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		<title>Contextual Math Problems Based on Field Trip to Lombokstraat , Utrecht, Netherlands</title>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Tue, 28 Jan 2014 18:23:18 +0000</pubDate>
				<category><![CDATA[Geometry]]></category>
		<category><![CDATA[Angle]]></category>
		<category><![CDATA[Contextual]]></category>
		<category><![CDATA[Estimation]]></category>
		<category><![CDATA[Math Problems]]></category>
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					<description><![CDATA[Decorative Lamp In the middle of Lombokstraat, Anton sees a decorative lamp, as shown in the pictures below. He is curious about the length of the wire used to form the lamps. If the distance between each lamp is 10 &#8230; <a href="https://lauradorindira.wordpress.com/2014/01/28/contextual-math-problems-based-on-field-trip-to-lombokstraat-utrecht-netherlands/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p><strong>Decorative Lamp</strong></p>
<p>In the middle of Lombokstraat, Anton sees a decorative lamp, as shown in the pictures below.</p>
<div data-shortcode="caption" id="attachment_381" style="width: 310px" class="wp-caption aligncenter"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg"><img aria-describedby="caption-attachment-381" data-attachment-id="381" data-permalink="https://lauradorindira.wordpress.com/2014/01/28/contextual-math-problems-based-on-field-trip-to-lombokstraat-utrecht-netherlands/1-a/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg" data-orig-size="647,205" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1390957717&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="1.a" data-image-description="" data-image-caption="&lt;p&gt;Front view&lt;/p&gt;
" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg?w=611" class="size-medium wp-image-381 " alt="Front view" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg?w=300&#038;h=95" width="300" height="95" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-a.jpg?w=150 150w" sizes="(max-width: 300px) 100vw, 300px" /></a><p id="caption-attachment-381" class="wp-caption-text">Front view</p></div>
<div data-shortcode="caption" id="attachment_382" style="width: 310px" class="wp-caption aligncenter"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg"><img aria-describedby="caption-attachment-382" data-attachment-id="382" data-permalink="https://lauradorindira.wordpress.com/2014/01/28/contextual-math-problems-based-on-field-trip-to-lombokstraat-utrecht-netherlands/1-b/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg" data-orig-size="670,208" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1390957804&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="1.b" data-image-description="" data-image-caption="&lt;p&gt;Front Side View&lt;/p&gt;
" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg?w=611" class="size-medium wp-image-382 " alt="Front Side View" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg?w=300&#038;h=93" width="300" height="93" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/1-b.jpg?w=150 150w" sizes="(max-width: 300px) 100vw, 300px" /></a><p id="caption-attachment-382" class="wp-caption-text">Front Side View</p></div>
<p>He is curious about the length of the wire used to form the lamps. If the distance between each lamp is 10 cm, calculate the length of the wire!</p>
<p><strong>Dutch House</strong></p>
<p>Dina takes a vacation in the Netherlands and visits Lombokstraat in Utrecht. She is curious about the height of a Dutch house with a funnel.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg"><img data-attachment-id="383" data-permalink="https://lauradorindira.wordpress.com/2014/01/28/contextual-math-problems-based-on-field-trip-to-lombokstraat-utrecht-netherlands/attachment/2/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg" data-orig-size="347,518" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1390957852&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="2" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg?w=201" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg?w=347" class="size-medium wp-image-383 aligncenter" alt="2" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg?w=200&#038;h=300" width="200" height="300" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg?w=200 200w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg?w=100 100w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/2.jpg 347w" sizes="(max-width: 200px) 100vw, 200px" /></a></p>
<p>Estimate the height of the house from the ground to the top of the funnel based on the picture above!</p>
<p><strong><span id="more-380"></span>Metal Frame of a Door</strong></p>
<p>When Lochen walks around Lombokstraat in Utrecht, he interests in a house which has a nice design of metal frame on the door of it as shown in the picture below.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg"><img loading="lazy" data-attachment-id="384" data-permalink="https://lauradorindira.wordpress.com/2014/01/28/contextual-math-problems-based-on-field-trip-to-lombokstraat-utrecht-netherlands/attachment/3/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg" data-orig-size="552,295" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1390957891&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="3" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg?w=552" class="size-medium wp-image-384 aligncenter" alt="3" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg?w=300&#038;h=160" width="300" height="160" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/3.jpg 552w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>Calculate the number of quadrilateral in the metal frame!</p>
<p><strong>Big Clock</strong></p>
<p>Have you ever seen a big clock of a church in Lombokstraat? Someday, Thomas walks around and takes pictures of the clock, as seen as bellow.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg"><img loading="lazy" data-attachment-id="385" data-permalink="https://lauradorindira.wordpress.com/2014/01/28/contextual-math-problems-based-on-field-trip-to-lombokstraat-utrecht-netherlands/attachment/4/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg" data-orig-size="585,326" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1390957965&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="4" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg?w=585" class="size-medium wp-image-385 aligncenter" alt="4" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg?w=300&#038;h=167" width="300" height="167" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/4.jpg 585w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>Find the measure of the smallest angle formed by the two hands of clock!</p>
<p><strong>Bridge</strong></p>
<p>When Sarah walks around Lombok, she finds a bridge and takes a photo of it.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg"><img loading="lazy" data-attachment-id="386" data-permalink="https://lauradorindira.wordpress.com/2014/01/28/contextual-math-problems-based-on-field-trip-to-lombokstraat-utrecht-netherlands/attachment/5/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg" data-orig-size="389,519" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1390958013&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="5" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg?w=225" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg?w=389" class="size-medium wp-image-386 aligncenter" alt="5" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg?w=224&#038;h=300" width="224" height="300" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg?w=224 224w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg?w=112 112w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/5.jpg 389w" sizes="(max-width: 224px) 100vw, 224px" /></a></p>
<p>While standing in front of the bridge, she confuses whether the bridge is higher than the building behind it. Can you help Sarah? Explain your answer!</p>
]]></content:encoded>
					
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		<title>Reflection Paper: The Didactical Use of Models in Realistic Mathematics Education (Part 1)</title>
		<link>https://lauradorindira.wordpress.com/2014/01/22/reflection-paper-the-didactical-use-of-models-in-realistic-mathematics-education-part-1/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Wed, 22 Jan 2014 06:02:22 +0000</pubDate>
				<category><![CDATA[Design Material]]></category>
		<category><![CDATA[characteristics (tenets) of RME]]></category>
		<category><![CDATA[Didactical Use of Models]]></category>
		<category><![CDATA[horizontal mathematizing]]></category>
		<category><![CDATA[mathematization]]></category>
		<category><![CDATA[model for]]></category>
		<category><![CDATA[model of]]></category>
		<category><![CDATA[RME]]></category>
		<category><![CDATA[vertical mathematizing]]></category>
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					<description><![CDATA[Referring to an article “The Didactical Use of Models in Realistic Mathematics Education: An Example from A Longitudinal Trajectory on Percentage” written by Marja van Den Heuvel-Panhuizen, I would like to share my understanding about general theory of RME and &#8230; <a href="https://lauradorindira.wordpress.com/2014/01/22/reflection-paper-the-didactical-use-of-models-in-realistic-mathematics-education-part-1/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p>Referring to an article “<i>The Didactical Use of Models in Realistic Mathematics Education: An Example from A Longitudinal Trajectory on Percentage</i>” written by Marja van Den Heuvel-Panhuizen, I would like to share my understanding about general theory of RME and the role of models within this theory.</p>
<p>Realistic Mathematics Education (RME) conceptualizes Freudenthal’s (1971) thought of “mathematics as a human activity”. It means that mathematics is an activity of organizing problems from real situation or mathematical matter &#8211; which is called <b>mathematization</b>&#8211; rather than just the body of mathematical knowledge. Treffers (1978, 1987), furthermore, distinguished the perspective of mathematizing into “horizontal” and “vertical” mathematizing. Mathematical tools, in term of horizontal mathematizing, are used to organize or solve a problem related to daily life. In the other hand, vertical mathematizing indicates for all kind of re-organization and operation within the mathematical system itself.</p>
<p>Regarding to mathematizing, the characteristics of RME is also historically related to the levels of <i>Van Hiele theory</i> and Freudenthal’s <i>didactical phenomenology</i>. As a result, those theory compose the five characteristics (tenets) of RME (de Lange, 1987; Gravemeijer, 1994) which is generally described as follows.</p>
<p><i>1. The use of contexts in phenomenological exploration</i>.</p>
<p>The starting point of mathematics learning in RME should use mathematics concept appearing in reality, such that pupils become immediately connected in contextual situation.</p>
<p><i>2. The use of models or bridging by vertical instruments</i></p>
<p>In this case, the term of “model” refers to situational and mathematical models developed by pupils themselves.</p>
<p><i>3. The use of pupils own creations and contributions</i></p>
<p>Using students’ own constructions and productions. Students’ own constructions and productions are meaningful for them, and so should be used as an essential part of instruction</p>
<p><i>4. The interactive character of the teaching process or interactivity</i></p>
<p>The interactivity principle of RME signifies that learning mathematics is not only an individual activity but also a social activity. Therefore, RME favors whole-class discussions and group work offering students opportunities to share their strategies and inventions with others. In this way, students can get ideas for improving their strategies. Moreover, interaction evokes reflection that enables students to reach a higher level of understanding.</p>
<p><i>5. The intertwining of various learning strands or units</i></p>
<p>The intertwinement principle means mathematical content domains such as number, geometry, measurement, and data handling are not considered as isolated curriculum chapters, but as heavily integrated. Students are offered rich problems in which they can use various mathematical tools and knowledge. This principle also applies within domains. For example, within the domain of number sense, mental arithmetic, estimation and algorithms are taught in close connection to each other</p>
<p><span id="more-378"></span>Since the second discussion within this article is talking about the role of models within RME, thus it much relates to the second characteristics of RME above. It is the use of models or bridging by vertical instruments.</p>
<p>Models within RME have different manifestation since it necessarily indicates essential aspects of mathematical concepts and structures that are relevant for the problem situation. It means that models within this theory can be served as materials, visual sketches, paradigmatic situations, schemes, diagrams, and even symbols in mathematics (Treffers and Goffree, 1985; Treffers 1987, 1991; Gravemeijer 1994). For instance, paradigmatic situation of repeated subtraction can be functioned as model that gives access to the formal long division algorithm.</p>
<p>The suitability of “models” to support the learning process accomplishes at least two important characteristics. First, they must be imaginable context or be originated in realistic. Second, it provides sufficient flexibility to be applied on more advance level. In brief, it means that a model acts as scaffolding which support the progression in vertical mathematizing without blocking the way back to the sources of original strategy.</p>
<p>In term of bridging role between informal and formal level, Streefland (1985) present that models can contribute to level rising by shifting from a “<i>model of</i>” to a “<i>model for</i>”. In brief, the “<i>model of</i>” is a model which represent close connection to the problem situation, while the “<i>model for</i>” is used to facilitate the organization of related and new problem situations as well as to reason mathematically. Freudhental (1970) also published the different between the two models above in his writings. He marked that “model of” is after-image of a piece of given reality, while “model for” is pre-image for a piece to be created reality.</p>
<p>In conclusion, the use of “models” in the implementation of RME at school should be observed carefully since it supports the progression in both horizontal and vertical mathematizing.</p>
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		<title>Hypothetical Learning Trajectory (HLT) within the Three Phases of Design Research</title>
		<link>https://lauradorindira.wordpress.com/2014/01/09/hypothetical-learning-trajectory-hlt-within-the-three-phases-of-design-research/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Thu, 09 Jan 2014 03:57:45 +0000</pubDate>
				<category><![CDATA[Design Material]]></category>
		<category><![CDATA[Design Research]]></category>
		<category><![CDATA[HLT]]></category>
		<category><![CDATA[Hypothetical Learning Trajectory]]></category>
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					<description><![CDATA[Hypothetical learning trajectory (HLT) regarded as an elaboration of Freudhental’s thought experiment provides design and research instrument that is very effective for design research. The HLT supply a marked connection between an instruction theory and a real teaching experiment in &#8230; <a href="https://lauradorindira.wordpress.com/2014/01/09/hypothetical-learning-trajectory-hlt-within-the-three-phases-of-design-research/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p>Hypothetical learning trajectory<b> </b>(HLT) regarded as an elaboration of Freudhental’s thought experiment provides design and research instrument that is very effective for design research. The HLT supply a marked connection between an instruction theory and a real teaching experiment in class. It monitors how the instructional activities perform in natural way appropriate with the didactical activities planned by teacher.</p>
<p>HLT, a key part of instructional learning design, consists of three components: the learning goal, plan for the learning activities, and the hypothesis of learning process. Regarding those components, Simon (1995) refer the learning goal to the possibility of the learning proceed as to the direction which is predicted by teacher. Furthermore, the term of hypothetical learning process relate to teacher’s prediction of how the students’ thinking and understanding will progress during learning activities.</p>
<p>In terms of design research, Bakker (2004) describe that HLT has different function depending on each phase as follows</p>
<ol>
<li>During preliminary design, HLT is formulated as to guidance of a sequence of instructional activities involving several conjectures of students’ thinking that had to be adapted.</li>
<li>During teaching experiment, HLT serves principles of what the teacher or researcher want to focus on in teaching, interviewing, and observing. It is possible that the HLT of the instructional activity during this phase needs to be adjusted for the next lesson. Several conditions that might caused the minor changes in HLT are anticipations that have not come true, strategies that have not been foreseen, activities that were too difficult, etc. However, it is very essential for teacher or researcher to report those kinds of changes supported by theoretical considerations.</li>
<li>During retrospective analysis, HLT is used as guideline in determining what teacher or researcher should focus on the analysis. The researcher can compare those anticipations with the observation during concrete learning in class. As the result, the HLT can be formed after the retrospective analysis in term of providing a guideline for the next design phase.</li>
</ol>
<p>To sum up, HLT is very useful in term of design and research instrument since it propose comprehensible step of teaching activities within all phases of design research. Moreover, the HLT helps teacher to reflect their teaching whether it needs any improvement or not. Therefore, the proceeding of HLT cannot be done in one design because it considers the feedback of each lesson as well as the implementation of its improvement in the next lesson.</p>
<p>References</p>
<p>Bakker, A. (2004). <i>Design Research in Statistics Education. On Symbolizing and Computer Tools. </i>Amersfoort: Wilco Press.<i></i></p>
<p>Simon, M. A. (1995). Reconstructing mathematics pedagogy from a constructivist perspective. <i>Journal for Research in Mathematics Education, 26</i>, 114-145.</p>
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		<title>Problems Related to Theorem of Thabit ibn Qurra</title>
		<link>https://lauradorindira.wordpress.com/2014/01/08/problems-related-to-theorem-of-thabit-ibn-qurra/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Wed, 08 Jan 2014 01:31:54 +0000</pubDate>
				<category><![CDATA[Euclidian Geometry]]></category>
		<category><![CDATA[Pythagorean Theorem]]></category>
		<category><![CDATA[Qurra's Theorem]]></category>
		<category><![CDATA[Theorem of Thabit ibn Qurra]]></category>
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					<description><![CDATA[See figure below Conditions: m ∠APC = m ∠BQC = m ∠ACB Sa, Sb, and Sc are squares on BC, AC, and AB respectively &#124;AB&#124; = c, &#124;BC&#124; = a, &#124;CA&#124; = b, and &#124;PC&#124; = d Problems Prove that &#8230; <a href="https://lauradorindira.wordpress.com/2014/01/08/problems-related-to-theorem-of-thabit-ibn-qurra/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p>See figure below</p>
<p align="center"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg"><img loading="lazy" data-attachment-id="370" data-permalink="https://lauradorindira.wordpress.com/2014/01/08/problems-related-to-theorem-of-thabit-ibn-qurra/capture1-3/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg" data-orig-size="510,296" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1389169236&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture1" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg?w=510" class="alignnone size-medium wp-image-370" alt="Capture1" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg?w=300&#038;h=174" width="300" height="174" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture11.jpg 510w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>Conditions:</p>
<ul>
<li>m ∠APC = m ∠BQC = m ∠ACB</li>
<li><i>S</i>a, <i>S</i>b, and <i>S</i>c are squares on BC, AC, and AB respectively</li>
<li>|AB| = c, |BC| = a, |CA| = b, and |PC| = d</li>
</ul>
<p>Problems</p>
<ol>
<li>Prove that : area <i>S</i>b = area <i>R</i>b and area <i>S</i>a = area <i>R</i>a</li>
<li>Prove that ab = cd</li>
</ol>
<p><span id="more-369"></span>Solutions</p>
<p>1.  The proof is as follows</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture22.jpg"><img loading="lazy" data-attachment-id="371" data-permalink="https://lauradorindira.wordpress.com/2014/01/08/problems-related-to-theorem-of-thabit-ibn-qurra/capture2-4/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture22.jpg" data-orig-size="476,527" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1389169342&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture2" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture22.jpg?w=271" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture22.jpg?w=476" class=" wp-image-371 aligncenter" alt="Capture2" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture22.jpg?w=540&#038;h=600" width="540" height="600" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture22.jpg?w=135 135w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture22.jpg?w=271 271w" sizes="(max-width: 540px) 100vw, 540px" /></a></p>
<p style="text-align:left;">2.   The proof is as follows</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture32.jpg"><img loading="lazy" data-attachment-id="372" data-permalink="https://lauradorindira.wordpress.com/2014/01/08/problems-related-to-theorem-of-thabit-ibn-qurra/capture3-3/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture32.jpg" data-orig-size="780,372" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1389169388&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture3" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture32.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture32.jpg?w=611" class="wp-image-372 aligncenter" alt="Capture3" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture32.jpg?w=500&#038;h=240" width="500" height="240" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture32.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture32.jpg?w=150 150w" sizes="(max-width: 500px) 100vw, 500px" /></a></p>
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		<title>Ceva’s Theorem</title>
		<link>https://lauradorindira.wordpress.com/2014/01/05/cevas-theorem/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Sun, 05 Jan 2014 23:42:45 +0000</pubDate>
				<category><![CDATA[Euclidian Geometry]]></category>
		<category><![CDATA[Ceva's theorem]]></category>
		<category><![CDATA[Euclidian]]></category>
		<category><![CDATA[Geometry]]></category>
		<category><![CDATA[Giovanni Ceva]]></category>
		<category><![CDATA[Triangle]]></category>
		<category><![CDATA[Yusuf Al-Mu’taman ibn Hud]]></category>
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					<description><![CDATA[This theorem is named after Giovanni Ceva (1647 &#8211; 1743), who proved it in his work De lineis rectis in 1678. However, this theorem was proved much earlier by Yusuf Al-Mu’taman ibn Hud at eleventh-century. Ceva’s theorem states that given &#8230; <a href="https://lauradorindira.wordpress.com/2014/01/05/cevas-theorem/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p>This theorem is named after Giovanni Ceva (1647 &#8211; 1743), who proved it in his work <i>De lineis rectis</i> in 1678. However, this theorem was proved much earlier by Yusuf Al-Mu’taman ibn Hud at eleventh-century.</p>
<p><b>Ceva’s theorem</b> states that given triangle ABC, and points D, E, and F along the triangle’s side, lines AE, BF, and DC are concurrent if only if</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture31.jpg"><img loading="lazy" data-attachment-id="362" data-permalink="https://lauradorindira.wordpress.com/2014/01/05/cevas-theorem/capture3-2/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture31.jpg" data-orig-size="169,61" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388992977&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture3" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture31.jpg?w=169" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture31.jpg?w=169" class="size-full wp-image-362 aligncenter" alt="Capture3" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture31.jpg?w=611"   srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture31.jpg 169w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture31.jpg?w=150&amp;h=54 150w" sizes="(max-width: 169px) 100vw, 169px" /></a></p>
<p>Illustration</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture.jpg"><img loading="lazy" data-attachment-id="358" data-permalink="https://lauradorindira.wordpress.com/2014/01/05/cevas-theorem/capture-4/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture.jpg" data-orig-size="673,529" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388987887&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture.jpg?w=611" class="wp-image-358 aligncenter" alt="Capture" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture.jpg?w=450&#038;h=352" width="450" height="352" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture.jpg?w=150 150w" sizes="(max-width: 450px) 100vw, 450px" /></a></p>
<p><span id="more-357"></span>The proof of Ceva’s Theorem is as follows</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture21.jpg"><img loading="lazy" data-attachment-id="359" data-permalink="https://lauradorindira.wordpress.com/2014/01/05/cevas-theorem/capture2-3/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture21.jpg" data-orig-size="466,530" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388988829&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture2" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture21.jpg?w=264" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture21.jpg?w=466" class=" wp-image-359 aligncenter" alt="Capture2" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture21.jpg?w=526&#038;h=600" width="526" height="600" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture21.jpg?w=132 132w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture21.jpg?w=264 264w" sizes="(max-width: 526px) 100vw, 526px" /></a></p>
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		<title>Pythagorean Theorem</title>
		<link>https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Fri, 03 Jan 2014 01:04:30 +0000</pubDate>
				<category><![CDATA[Euclidian Geometry]]></category>
		<category><![CDATA[History of Mathematics]]></category>
		<category><![CDATA[Euclid's Proof]]></category>
		<category><![CDATA[Proof]]></category>
		<category><![CDATA[Pythagorean Theorem]]></category>
		<category><![CDATA[Rearrangement]]></category>
		<category><![CDATA[Similar Triangle]]></category>
		<category><![CDATA[Similarity]]></category>
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					<description><![CDATA[The Pythagorean Theorem is named for the ancient Greek mathematician Pythagoras (569 B.C – 500 B.C). It states the area of the square built upon of a right triangle is equal to the sum of the areas of the squares &#8230; <a href="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p>The Pythagorean Theorem is named for the ancient Greek mathematician Pythagoras (569 B.C – 500 B.C). It states the area of the square built upon of a right triangle is equal to the sum of the areas of the squares upon the remaining sides.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg"><img loading="lazy" data-attachment-id="346" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture1-2/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg" data-orig-size="582,488" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388734637&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture1" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg?w=582" class="size-medium wp-image-346 aligncenter" alt="Capture1" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg?w=300&#038;h=251" width="300" height="251" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture1.jpg 582w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>Thus, according to the illustration above, the Pythagorean Theorem can be written algebraically as an equation</p>
<p align="center">Q<sub>1</sub> + Q<sub>2 </sub>= Q<sub>3</sub></p>
<p>The statement revealed that Greeks used <b><i>area</i></b> as a fundamental concept, not the <b><i>length</i></b>. As consequence, we have square root the area to define the length.</p>
<p>The theorem has numerous proofs including geometric proofs and algebraic proofs with some dating back thousands of years. Some of the proofs are described below.</p>
<p><b>Proof 1 (Using Rearrangement)</b></p>
<p>To begin with, I have a large square that contain four identical triangles as shown in the figure 1. After that, I want to move all of the triangles to another identical square. As shown in the figure 2.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg"><img loading="lazy" data-attachment-id="347" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture2-2/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg" data-orig-size="944,432" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388734690&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture2" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg?w=611" class="size-medium wp-image-347 aligncenter" alt="Capture2" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg?w=300&#038;h=137" width="300" height="137" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture2.jpg?w=150 150w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>Therefore, the white space within each of the two large squares must have equal area.</p>
<p><b><span id="more-345"></span>Proof 2 (Using Similar Triangles)</b></p>
<p>Let ABC is a right triangle with right angle at C as shown on the figure below.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg"><img loading="lazy" data-attachment-id="348" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture3/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg" data-orig-size="713,398" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388734736&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture3" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg?w=611" class="size-medium wp-image-348 aligncenter" alt="Capture3" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg?w=300&#038;h=167" width="300" height="167" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture3.jpg?w=150 150w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>We draw the altitude from point <i>C</i>, and call <i>D</i> its intersection with the side <i>AB.</i></p>
<p>The proof of the Pythagorean Theorem is as follows</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg"><img loading="lazy" data-attachment-id="349" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture4/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg" data-orig-size="652,498" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388734807&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture4" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg?w=611" class=" wp-image-349 aligncenter" alt="Capture4" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg?w=600&#038;h=458" width="600" height="458" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture4.jpg 652w" sizes="(max-width: 600px) 100vw, 600px" /></a></p>
<p>Euclid was eager to place this result on his book, but he came up with another way to prove the Pythagorean Theorem. One speculation is that the proof by similar triangles involved a theory of proportions, a topic not discussed until later in the <i>Elements</i>. Furthermore, the theory of proportions needed further development at that time.</p>
<p><b>Proof 3 (Euclid’s Proof)</b></p>
<p>The following is a summation of the proof by Euclid that can be found in Book I of Euclid’s <i>Elements</i> specifically in Proposition 47 and 48.</p>
<p>Proposition: In right- angled triangle, the square on the hypotenuse is equal to the sum of the squares on the legs.</p>
<p>Euclid begin with the Pythagorean configuration shown in figure below.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg"><img loading="lazy" data-attachment-id="350" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture5/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg" data-orig-size="518,459" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388734852&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture5" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg?w=518" class="size-medium wp-image-350 aligncenter" alt="Capture5" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg?w=300&#038;h=265" width="300" height="265" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture5.jpg 518w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>Then, he constructed a perpendicular line from B to the segment HI on the square of the hypotenuse. Both points J and K are the intersection of the perpendicular line and the sides of the square on the hypotenuse. The line also lies along the altitude to the right triangle ABC.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg"><img loading="lazy" data-attachment-id="351" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture6/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg" data-orig-size="615,478" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388734877&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture6" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg?w=611" class="size-medium wp-image-351 aligncenter" alt="Capture6" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg?w=300&#038;h=233" width="300" height="233" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture6.jpg?w=150 150w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>Furthermore, Euclid showed that the area of rectangle CHKJ is equal to the area of square on side BC and that the area of rectangle JKIA is equal to area of square on side AB. To prove it, he constructed segment BI and CF, to form the triangles AEC and AIB</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg"><img loading="lazy" data-attachment-id="352" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture7/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg" data-orig-size="683,526" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388734905&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture7" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg?w=611" class="size-medium wp-image-352 aligncenter" alt="Capture7" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg?w=300&#038;h=231" width="300" height="231" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture7.jpg?w=150 150w" sizes="(max-width: 300px) 100vw, 300px" /></a></p>
<p>The proof is as follows.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg"><img loading="lazy" data-attachment-id="353" data-permalink="https://lauradorindira.wordpress.com/2014/01/03/pythagorean-theorem/capture8/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg" data-orig-size="538,533" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388735042&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture8" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg?w=538" class=" wp-image-353 aligncenter" alt="Capture8" src="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg?w=600&#038;h=594" width="600" height="594" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2014/01/capture8.jpg 538w" sizes="(max-width: 600px) 100vw, 600px" /></a></p>
<p>Similarly, it can be shown that rectangle CHJK must have the same area as square BCGF = BC × BC = BC<sup>2</sup>. Then, adding two results, AB<sup>2 </sup>+ BC<sup>2</sup> = (AI × AJ) + (JK × JC). Since AI = AB<sup>2</sup>, then we have AB<sup>2 </sup>+ BC<sup>2</sup> = (AI × AJ) + (AI × JC) = AI × (AJ + JC) = AI × AC = AC × AC = AC<sup>2</sup>. So, in conclusion, we obtain that AB<sup>2 </sup>+ BC<sup>2</sup>= AC<sup>2</sup>.</p>
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		<title>Being Acquainted with the van Hiele Model of Geometric Thought</title>
		<link>https://lauradorindira.wordpress.com/2013/12/29/being-acquainted-with-the-van-hiele-model-of-geometric-thought/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Sun, 29 Dec 2013 23:35:17 +0000</pubDate>
				<category><![CDATA[Uncategorized]]></category>
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					<description><![CDATA[The van Hiele model, in mathematics education, is a theory involving levels of geometrical thinking that can be used to assess students’ abilities as well as lead instruction. This theory was completed simultaneously by Dina van Hiele-Geldof and Pieree van &#8230; <a href="https://lauradorindira.wordpress.com/2013/12/29/being-acquainted-with-the-van-hiele-model-of-geometric-thought/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p>The van Hiele model, in mathematics education, is a theory involving levels of geometrical thinking that can be used to assess students’ abilities as well as lead instruction. This theory was completed simultaneously by Dina van Hiele-Geldof and Pieree van Hiele and was published in their doctoral dissertation at Utrecht University in 1957.</p>
<p>The model is made of five levels that describe how children’ thinking process to reason in geometry. The children will pass through these levels as they improve from merely recognizing geometric figures to being able to work with in a variety of axiomatic systems.</p>
<p>The levels of van Hiele are described as follows:</p>
<p><strong>Level 0. Visualization</strong></p>
<p>At this initial level, students are able to recognize geometric concepts by only their holistic appearance. Geometrical figures, in addition, are identified based on their visual prototypes rather than their components or attributes. Students, for example, recognize that squares in figure (a) are completely different with rectangles in figure (b) in theirs’ mind.</p>
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<p>A student at this level, however, would not distinguish that all of the figures above have right angle or that opposite sides are parallel.<span id="more-341"></span></p>
<p><strong>Level 1. Analysis</strong></p>
<p>Students at this level begin to analyze the properties of geometric figures. Through observation and experimentation, for instance, a student might declare “A square has 4 equal sides and 4 equal angles. Its diagonals are congruent, perpendicular, and bisect each other”. Even though students are able to list all properties of given geometric figures, they do not tolerate the properties to overlap because each property is separate from others. For example, they still insist that “a square” different with “a rectangle” because they may propose extraneous properties to bear such beliefs, such as rectangle has one pair of sides longer than the other pair of sides. As a result, interconnections between figures are still not discerned and the definitions are not yet appreciated.</p>
<p><strong>Level 2. Abstraction (Informal Deduction)</strong></p>
<p>At this stage, students can ascertain the interconnections of properties within figures as well as among figures itself. Learners, for example, begin to recognize that all squares are rectangle, but not all rectangles are squares based on their comprehension of the properties of each. Furthermore, both meaningful definitions and informal arguments of geometric concept can be generated to rationalize their reasoning. Even if formal proofs can be grasped, however, they do not see how the logical order could be altered nor do they see how different premises could be starting construction of a proof.</p>
<p><strong>Level 3. Deduction</strong></p>
<p>Either axiomatic system or the role of undefined terms, axioms, postulates, definitions, theorems, and proof is well understood. A learner at this stage is able to construct, not just memorize, geometric proofs and understand their meaning. Students at this level, however, judge that axioms and definitions are fixed rather than arbitrary. Therefore, they cannot yet consider non-Euclidian geometry because they geometric ideas are still accepted as objects in the Euclidian plane.</p>
<p><strong>Level 4. Rigor</strong></p>
<p>This geometry level is well known as the level of mathematician. Students at this level understand that the object of thought is deductive geometric systems, for which they associate axiomatic systems. In addition, learners can study non-Euclidian geometries with fully understanding.</p>
<p>&nbsp;</p>
<p>This post will be continued with issue “properties of the levels” as soon as possible.</p>
<p>&nbsp;</p>
<p>Reference:</p>
<p>Crowley, Mary L. “<b>The van Hiele Model of the Development of Geometric Thought</b>.” In Learning and Teaching<b><i> </i></b>Geometry, <b><i>K-12, </i></b>1987 Yearbook of the National Council of Teachers of Mathematics, edited by Mary Montgomery Lindquist, pp.1-16. Reston, Va.: National Council of Teachers of Mathematics, 1987.</p>
<p>Mason, Marguerite. “<b>The van Hiele Levels of Geometric Understanding</b>”. Virginia: McDougal Little Inc</p>
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		<title>Angle Bisectors of a Triangle are Concurrent</title>
		<link>https://lauradorindira.wordpress.com/2013/12/28/angle-bisectors-of-a-triangle-are-concurrent/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Sat, 28 Dec 2013 22:44:30 +0000</pubDate>
				<category><![CDATA[Euclidian Geometry]]></category>
		<category><![CDATA[Task 2 (Slideshare)]]></category>
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					<description><![CDATA[Angle bisectors of triangle are concurrent from Laurado Sabatini]]></description>
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<div style="margin-bottom:5px"> <strong> <a href="https://www.slideshare.net/LolaLaurandz/angle-bisectors-of-triangle-are-concurrent" title="Angle bisectors of triangle are concurrent" target="_blank">Angle bisectors of triangle are concurrent</a> </strong> from <strong><a href="https://www.slideshare.net/LolaLaurandz" target="_blank">Laurado Sabatini</a></strong> </div>
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		<title>The Three Perpendicular Bisectors of A Triangle are Concurrent</title>
		<link>https://lauradorindira.wordpress.com/2013/12/27/the-three-perpendicular-bisectors-of-triangle-are-concurrent/</link>
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		<dc:creator><![CDATA[lauradorindira]]></dc:creator>
		<pubDate>Fri, 27 Dec 2013 23:06:58 +0000</pubDate>
				<category><![CDATA[Euclidian Geometry]]></category>
		<category><![CDATA[Task 2 (Slideshare)]]></category>
		<guid isPermaLink="false">http://lauradorindira.wordpress.com/?p=335</guid>

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		<title>Does Comparing Fractions with “Cross-Multiply Trick” really MAKE SENSE for third graders?</title>
		<link>https://lauradorindira.wordpress.com/2013/12/26/does-comparing-fractions-with-cross-multiply-trick-really-make-sense-for-third-graders/</link>
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		<pubDate>Thu, 26 Dec 2013 23:20:18 +0000</pubDate>
				<category><![CDATA[Reflection Paper]]></category>
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					<description><![CDATA[That question came up when my little brother, who is in 3rd grade, asked for her math homework, specifically how to compare fractions with like or unlike denominators. Before further discussion to answer the question arises, to begin with, I &#8230; <a href="https://lauradorindira.wordpress.com/2013/12/26/does-comparing-fractions-with-cross-multiply-trick-really-make-sense-for-third-graders/">Continue reading <span class="meta-nav">&#8594;</span></a>]]></description>
										<content:encoded><![CDATA[<p>That question came up when my little brother, who is in 3<sup>rd</sup> grade, asked for her math homework, specifically how to compare fractions with like or unlike denominators.</p>
<p>Before further discussion to answer the question arises, to begin with, I would like to describe how he understands the term of comparing fractions with like denominators. Spontaneously, the idea of utilizing “bar modeling” and “number line” came to my mind and it worked.</p>
<p>In the term of “bar modeling”, I introduce two equal bars in size which are representing two fractions that being compared. Let’s say I want to compare  1/4 and 3/4 . The first bar models the first fraction 1/4, while the second bar be 3/4 as shown in the illustration below.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg"><img loading="lazy" data-attachment-id="329" data-permalink="https://lauradorindira.wordpress.com/2013/12/26/does-comparing-fractions-with-cross-multiply-trick-really-make-sense-for-third-graders/capture/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg" data-orig-size="691,222" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388124752&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg?w=611" class=" wp-image-329 aligncenter" alt="Capture" src="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg?w=400&#038;h=128" width="400" height="128" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg?w=400 400w, https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture.jpg 691w" sizes="(max-width: 400px) 100vw, 400px" /></a></p>
<p>It is obvious from the illustration above that  1/4 &lt; 3/4</p>
<p>Furthermore, considering “number line” as the alternative, I asked him to put fractions (1/4, 2/4, 3/4 and 4/4) on the given number line in the correct order.</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg"><img loading="lazy" data-attachment-id="330" data-permalink="https://lauradorindira.wordpress.com/2013/12/26/does-comparing-fractions-with-cross-multiply-trick-really-make-sense-for-third-graders/capture1/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg" data-orig-size="974,124" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388124814&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture1" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg?w=611" class=" wp-image-330 aligncenter" alt="Capture1" src="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg?w=600&#038;h=76" width="600" height="76" srcset="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg?w=300 300w, https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg?w=600 600w, https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg?w=150 150w, https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg?w=768 768w, https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture1.jpg 974w" sizes="(max-width: 600px) 100vw, 600px" /></a></p>
<p><span id="more-328"></span>After that, I have him compare which one is bigger 1/4 or 3/4 and without any hesitation, he positively answered that  1/4 &lt;  3/4. In addition, he concluded that the larger fraction is the one with the greater numerator as long as the fractions have like denominators.</p>
<p>In the other hand, different situation happened when he struggled with unlike denominators. I started to explain about equivalent fractions and common denominators but he was confused with it. Curiously, I checked his math textbook and found “multiply-cross trick” without giving any prior explanation about equivalent fractions or common denominators. Then, spontaneously, I asked for myself “Does the trick really make sense for third grader?” and my answer is “I don’t think so”.</p>
<p>Before I elaborate the best way to teach this material, in a brief, here I show you the rule of “multiply-cross trick” that I found on the textbook:</p>
<ol>
<li>Multiply the numerator of the first fraction by the denominator of the second fraction and write down the answer under the first fraction.</li>
<li>Multiply the numerator of the second fraction by the denominator of the first fraction and write down the answer under the second fraction.</li>
</ol>
<p>For example, suppose you are asked to compare these two fractions 4/7 and 3/9</p>
<p>By applying the rule, then you obtain</p>
<p style="text-align:center;"><a href="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture2.jpg"><img loading="lazy" data-attachment-id="331" data-permalink="https://lauradorindira.wordpress.com/2013/12/26/does-comparing-fractions-with-cross-multiply-trick-really-make-sense-for-third-graders/capture2/" data-orig-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture2.jpg" data-orig-size="353,147" data-comments-opened="1" data-image-meta="{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;Rindira&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;1388124842&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}" data-image-title="Capture2" data-image-description="" data-image-caption="" data-medium-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture2.jpg?w=300" data-large-file="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture2.jpg?w=353" class=" wp-image-331 aligncenter" alt="Capture2" src="https://lauradorindira.wordpress.com/wp-content/uploads/2013/12/capture2.jpg?w=200&#038;h=93" width="200" height="93" /></a></p>
<p>Since 36 is greater than 21, then the fraction 4/7 is greater than 3/9.</p>
<p>This trick, perhaps, is going to be easy for my brother or others third graders as long as they understand about multiplication. However, the idea of presenting this trick in his textbook disturbs my mind since it fails in providing any REASONING to understand the concepts.</p>
<p>&nbsp;</p>
<p>So, what should we do to scaffold that idea such that it become more makes sense?”</p>
<p>(The answer of that question will be continued on this post as soon as possible)</p>
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