Constructions and Tilings | IT

Question 26

In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines?

Question diagram 1
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Solution

We can use the angles around the center of the hexagon to show these lines are straight.

Step 1 — Understanding the Hexagon

A regular hexagon has six equal sides. It also has six equal interior angles. The center point is O. We can draw lines from O to each vertex. This divides the hexagon into six triangles. These six triangles are all equilateral triangles. So, each angle at the center is 6060^\circ.

Step 2 — Checking Line AOD

We want to see if AOD is a straight line. A straight line has an angle of 180180^\circ. Let us look at the angles around point O. The angle AOB\angle AOB is 6060^\circ. The angle BOC\angle BOC is 6060^\circ. The angle COD\angle COD is 6060^\circ. We add these angles to find AOD\angle AOD.

AOD=AOB+BOC+COD\angle AOD = \angle AOB + \angle BOC + \angle COD

=60+60+60= 60^\circ + 60^\circ + 60^\circ

180\boxed{\mathbf{180^\circ}}

Since AOD\angle AOD is 180180^\circ, AOD is a straight line.

Diagram 1

Step 3 — Checking Line BOE

Now we check if BOE is a straight line. We look at the angles around point O again. The angle BOC\angle BOC is 6060^\circ. The angle COD\angle COD is 6060^\circ. The angle DOE\angle DOE is 6060^\circ. We add these angles to find BOE\angle BOE.

BOE=BOC+COD+DOE\angle BOE = \angle BOC + \angle COD + \angle DOE

=60+60+60= 60^\circ + 60^\circ + 60^\circ

180\boxed{\mathbf{180^\circ}}

Since BOE\angle BOE is 180180^\circ, BOE is a straight line.

Step 4 — Checking Line COF

Finally, we check if COF is a straight line. We look at the angles around point O one more time. The angle COD\angle COD is 6060^\circ. The angle DOE\angle DOE is 6060^\circ. The angle EOF\angle EOF is 6060^\circ. We add these angles to find COF\angle COF.

COF=COD+DOE+EOF\angle COF = \angle COD + \angle DOE + \angle EOF

=60+60+60= 60^\circ + 60^\circ + 60^\circ

180\boxed{\mathbf{180^\circ}}

Since COF\angle COF is 180180^\circ, COF is a straight line.

Answer

AOD, BOE, and COF are straight lines because the sum of the angles around the center O for each line is 180180^\circ.

More questions in IT

Q1

How do we find such A and B?

From X and Y, draw arcs above and below XY, with the same radii. The two points at which the arcs meet, above and below XY, give us A and B, respectively.

Use this to construct an eye.

Q2

In Fig. 6.1, join A and B with a line. Where does AB intersect XY, and what is the angle formed between them?

Q3

Will the line joining the two points at which the arcs meet, above and below XY, always be the perpendicular bisector of XY, i.e., when XY is of any length, and the arcs are drawn using a radius of any length?

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

Justify the following statement using the facts that we have established.

Any point that has the same distance from X and Y lies on the perpendicular bisector of XY.

Q8

Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and a compass?

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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Q14

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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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The supporting lines are just two line segments of equal length.

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

In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines?

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

Are the six triangles forming the 6 points of the star — Δ\DeltaAGH, Δ\DeltaBHI, Δ\DeltaCIJ, Δ\DeltaDJK, Δ\DeltaELK, Δ\DeltaFLG — equilateral? Why?

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Q34

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Q35

Can a 4×74 \times 7 grid be tiled using 2×12 \times 1 tiles?

Q36

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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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