Difference between revisions of "Concepts in Action: Equivalence between two rhombi"
From wikisori
(New page: === Age === <br> === Materials === <br> === Preparation === <br> === Presentation === <br> === Control Of Error === <br> === Points Of Interest === <br> === Purpose...) |
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| Line 1: | Line 1: | ||
=== Age === | === Age === | ||
| − | <br> | + | 6-9.<br> |
=== Materials === | === Materials === | ||
| − | <br> | + | *From the small hexagonal box: two red obtuse-angled and two red equilateral triangles |
| + | *From the triangular box: grey equilateral, green right- angled 2/2 triangle fraction insets (optional)<br> | ||
=== Preparation === | === Preparation === | ||
| − | <br> | + | <br> |
=== Presentation === | === Presentation === | ||
| − | <br> | + | #Join each pair of red triangles along the black lines to form two rhombi. |
| + | #Superimpose one rhombus on the other to show congruency. | ||
| + | #Demonstrate that each rhombus was divided into two equal pieces. | ||
| + | #Since each triangle has the value of 1/2 of the same rhombus, all of the triangles are equivalent. | ||
| + | #Demonstrate this equivalence sensorially by superimposing the two metal inset pieces on first one red triangle, and then in another arrangement on the other. | ||
| + | #We know that the red equilateral triangle has the value of 1/4 T1. | ||
| + | #Since the red obtuse-angled triangle is equivalent, it must also have the value of 1/4. | ||
| + | #Therefore, 1/4 + 1/4 = 1/2 T1. | ||
| + | #Superimpose the green right-angled triangle as a proof. | ||
| + | #Following this line, the grey triangle can be constructed with two rhombi. | ||
| + | #Also this 1/2 is the difference between the two hexagons in the form of a rhombus or this right-angled scalene triangle.<br> | ||
=== Control Of Error === | === Control Of Error === | ||
| − | <br> | + | <br> |
=== Points Of Interest === | === Points Of Interest === | ||
| − | <br> | + | <br> |
=== Purpose === | === Purpose === | ||
| − | <br> | + | <br> |
=== Variation === | === Variation === | ||
| − | <br> | + | <br> |
=== Links === | === Links === | ||
| − | <br> | + | <br> |
=== Handouts/Attachments === | === Handouts/Attachments === | ||
| − | <br> | + | <br> |
| − | [[Category:Mathematics]] | + | [[Category:Mathematics]] [[Category:Mathematics_6-9]] |
Latest revision as of 05:29, 11 August 2009
Contents
Age
6-9.
Materials
- From the small hexagonal box: two red obtuse-angled and two red equilateral triangles
- From the triangular box: grey equilateral, green right- angled 2/2 triangle fraction insets (optional)
Preparation
Presentation
- Join each pair of red triangles along the black lines to form two rhombi.
- Superimpose one rhombus on the other to show congruency.
- Demonstrate that each rhombus was divided into two equal pieces.
- Since each triangle has the value of 1/2 of the same rhombus, all of the triangles are equivalent.
- Demonstrate this equivalence sensorially by superimposing the two metal inset pieces on first one red triangle, and then in another arrangement on the other.
- We know that the red equilateral triangle has the value of 1/4 T1.
- Since the red obtuse-angled triangle is equivalent, it must also have the value of 1/4.
- Therefore, 1/4 + 1/4 = 1/2 T1.
- Superimpose the green right-angled triangle as a proof.
- Following this line, the grey triangle can be constructed with two rhombi.
- Also this 1/2 is the difference between the two hexagons in the form of a rhombus or this right-angled scalene triangle.
Control Of Error
Points Of Interest
Purpose
Variation
Links
Handouts/Attachments
