles of the triangles later, highlight the edges of each triangle using a black (or any dark colored) marker.

3. Using a pair of scissors or cutter, cut the sector/region of the angles of each triangle.

4. For each of the triangles, arrange the regions of the angles in such a way that they are adjacent to each other. Notice that in this case, the lower part of the angles form a straight line.

5. In each of the figures formed, the angles formed a straight line at the bottom part. Recall that the sum of the angles that form a straight line is 180 degrees.

CONCLUSION: The sum of the measures of the angles of any triangle is 180 degrees.

For educators and parents, I have made a worksheet for this activity for your class/children. You are free to download and print it. You may group your students and let each group work on a triangle or all the triangles. I hope this will be helpful.

Your comments and suggestions are welcome here. Write them in the comment box below. Thank you!

1 = 2?

noreply@blogger.com (Mathrealm) — Fri, 21 Jun 2013 07:58:00 +0000

Maybe you are wondering why or how can 1 be equal to 2. Let us look at the following proof:

First, let us have two real numbers a and b where

Using the addition property of equality, let us add b on both sides.

If we simplify both sides by combining like terms, it will become

Now let us subtract 2a on both sides

Combining like terms, we will arrive at

The right side has a common factor, which is 2. Using distributive property, it can be rewritten as

If you notice, both sides has a common factor, which is b - a. To simplify the equation, let us divide both sides by the common factor.

The process will arrive at

Are you convinced? No no no...

Seems like the proof is valid but there is something wrong with it. Look over the proof once again. Can you identify which of the process is not valid?

There is nothing wrong with the given. Real numbers can be equal. There is nothing wrong also with adding b and subtracting 2a on both sides. Likewise, there is nothing wrong with combining like terms on both sides. Then, where is the mistake?

There is nothing wrong with dividing both sides by any number but in this case it becomes invalid. The reason is that b - a = 0 since a = b. Subtracting equal numbers will yield 0. Since b - a = 0, then the result will be UNDEFINED. We cannot also cancel out b - a on both sides because of that.

Therefore, 1 is not equal to 2.

Here is a copy of the proof in pdf form. You may download and print it for educational purposes. You may share it to your friends. Your comments and suggestions are also welcome here.

THE GREEK ALPHABET

noreply@blogger.com (Mathrealm) — Sun, 14 Apr 2013 08:39:00 +0000

The letters of the Greek alphabet is commonly used to represent variables, constants, functions and more. They are used in different areas of mathematics and sciences. Here is a downloadable copy of the Greek alphabet.

DAMATH BOARDS

noreply@blogger.com (Mathrealm) — Thu, 11 Apr 2013 16:43:00 +0000

Here are the variations of the DAMATH board available for download. You may choose which one is applicable for your class/school.

If you have any suggestions regarding the form or appearance of the board, please let me know. Just write it in the comment box below. Thank you!

GRAPHING PAPER (SIZE A4) FOR DOWNLOAD

noreply@blogger.com (Mathrealm) — Thu, 04 Apr 2013 15:28:00 +0000

This is the second version of the downloadable graphing paper that can be used for math and science classes. The size of this grid paper is a4 (8.27in. x 11.69 in or 210mm x 297 mm). You may directly print it also.

You are also free to write your comments and suggestions about this below (using the comment box). Thank you!

GRAPHING PAPER (LETTER SIZE) FOR DOWNLOAD

noreply@blogger.com (Mathrealm) — Thu, 04 Apr 2013 15:09:00 +0000

This is one of the useful materials in Mathematics - the graphing paper or grid paper. The following graphing paper is designed for classroom use. It may be used not only for mathematics subjects but also for science subjects like Physics.

This is a graphing paper in letter size (8.5 in. x 11 in.). You can download this pdf file using the download arrow below. You can directly print it also.

I hope this may help you in any way. If there are any suggestions or comments about this graphing paper, please inform me through the comment box below. Thank you!

HELLO WORLD!

noreply@blogger.com (Mathrealm) — Thu, 28 Mar 2013 05:49:00 +0000

Welcome to a new site of Mathematics Realm. This is an extension of the realm where you can find discussions, tutorials, and downloads for the different branches of mathematics.

It will include basic to complex topics in mathematics. First few topics will focus on the basic concepts. This is to ensure clarity and understanding on the foundations.

Join me as we start our journey with mathematics. I hope eventually you will learn how to love mathematics.

Comments will be accepted from readers especially if there are corrections or suggestions. I would be glad to hear them from you.

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