Constructions and Tilings | IT

Question 24

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

Question diagram 1
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Solution
Understand the Question
  • Each interior angle of an equilateral triangle is 6060^\circ, and all its sides are equal in length.
  • For shapes to fit together around a common vertex without any gaps or overlapping, the sum of the angles meeting at that central vertex must equal 360360^\circ.
  • A polygon is classified as a regular polygon if it is both equilateral (all sides equal) and equiangular (all interior angles equal).

Step 1 · Check Angles at the Center

Each interior angle of an equilateral triangle measures 6060^\circ.Diagram 1

Sum of angles around the central vertex OO

6×60=360\begin{aligned} 6 \times 60^\circ &= 360^\circ \end{aligned}

Since the sum of angles at point OO is exactly 360360^\circ, six congruent equilateral triangles fit together perfectly without any gaps or overlaps.

Step 2 · Verify Properties of the Resulting Polygon

  • Equal Sides: Since the six equilateral triangles are congruent, all their outer sides are equal in length: AB=BC=CD=DE=EF=FA\text{AB} = \text{BC} = \text{CD} = \text{DE} = \text{EF} = \text{FA}

  • Equal Angles: Each interior angle of the resulting hexagon is formed by two adjacent angles of the equilateral triangles:

60+60=120\begin{aligned} 60^\circ + 60^\circ &= 120^\circ \end{aligned}

All 66 interior angles are equal to 120120^\circ.

Since all 66 sides are equal and all 66 interior angles are equal, the shape is a regular hexagon.

Answer

Yes, six congruent equilateral triangles can be placed together, and the resulting shape is a regular hexagon.

Common Mistakes
  • Overlooking the Central Angle Condition: For polygons to tile together around a single vertex without overlapping or leaving empty space, the sum of angles meeting at that vertex must be exactly 360360^\circ.
  • Incomplete Regularity Check: Assuming a polygon is regular just because its sides are equal. Both conditions—all sides equal and all interior angles equal—must be verified.

More questions in IT

Q1

How do we find such AA and BB?

From XX and YY, draw arcs above and below XYXY, with the same radii. The two points at which the arcs meet, above and below XYXY, give us AA and BB, respectively.

Use this to construct an eye.

Q2

In Fig. 6.1, join AA and BB with a line. Where does ABAB intersect XYXY, and what is the angle formed between them?

Q3

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

Q4

Which two triangles should be congruent for ABAB to be the perpendicular bisector of XYXY (that is, OO is the midpoint of XYXY and ABAB is perpendicular to XYXY)?

Q5

How do we get these different shapes? Try!

Q6

Will CC and DD lie on the perpendicular bisector AB\text{AB}?

Q7

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

Any point that has the same distance from XX and YY lies on the perpendicular bisector of XY\text{XY}.

Q8

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

Q9

Can we extend the method of constructing the perpendicular bisector to construct a 9090^\circ angle at any point on a line? Draw a line and mark a point OO on it. Construct a 9090^\circ angle at point OO.

Q10

Find a segment of this line for which OO is the midpoint.

Q11

How do we construct this figure?

Q12

What is the angle between two adjacent lines?

Q13

How do we construct a 4545^\circ angle using only a ruler and a compass?

Q14

Construct the following figure.

Q15

Draw an angle. Create a copy of this angle using only a ruler and compass.

Q16

How do we implement this idea using a ruler and a compass?

Q17

How did they make these arches?

Q18

Construct this arch shape on a piece of paper.

Let us think about the support lines this figure will need.

For symmetry, we should have AB=CDAB = CD, and BAD=CDA\angle BAD = \angle CDA. How would you construct these support lines?

Q19

Use these support lines to construct an arch. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing.

Q20

How do we construct this shape?

What supporting lines will you use to draw this arch?

Remember 'Wavy Wave' from the Grade 6 Textbook?

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

How do we construct a regular pentagon (5-sided figure) and a regular hexagon (6-sided figure)? To begin with, try to construct a pentagon and hexagon with equal sidelengths.

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 7070^\circ angle fit into the gap? What is the gap angle AOI\angle AOI?

We have, 40+60+50+30+40+90+gap angle=36040^\circ + 60^\circ + 50^\circ + 30^\circ + 40^\circ + 90^\circ + \text{gap angle} = 360^\circ

Use this to determine whether the 7070^\circ angle fits the gap.

Q26

In Fig. 6.12 can you explain why AODAOD, BOEBOE and COFCOF are straight lines?

Q27

Construct a regular hexagon with a sidelength 4 cm4\text{ cm} using a ruler and a compass.

Q28

Context: We can construct a regular hexagon more directly if we can construct a 120120^\circ 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 cm5\text{ cm}.

Q31

How will you construct 3030^\circ and 1515^\circ angles?

Q32

Construct the following 6-pointed star. Note that it has a rotational symmetry.

Q33

Are the six triangles forming the 6 points of the star — ΔAGH\Delta \text{AGH}, ΔBHI\Delta \text{BHI}, ΔCIJ\Delta \text{CIJ}, ΔDJK\Delta \text{DJK}, ΔELK\Delta \text{ELK}, ΔFLG\Delta \text{FLG} — equilateral? Why?

[Hint: Find the angles.]

Q34

Can a 4×64 \times 6 grid be tiled using multiple copies of 2×12 \times 1 tiles? We are allowed to rotate a 2×12 \times 1 tile and use it.

Q35

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

Q36

What about a 5×75 \times 7 grid?

Complete the justification.

Q38

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

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 5×35 \times 3 grid, makes it non-tileable?

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