Constructions and Tilings | IT

Question 23

Can we break a regular hexagon into smaller pieces that can be constructed?

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Solution

A regular hexagon can be divided into six identical equilateral triangles, which are easy to construct.

Step 1 — Understanding the Hexagon's Structure

A regular hexagon has six equal sides. It also has six equal interior angles. We can find each interior angle's measure. Let 'n' be the number of sides. The sum of interior angles is (n2)×180(n-2) \times 180^\circ.

For a hexagon, 'n' is 6. Let us calculate the sum of its interior angles.

Sum of angles=(62)×180\text{Sum of angles} = (6 - 2) \times 180^\circ

=4×180= 4 \times 180^\circ

=720= \mathbf{720^\circ}

Each interior angle is the sum divided by the number of angles.

Each angle=720÷6\text{Each angle} = 720^\circ \div 6

120\boxed{\mathbf{120^\circ}}

A regular hexagon can be divided into six equilateral triangles. These triangles meet at the center of the hexagon. All sides of these triangles are equal. The side length of the hexagon is equal to the side length of these triangles. Each angle in an equilateral triangle is 6060^\circ.

Diagram 1

Step 2 — Constructing an Equilateral Triangle

An equilateral triangle is a constructible piece. We need a ruler and a compass for construction. First, draw a line segment. Let its length be 's'. This segment will be one side of our triangle.

Place the compass point on one end of the segment. Open the compass to the length 's'. Draw an arc above the segment. Move the compass point to the other end of the segment. Keep the compass opening the same. Draw another arc that intersects the first arc. Let us call the intersection point 'P'. Connect point 'P' to both ends of the original segment. This forms an equilateral triangle. All three sides of this triangle are equal to 's'. All three angles are 6060^\circ.

Diagram 2

Step 3 — Breaking the Hexagon into these Pieces

We have shown how to construct an equilateral triangle. A regular hexagon is made up of six such triangles. Imagine placing six identical equilateral triangles together. They meet at a central point. Each triangle contributes a 6060^\circ angle to the center. Six triangles make 360360^\circ around the center. The outer edges of these triangles form the hexagon. So, yes, a regular hexagon breaks into constructible pieces. The pieces are equilateral triangles.

Answer

(i) Yes, a regular hexagon can be broken into smaller pieces that can be constructed. (ii) The smaller pieces are six identical equilateral triangles. (iii) An equilateral triangle can be constructed using a ruler and a compass.

More questions in IT

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Q2

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Q3

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Q4

Which two triangles should be congruent for AB to be the perpendicular bisector of XY (that is, O is the midpoint of XY and AB is perpendicular to XY)?

Q5

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Q6

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Q7

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Q8

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Q9

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Q10

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Q11

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Q12

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Q13

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Q15

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Q16

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Q17

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Q18

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Q19

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Q20

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Q21

If their midpoints are marked, will you be able to construct a pointed arch?

Q22

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Q23

Can we break a regular hexagon into smaller pieces that can be constructed?

Q24

Can six congruent equilateral triangles be placed together as in Fig. 6.12? If yes, will it result in a regular hexagon?

Q25

Consider this figure. Will the 70° angle fit into the gap? What is the gap angle AOI\angle AOI?

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Use this to determine whether the 70° angle fits the gap.

Q26

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Q27

Construct a regular hexagon with a sidelength 4 cm using a ruler and a compass.

Q28

Context: We can construct a regular hexagon more directly if we can construct a 120° angle using a ruler and a compass.

Q. How do we do it?

Q29

Why is CAX=60\angle\text{CAX} = 60^\circ? Is there an equilateral triangle here?

Q30

Construct a regular hexagon of sidelength 5 cm.

Q31

How will you construct 30° and 15° angles?

Q32

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Q33

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[Hint: Find the angles.]

Q34

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Q35

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Q36

What about a 5×75 \times 7 grid?

Complete the justification.

Q38

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Q39

Is an m×nm \times n grid tileable with 2×12 \times 1 tiles, if one of mm and nn is even and the other is odd? If yes, come up with a general strategy to tile it.

Q40

Is an m×nm \times n grid tileable with 2×12 \times 1 tiles, if both mm and nn are odd? Give reasons.

Q41

Here is a 5×35 \times 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2×12 \times 1 tiles?

Q42

Is the following region tileable with 2×12 \times 1 tiles?

Q43

Context: We are considering whether a given region can be tiled using 2×12 \times 1 tiles.

Q. What about this one?

Q44

Were you able to tile this? How can we be sure that this is not tileable? Can you find another unit square that, when removed from a 5×35 \times 3 grid, makes it non-tileable?

Q45

If the plain grid is tileable, is the black-and-white-grid tileable?

Q46

If the black-and-white grid is tileable, is the plain grid tileable?

Q47

Use this idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?

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