Constructions and Tilings Class 7 Free Notes and Mind Map (Free PDF Download)

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Constructions

This chapter teaches you how to construct geometric figures using only a ruler and compass, including perpendicular bisectors, angle bisectors, parallel lines, and regular polygons. You will also learn about tiling patterns and how shapes can cover a plane without gaps or overlaps. Get free mind map from link below.

Geometric Constructions

Eyes

  • Do you recall the ‘Eyes’ construction we did in Grade 6?
  • Eyes can be drawn freehand, but we wanted to construct them so that the lower arc and upper arc of each eye look symmetrical.
  • We relied on our spatial estimation to determine the two centres, A and B, from which we drew the lower arc and upper arc respectively.
  • The arcs define a line XY that ‘supports’ the drawing though it is not part of the final figure.
  • For the eye to be symmetrical, or for the supporting line to be the line of symmetry, the upper and lower arcs should have the same radius.
  • In other words, we must have AX = BX.
  • Since AX = AY and BX = BY, this means AX = AY = BX = BY.

Question: How do we find such A and B?

Solution:

  • 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.

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

Solution:

  • We observe that AB passes through the midpoint of XY, and is also perpendicular to it.

Definition of Bisection and Perpendicular Bisector

  • A division of a line, or any geometrical object, into two identical parts is called bisection.
  • A line that bisects a given line and is perpendicular to it, is called the perpendicular bisector.

Question: 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?

Solution:

  • This can be answered through congruence.
  • Let us consider a line segment XY. Find points A and B such that AX = AY = BX = BY.
  • Draw the lines AB, AX, AY, BX and BY. Let O be the point of intersection between AB and XY.

Question: 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)?

Solution:

  • If we show that ΔAOX ≅ ΔAOY, then OX = OY, and ∠AOX = ∠AOY because they are corresponding parts of congruent triangles.
  • Since ∠AOX and ∠AOY together form a straight angle, we have ∠AOX + ∠AOY = 180°.
  • Thus, ∠AOX = ∠AOY = 90°.
  • This establishes that O is the midpoint of XY and AB is perpendicular to XY.

Proof of Congruence:

  • In ΔAOX and ΔAOY, we already know that AX = AY, and AO is common to both triangles.
  • If we can show that ∠XAO = ∠YAO then, by the SAS congruence condition, we can conclude that the triangles are congruent.
  • To show this, we observe that ΔABX ≅ ΔABY.
  • This is so because AX = AY, BX = BY, and AB is common to both the triangles.
  • Thus, we have ∠XAB = ∠YAB, or ∠XAO = ∠YAO because they are corresponding parts of congruent triangles.
  • Hence, AB is the perpendicular bisector of XY.

Eyes of Different Shapes

Question: How do we get these different shapes? Try!

Solution:

  • One way is to choose two other points C and D such that CX = CY = DX = DY.
  • An eye of a different shape can be drawn using these points.

Question: Will C and D lie on the perpendicular bisector AB?

Solution:

  • The points C and D are at the same distance from both X and Y.
  • We have just seen that joining any two such points gives the perpendicular bisector of XY.
  • Since XY has only one perpendicular bisector, which is the line AB, the points C and D must lie on the line AB.

Question: 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.

Solution:

  • We have established that if a point is equidistant from X and Y, it must lie on the perpendicular bisector of XY.
  • Thus, eyes of different shapes can be drawn by suitably choosing different pairs of points on the perpendicular bisector as centres to construct the upper and lower arcs of the eyes.

Construction of Perpendicular Bisector

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

Steps:

  1. Taking some fixed radius, from X and then Y, construct two sufficiently long arcs above XY. Name the point where the arcs meet as A.
  2. Using the same radius, from X and then Y, construct two sufficiently long arcs below XY. Name the point where the arcs meet as B.
  3. AB is the required perpendicular bisector.

Note:

  • Thus, the perpendicular bisector can be constructed using the simplest geometric tools — an unmarked ruler and a compass.
  • We will use only these two tools for all the other geometric constructions in this chapter, unless there is a need for drawing lines of specific lengths in standard units.

Figure it Out

Question 1: When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore this through construction, and then justify your answer.

Solution:

  • Hint 1: Any point that is of the same distance from X and Y lies on the perpendicular bisector.
  • Hint 2: We can draw the whole line if any two of its points are known.
  • Answer: No, it is not necessary to have the same radius for arcs above and below XY. As long as the arcs from X and Y above XY intersect at a point A, and the arcs from X and Y below XY intersect at a point B, both A and B will be equidistant from X and Y, and thus lie on the perpendicular bisector.

Question 2: Is it necessary to construct the pairs of arcs above and below XY? Instead, can we construct both the pairs of arcs on the same side of XY? Explore this through construction, and then justify your answer.

Solution:

  • Answer: No, it is not necessary to construct arcs on both sides. We can construct two pairs of intersecting arcs on the same side of XY to get two points equidistant from X and Y. Joining these two points will give the perpendicular bisector.

Question 3: While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them? Explore this through construction, and then justify your answer.

Solution:

  • Answer: Yes, it is necessary to use the same radii for both arcs (from X and Y) to ensure that the point of intersection is equidistant from X and Y.

Question 4: Recreate this design using only a ruler and compass.

Solution:

  • Use the construction techniques learned (perpendicular bisectors, angle bisectors, arcs) to recreate the design.
  • After completing the design, you can use a colour pencil with a ruler or compass to trace its boundary to make the design stand out from the supporting lines and arcs.

Note:

  • This method of constructing the perpendicular bisector is not only geometrically exact but also a practical way to construct it accurately.
  • This method to find the midpoint of a line segment is more accurate than measuring the length using a marked scale.

Construction of a 90° Angle at a Given Point

Question: Can we extend the method of constructing the perpendicular bisector to construct a 90° angle at any point on a line? Draw a line and mark a point O on it. Construct a 90° angle at point O.

Steps:

Question: Find a segment of this line for which O is the midpoint.

Solution:

  • Extend the line on either side of O.
  • Using a compass, mark two points X and Y at equal distance from O, so that O is the midpoint of XY.
  • The perpendicular bisector of XY will pass through O and is perpendicular to the given line.

Question: In this case, do we need to draw two pairs of intersecting arcs to get the perpendicular bisector of XY?

Solution:

  • No, we don’t. We already have one point, O, lying on the perpendicular bisector.
  • We only need to construct one pair of arcs (either above or below the line) from X and Y to get another point on the perpendicular bisector.

Method:

  1. Mark points X and Y at equal distances from O on the line.
  2. Using a compass with any radius, draw arcs from X and Y on one side of the line (either above or below).
  3. Let the arcs intersect at point A.
  4. Join O and A. The line OA is perpendicular to XY at O, forming a 90° angle.

Construction Methods in Śulba-Sūtras

  • Ancient mathematicians from different civilizations, including India, knew exact procedures to construct perpendiculars and perpendicular bisectors.
  • In India, the earliest known texts containing these methods are the Śulba-Sūtras.
  • These are geometric texts of Vedic period dealing with the construction of fire altars for rituals.
  • The Śulba-Sūtras are part of one of the six Vedāngas (a term that literally means ‘limbs of the Vedas’).
  • The Sulbas contain the methods that we developed earlier to construct a perpendicular and the perpendicular bisector.
  • All the construction methods in the Śulba-Sūtras make use of a different kind of compass from what you would have used — a rope.
  • A rope can be used to draw circles or arcs. It can also be stretched to form a straight line.

Construction of Perpendicular Bisector Using a Rope (Kātyāyana-Śulbasūtra 1.2)

Let XY be the given line segment, drawn on the ground, for which we need to construct a perpendicular bisector.

Steps:

  1. Fix a small pole or peg vertically into the ground at each point X and Y.
  2. Take a sufficiently long rope. Make two loops at its ends. Without taking into account the parts of the rope that has gone into the loops, fold the rope into half to find and mark its midpoint.
  3. Fasten the two loops at the ends of the rope to the poles at X and Y.
  4. Pull the midpoint of the rope above XY, as shown in the figure, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as A.
  5. Now similarly pull the midpoint of the rope below XY, as shown in the figure, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as B.
  6. AB is the perpendicular bisector.

Question: Justify why AB in the figure is the perpendicular bisector.

Solution:

  • When the rope is folded in half and stretched with loops at X and Y, the midpoint of the rope is equidistant from X and Y.
  • Both points A and B (above and below) are equidistant from X and Y.
  • Therefore, joining A and B gives the perpendicular bisector of XY.

Question: Can you think of different methods to construct a 90° angle at a given point on a line using a rope?

Solution:

  • Use the same rope method as above to find the perpendicular bisector, which will form a 90° angle at the given point.

Angle Bisection for a Design

Question: How do we construct this figure?

  • The supporting lines for this figure will look like a star with 8 equally spaced lines radiating from a center.

Question: What is the angle between two adjacent lines?

Solution:

  • We need the angle between every pair of adjacent lines to be equal.
  • Since 360° is equally divided into 8 parts, every angle is 360° ÷ 8 = 45°.

Question: How do we construct a 45° angle using only a ruler and a compass?

Solution:

  • We know how to construct a 90° angle.
  • If we can divide it into two equal parts, or bisect it, then we get a 45° angle.

General Method to Bisect Any Angle

Consider an angle ∠XOY. We can bisect it if we can draw two congruent triangles ΔOBC and ΔOAC as shown in the figure. Then ∠BOC = ∠AOC.

Question: How do we construct these congruent triangles, given the angle?

Solution:

  • If A and B are marked such that OA = OB, and if C is chosen such that BC = AC, then by the SSS congruence condition, ΔOBC ≅ ΔOAC.
  • So we can bisect an angle as follows.

Steps for Angle Bisection

  1. Mark points A and B such that OA = OB (using the same radius from O on both arms of the angle).
  2. Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius the same. Mark the point of intersection as C.
  3. OC bisects ∠AOB.

Note:

  • So, a 45° angle can be constructed by constructing a 90° angle and then bisecting it.

Figure it Out

Question 1: Construct at least 4 different angles. Draw their bisectors.

Question 2: Construct the 8-petalled figure shown.

Solution:

  • Start by constructing a 90° angle at a point.
  • Bisect it to get 45° angles.
  • Construct another 90° angle perpendicular to the first, then bisect each to create 8 equal angles of 45° each.
  • Draw arcs from the center to create the 8-petalled design.

Question 3: In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.

Solution:

  • Yes, the line OC will still be an angle bisector.
  • The point C will still be equidistant from A and B, ensuring that ΔOAC ≅ ΔOBC, so OC bisects the angle.

Question 4: What are the other angles that can be constructed using angle bisection? Can you construct 65.5° angle?

Solution:

  • Using angle bisection, we can construct angles like 45°, 22.5°, 11.25°, etc. (by successive bisections of 90°).
  • We can construct 60° (using an equilateral triangle), then bisect to get 30°, 15°, 7.5°, etc.
  • To construct 65.5°, we would need to construct 60° + 5.5°, but 5.5° cannot be constructed exactly using ruler and compass alone, so 65.5° cannot be constructed.

Question 5: Come up with a method to construct the angle bisector using a rope.

Solution:

  • Mark two points A and B at equal distances from the vertex O on the two arms of the angle.
  • Use a rope with a fixed length to find a point C that is equidistant from A and B (similar to perpendicular bisector method).
  • Join O and C to get the angle bisector.

Question 6: Construct the following figure. How do we construct the petals so that they are of the maximum possible size within a given square?

Solution:

  • Draw a square.
  • Draw arcs from each corner of the square with radius equal to half the side of the square.
  • The arcs will form petals that touch the midpoints of the sides, creating the maximum possible size.

Repeating Units and Repeating Angles

Question: Construct the following figure.

  • In this figure, there is a single unit repeating itself. To construct this figure, we need to make exact copies of this unit in two different orientations.
  • In order to make exact copies, all the units must have the same arm lengths and the same angle between the arms.
  • We can ensure equal arm lengths using a compass, but how do we ensure equal angles?
  • Let us develop a method to create an exact copy of a given angle.

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

Steps of Construction to Copy an Angle

  1. Draw an arc from A (the vertex of the angle). This gives us three points that form the isosceles triangle ΔABC.
  2. Draw an arc of the same radius from X (the vertex where we want to copy the angle).
  3. Measure BC using a compass. Transfer this length on the arc from Z to get YZ = BC.
  4. By the SSS congruence condition, ΔABC ≅ ΔXYZ. So, ∠A = ∠X.

Figure it Out

Question 1: Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.

Question 2: Construct the given figure.

Solution:

  • Use the method of copying angles to construct repeating units in different orientations.

Construction of a Line Parallel to the Given Line

  • Recall that in the construction using a ruler and a set square, we constructed equal corresponding angles to get parallel lines.

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

Steps:

  • Suppose there is a line m to which we need to draw a parallel line.
  • We construct a line l that intersects m. Line l will serve as a transversal to line m and to the line parallel to m that we are going to construct.
  • Let us choose a point B on l through which we are going to draw the parallel line.
  • This parallel line must make the same corresponding angle, as shown in the figure.
  • This can be done by copying the angle between m and l.

Detailed Steps:

  1. Draw a transversal line l that intersects line m at point A.
  2. Mark a point B on line l through which the parallel line will pass.
  3. Using the angle copying method:
    • Construct arcs of equal radius from A and B on the transversal l.
    • Measure the distance between the points where the arc from A intersects the two lines.
    • Transfer this length to the arc from B to get point E.
  4. Draw a line through B and E. This line is parallel to m.

Figure it Out

Question 1: Construct 4 pairs of parallel lines in different orientations.

Question 2: Construct the following figure.

Solution:

  • Use the construction of parallel lines and copying angles to recreate the given geometric figure.

Arch Designs

Trefoil Arch

  • Have you seen this kind of beautiful arch?
  • Examples: Diwan-i-Aam at Red Fort (Delhi), Central Park (New York City).

Question: How did they make these arches?

Solution:

  • The first step is to be able to draw them on a plane surface such as paper or stone.

Question: Construct this arch shape on a piece of paper.

Solution:

  • Let us think about the support lines this figure will need.
  • For symmetry, we should have AB = CD, and ∠BAD = ∠CDA.

How would you construct these support lines?

Steps:

  1. Draw a horizontal line AD.
  2. Construct equal angles at A and D (using angle copying or angle construction methods).
  3. Mark B and C such that AB = CD.
  4. Use these support lines to construct an arch by drawing arcs from appropriate centers.
  5. If required, adjust the radii of the arcs to make the arch look more aesthetically pleasing.

A Pointed Arch

  • Some arches look like a pointed shape.

Question: How do we construct this shape? What supporting lines will you use to draw this arch?

Solution:

  • Remember “Wavy Wave” from the Grade 6 Textbook?
  • The supporting lines are just two line segments of equal length.

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

Solution:

  • Yes, mark the midpoints of two equal line segments placed side by side.
  • Draw arcs from these midpoints to create the pointed arch shape.

Figure it Out

Question 1: Use support lines to construct a pointed arch. Make different arches, by changing the radius of the arcs.

Question 2: Make your own arch designs.

Regular Hexagons

  • Recall that a regular polygon has equal sides and equal angles.
  • A regular polygon with 3 sides is an equilateral triangle, and a regular polygon with 4 sides is a square.
  • We have constructed these figures earlier.

Question: 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.

Note:

  • To construct a regular pentagon, we first need to have a better understanding of triangles and pentagons. We will discuss this in later years.
  • However, constructing a regular hexagon is within our reach!

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

Regular Hexagon and Equilateral Triangles

Question: What happens when we join the ‘opposite’ points of a regular hexagon? Since a regular hexagon has equal sides and angles, can we expect a figure like this? Will all the triangles in the figure be equilateral triangles?

Solution:

  • To answer these questions, we will reverse our approach.

Question: Can six congruent equilateral triangles be placed together as shown? If yes, will it result in a regular hexagon?

Solution:

  • If six congruent equilateral triangles can indeed be placed together, then the sides of the resulting hexagon are equal, and their angles are 60° + 60° = 120°.
  • So what we really need to examine is whether 6 congruent equilateral triangles can fit this way without overlapping and without leaving any gaps around the centre.
  • We have defined a degree by taking the complete angle around a point to be 360°.
  • So all the angles around the centre should add up to 360°.

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

Solution:

  • We have: 40° + 60° + 50° + 30° + 40° + 90° + gap angle = 360°.
  • Gap angle = 360° – (40° + 60° + 50° + 30° + 40° + 90°) = 360° – 310° = 50°.
  • So the 70° angle will not fit the gap (since the gap is only 50°).

Conclusion:

  • Thus, if there are angles that add up to 360°, their vertices can be joined together at a single point such that:
    • (a) the angles do not overlap, and
    • (b) they completely cover the region around the point.
  • Since each angle in an equilateral triangle is 60°, six such angles add up to 360° (6 × 60° = 360°).
  • Therefore, six congruent triangles can be arranged to form a regular hexagon.

Question: In the figure, can you explain why AOD, BOE and COF are straight lines?

Solution:

  • Since the six equilateral triangles are arranged symmetrically around point O, the opposite vertices form straight lines through O.
  • Each pair of opposite triangles forms a 180° angle at O, making AOD, BOE, and COF straight lines.

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

Solution:

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

Question: How do we do it?

Solution:

  • This can be done if we can construct a 60° angle.
  • If we construct 60°, we also get 120° (since 180° – 60° = 120°).

Construction of a 60° Angle

Question: How do we construct a 60° angle?

Solution:

  • We get a 60° angle if we construct an equilateral triangle!
  • We can use the following steps for this.

Steps:

  • Suppose we need a 60° angle at point A on a line segment AX.

Step 1: Construct an arc with centre A and any radius. Let it intersect AX at B.

Step 2: With the same radius, draw an arc with centre B that intersects the first arc at C.

Step 3: Join A and C. We have ∠CAX = 60°.

Explanation:

  • ΔABC is an equilateral triangle because AB = AC = BC (all equal to the radius).
  • Therefore, all angles of the triangle are 60°.

Question: Construct a regular hexagon of sidelength 5 cm.

Solution:

  1. Draw a line segment AB = 5 cm.
  2. At A, construct a 60° angle using the method above.
  3. Mark AF = 5 cm on the new line.
  4. At B, construct a 60° angle on the other side.
  5. Mark BC = 5 cm.
  6. Continue this process to construct all six sides, each of 5 cm, with 60° angles between adjacent sides.
  7. The six vertices form a regular hexagon.

Construction of 30° and 15° angles

Question: How will you construct 30° and 15° angles?

Solution:

  • Construct a 60° angle, then bisect it to get 30°.
  • Bisect the 30° angle to get 15°.

6-Pointed Star

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

Hint: Do you see a hexagon inside?

Solution:

  • Construct a regular hexagon.
  • Extend alternate sides of the hexagon to form equilateral triangles pointing outward.
  • This creates a 6-pointed star.

Question: Are the six triangles forming the 6 points of the star — ΔAGH, ΔBHI, ΔCIJ, ΔDJK, ΔELK, ΔFLG — equilateral? Why?

Solution:

  • Yes, they are equilateral.
  • [Hint: Find the angles. Each angle of the hexagon is 120°, so the triangles formed have angles of 60° each.]

Figure it Out

Question 1: Construct the following figures:

(a) An Inflexed Arc

Solution:

  • The fun part about this figure is that it can also be constructed using only a compass! Can you do it?

(b) Other figures as shown

Question 2: Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.

Solution:

  • Optical illusions often use patterns, angles, and symmetry to create visual effects.
  • Recreate the figure by carefully following the construction steps and observing the patterns.

Question 3: Construct this figure.

Hint: Find the angles in this figure.

Solution:

  • Analyze the angles in the figure (likely multiples of 60°, 90°, or 45°).
  • Use angle construction and bisection methods to recreate the figure.

Question 4: Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P.

Hint: Find a line segment on l whose perpendicular bisector passes through P.

Solution:

  1. From point P, draw two arcs with the same radius that intersect line l at two points, say A and B.
  2. P is equidistant from A and B.
  3. Find another point Q (on the opposite side of l from P) that is also equidistant from A and B.
  4. Join P and Q. The line PQ is the perpendicular bisector of AB and is perpendicular to l.

Tiling

  • Tangrams are puzzles that originated in China.
  • They make use of 7 pieces obtained by dividing a square as shown.
  • For the problems ahead, we need these 7 tangram pieces. These are provided at the end of the book.
  • Or, by looking at the figure, you could make cardboard cutouts of the pieces.

The 7 tangram pieces:

  • A (large triangle)
  • B (large triangle)
  • C (medium triangle)
  • D (small triangle)
  • E (small triangle)
  • F (square)
  • G (parallelogram)

Example:

  • We can form interesting shapes by rearranging the tangram pieces. Here is an arrow.

Figure it Out

Question: How can the tangram pieces be rearranged to form each of the following figures?

Solution:

  • Experiment with different arrangements of the 7 tangram pieces to form the given shapes.

Definition of Tiling

  • Covering a region using a set of shapes, without gaps or overlaps, is called tiling.

Tiling with Rectangular Grids

Consider a rectangular grid made of unit squares.

Example: A 4 × 6 grid has 4 rows and 6 columns.

Question: Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles?

Solution:

  • Yes. We are allowed to rotate a 2 × 1 tile and use it.
  • Here is one way: Use vertical tiles or horizontal tiles to cover the grid.
  • Obviously, this is not the only tiling possible.

Question: Can a 4 × 7 grid be tiled using 2 × 1 tiles?

Solution:

  • A 4 × 7 grid has 4 × 7 = 28 unit squares.
  • Each 2 × 1 tile covers 2 unit squares.
  • Number of tiles needed = 28 ÷ 2 = 14 tiles.
  • Since 28 is even, a 4 × 7 grid can be tiled using 2 × 1 tiles.

Question: What about a 5 × 7 grid?

Solution:

  • A 5 × 7 grid has 5 × 7 = 35 unit squares.
  • To tile this grid using 2 × 1 tiles, we need each tile to cover exactly 2 unit squares.
  • Since 35 is odd, it is impossible to tile a 5 × 7 grid using 2 × 1 tiles (because 35 cannot be evenly divided by 2).

Question: Complete the justification.

Solution:

  • To tile a grid, the total number of unit squares must be even (since each tile covers 2 squares).
  • 5 × 7 = 35 is odd, so tiling is not possible.

Question: Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even? If yes, come up with a general strategy to tile it.

Solution:

  • Yes, if both m and n are even, the grid has an even number of unit squares (m × n is even).
  • General strategy: Cover each column with vertical tiles. This is possible because the number of rows is even.

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

Solution:

  • Yes, if one of m or n is even, then m × n is even.
  • General strategy: If m is even, use vertical tiles to cover all columns. If n is even, use horizontal tiles to cover all rows.

Question: Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd? Give reasons.

Solution:

  • No, if both m and n are odd, then m × n is odd.
  • Since each tile covers 2 unit squares, we need an even number of unit squares to tile the grid.
  • Therefore, a grid with an odd number of unit squares cannot be tiled using 2 × 1 tiles.

Tiling with a Missing Square

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

Solution:

  • A 5 × 3 grid has 15 unit squares. Removing 1 square leaves 14 unit squares (which is even).
  • However, even though the number of squares is even, the grid may not be tileable depending on which square is removed.

Question: Is the region tileable with 2 × 1 tiles?

Solution:

  • Depends on the specific configuration. Some configurations may be tileable, others may not.

Question: What about this one?

Solution:

  • Need to analyze the specific configuration.

Question: 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 × 3 grid, makes it non-tileable?

Solution:

  • There is an interesting way to look at these questions using a black-and-white coloring method.

Black-and-White Grid Method

  • For any tiling problem of this kind, we can create a similar problem with the unit squares coloured black and white so that:
    • Black squares have only white neighbours.
    • White squares have only black neighbours (like a chessboard).

Example:

  • For the tiling problem in the figure, we get a black-and-white grid.
  • In the black-and-white region, the problem is to tile the region with the 2 × 1 black-and-white tiles so that:
    • Each black square of a tile sits on a black square of the grid.
    • Each white square sits on a white square.

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

Solution:

  • Yes, if the plain grid can be tiled, the corresponding black-and-white grid can also be tiled.

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

Solution:

  • Yes, if the black-and-white grid can be tiled, the plain grid can also be tiled.

Conclusion:

  • It can be seen that the answer to both the questions is yes.

Question: Is the black-and-white region in the figure tileable?

Solution:

  • Any region tiled with black-and-white tiles must have an equal number of black tiles and white tiles.
  • Since the black-and-white region in the figure has 8 white squares and 6 black squares, it can never be tiled with these tiles!

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

Solution:

  • In a 5 × 3 grid with a chessboard coloring, there are 8 squares of one color and 7 squares of the other color.
  • If we remove a square of the color that has 8 squares, we get 7 squares of each color, and the grid can be tiled.
  • If we remove a square of the color that has 7 squares, we get 8 squares of one color and 6 squares of the other, and the grid cannot be tiled.

Note:

  • Isn’t it surprising how, by introducing colours and making the problem more complicated, it becomes easier to tackle? What a creative way of looking at the problem!

Figure it Out

Question: Are the following tilings possible?

1. Region to be tiled with a given tile:

Solution:

  • Use the black-and-white method to check if the number of black and white squares in the region matches the number of black and white squares required by the tiles.

2. Region to be tiled with a T-shaped tile:

Solution:

  • Analyze the specific configuration and use logical reasoning or the black-and-white method to determine if tiling is possible.

Tiling the Entire Plane

  • So far we have seen how to tile a given region. What about tiling the entire plane?

Question: Can you think of a shape whose copies can tile the entire plane?

Solution:

  • Clearly, squares can tile the entire plane.

Tiling with Regular Polygons

Question: Are there other regular polygons that can tile the plane?

Solution:

  • Equilateral triangles can tile the plane.

Explanation:

  • Tiling with equilateral triangles shows the possibility of tiling with another regular polygon.

Other Regular Polygons:

  • A plane can be tiled using regular hexagons as well.

Note:

  • A plane can also be tiled using more than one shape, and using non-regular polygons.
  • People such as the great Dutch artist Escher (1898–1972) — whose works explored mathematical themes such as tiling — have come up with creative ways of tiling a plane with animal shapes!

Examples of Escher-like Tilings:

  • (a) Fish tessellation
  • (b) Bird tessellation
  • (c) Lizard tessellation
  • (d) Abstract pattern tessellation

Recent Discovery:

  • Mathematicians are still exploring various ways of tiling the plane!
  • Tiling (b) was found as recently as 2023.

Tilings in Daily Life

  • Have you seen tilings in daily life? They are often used in buildings and in designs.
  • Tilings are found in nature too. The front face of bee hives and some wasp nests are tiled using hexagonal cells!
  • These cells are used by the insects to keep their eggs, larvae and pupae safe, as well as to store food.
  • Because the region is tiled, no space is wasted.
  • Scientists still wonder how bees and wasps are able to make hexagonal cells.
  • Next time you see any tiling, pay closer attention to it!

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