# Concepts in Action: Large hexagonal box

From wikisori

## Contents

### Age

6-9.

### Materials

### Preparation

### Presentation

- Empty the contents of the box.
- Sorting the pieces by their shape, the child observes that all but one of the pieces are congruent obtuse-angled isosceles triangles.
- Sorting them by color and uniting them as always, the child identifies the four figures: hexagon, triangle, rhombus, parallelogram.
- Isolate the figures in yellow.
- Turn the three triangles of the hexagon over as if on hinges so that they are superimposed wrong side up on the unit triangle.
- Thus the three outside triangles are congruent to the unit triangle, and the hexagon is equal to twice the equilateral formed by the three yellow triangles nearby.
- Bring back the other pairs of triangles and determine the value of each figure in relation to the unit triangle.
- Invite the child to experiment to try to find another figure - the arrowhead. determine its value. conclude: since the rhombus = 2/3 unit triangle, and the arrowhead = 2/3 unit triangle, then, rhombus = the parallelogram = the arrowhead.
- Invite the child to experiment with three obtuse- angled isosceles triangles to make all of the possible figures.
- There are four: two concave pentagons, and obtuse-angled trapezoid and an isosceles trapezoid.
- The value of each of these figures is 3/3, thus they are equivalent to the unit triangle and equivalent to each other.
- Again the child forms the various figures by tracing, cutting, pasting or drawing.
- As before, the lines of the various figures are compared first with the guide triangle, then among themselves.
- The same is done with figures formed of three triangles.
- Form the hexagon again, observing that the inscribed figure is 1/2 the circumscribing figure.
- The three black lines of the hexagon are special diagonals (recall how diagonals are formed).
- These connect a vertex with the first non-successive vertex it meets.
- At this point we examine the equivalence between one figure (hexagon) and the sum of several congruent figures (rhombi).
- Isolate the hexagon and the triangle (the figures formed by the yellow pieces).
- Open the hatch of the hexagon and remove the inscribed triangle, replacing it with the triangle made of three parts, and reclose the hatch.
- Notice that the hexagon still has the same value, though it was made of four pieces and is now made of six.
- Divide the hexagon into three rhombi to show that this one figure is equivalent to the sum of these three congruent figures.
- As usual, examine the relationship between the lines of the hexagon and the equilateral triangle; the lines of the hexagon and those of the rhombus; the lines of the rhombus and the equilateral triangle (the diagonals are the only lines considered).

### Control Of Error

### Points Of Interest

This first hexagon will be called H1. The unit triangle is congruent to that of the first box and therefore is called T1. Here we have found that T1 = 1/2 H1 and T1 is inscribed in H1.

### Purpose

### Variation

### Links

### Handouts/Attachments