Constructions and Tilings | IT

Question 14

Construct the following figure.

Question diagram 1
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Solution
Understand the Question
  • The given figure consists of 5 identical rhombuses arranged around a central vertex AA.
  • A complete rotation around a point is 360360^\circ. A full circular pattern would have 8 identical rhombuses, making the central angle of each rhombus: 3608=45\dfrac{360^\circ}{8} = 45^\circ
  • To construct the figure, we construct a single rhombus of side length ss with a 4545^\circ vertex angle, and then replicate it 5 times consecutively around the central vertex, alternating between shaded and unshaded.

Step 1 · Construct a 4545^\circ Angle

Diagram 1

  1. Draw a straight line and mark a point AA on it (the common vertex).
  2. With AA as center, draw a semicircle intersecting the line at points BB and CC.
  3. With centers BB and CC and equal radius, draw intersecting arcs above the line at point DD.
  4. Draw ray ADAD perpendicular to BCBC, giving DAC=90\angle DAC = 90^\circ.
  5. Let the semicircle intersect ray ADAD at EE. With centers CC and EE and equal radius, draw arcs intersecting at FF.
  6. Draw ray AFAF bisecting DAC\angle DAC to get: FAC=45\angle FAC = 45^\circ

Step 2 · Construct One Rhombus

Diagram 2

  1. Choose a convenient side length ss.
  2. With center AA and radius ss, draw an arc cutting ray ACAC at GG and ray AFAF at HH, so: AG=AH=sAG = AH = s
  3. With centers GG and HH and radius ss, draw two arcs intersecting at point II.
  4. Join GIGI and HIHI.

AGIHAGIH is a rhombus with side length ss and vertex angle GAH=45\angle GAH = 45^\circ.

Step 3 · Replicate the Rhombus to Complete the Figure

Diagram 3

  1. Trace and cut out a template of rhombus AGIHAGIH.
  2. Draw around the template on a fresh sheet to create the first rhombus, and shade it completely.
  3. Rotate the template by 4545^\circ around vertex AA sharing an adjacent edge, and draw the second rhombus (leave it unshaded).
  4. Repeat the rotation and drawing process for a total of 5 rhombuses.
  5. Shade the 1st, 3rd, and 5th rhombuses, leaving the 2nd and 4th unshaded.
Answer

The figure of 5 arranged rhombuses with vertex angle 4545^\circ each is constructed, with the 1st, 3rd, and 5th rhombuses shaded.

Common Mistakes
  • Angle Calculation Error: Calculating the vertex angle incorrectly instead of dividing the full circular angle 360360^\circ by 88 to get 4545^\circ.
  • Inconsistent Side Lengths: Changing the compass radius between constructing adjacent sides of the rhombus, which results in a general parallelogram rather than a rhombus.
  • Incorrect Shading Pattern: Forgetting to alternate shading, which must follow the pattern: shaded, unshaded, shaded, unshaded, shaded.

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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