Constructions and Tilings | CBSE Class 7 Maths Notes

This note covers geometric constructions, perpendicular bisectors, angle bisection, copying angles, parallel lines, arch designs, regular hexagons, tangrams, rectangular grids, colouring arguments and tiling the plane.

How does an eye design lead to a perpendicular bisector?

A geometric construction makes a figure using specified tools and geometric relationships. An unmarked ruler draws straight lines, while a compass draws circles or arcs. An arc is part of a circle; its radius is the distance from its centre to the circle.

For an eye design, label its two ends X and Y. The supporting line XY guides the construction but is not part of the finished eye. The notation XY names the segment joining X and Y, or its length when used in an equality.

Choose A as the centre of the lower arc and B as the centre of the upper arc. For the supporting line to be a line of symmetry, dividing the eye into matching reflected halves, the two arcs must have equal radii.

Thus AX = AY = BX = BY, where the equals sign means that the lengths are equal. Each centre is equally far from both ends of the eye. Draw equal-radius arcs from X and Y to locate A above XY and B below it.

What does joining the arc intersections reveal?

Join A and B. Their joining line crosses XY at its midpoint, the point dividing the segment into equal lengths. It also meets XY at a right angle, an angle of 90°. The symbol ° means degrees, the unit used to measure angles.

Definition: Bisection divides a line segment or geometrical quantity into two identical parts. A perpendicular bisector passes through a segment's midpoint and meets the segment at a right angle.

X and Y are the segment's endpoints. Equal-radius arcs intersect at A above the segment and B below it. Joining these intersections gives the perpendicular bisector.

See Fig. 6.1 in your NCERT textbook

Why does the perpendicular-bisector construction work?

Let O be the intersection of AB and XY in the eye construction, with A and B on opposite sides of XY and AX = AY = BX = BY. Congruent triangles have the same shape and size, so their matching sides and angles are equal.

The symbol Δ names a triangle, and ≅ means “is congruent to”. The symbol ∠ names an angle: ∠XAB is the angle between AX and AB, with its vertex, or meeting point, at A.

Property: Equal distances identify the perpendicular bisector

Any point at the same distance from X and Y lies on the perpendicular bisector of XY. The construction can be justified using two congruence conditions. SSS means three matching sides are equal. SAS means two matching sides and the angle between them are equal.

1. Compare ΔABX and ΔABY. They have AX = AY, BX = BY and the common side AB. They are congruent by SSS.
2. The matching angles ∠XAB and ∠YAB are equal. Since O lies on AB between A and B, ∠XAO = ∠YAO.
3. Now compare ΔAOX and ΔAOY. They have AX = AY, the common side AO and equal angles between these sides. They are congruent by SAS.
4. Consequently, OX = OY and ∠AOX = ∠AOY. These two angles together form a straight angle of 180°, so each is 90°.

Worked example 1. Two equal adjacent angles together form a straight angle of 180°. Find each angle.

Answer: Equal angles divide the total equally. Each angle is 180° ÷ 2 = 90°, where ÷ means division. In the construction, this proves perpendicularity, while OX = OY proves bisection.

Which changes preserve the construction?

The arcs in each intersecting pair must have equal radii to guarantee equal distances from X and Y. Different pairs may use different radii. Two distinct intersections on the same side also determine the perpendicular bisector, provided each is equally distant from X and Y.

Different suitable pairs of centres on this perpendicular bisector produce different eye shapes. For an eye symmetrical about XY, retain equal radii for its upper and lower arcs.

How can we construct a right angle with a compass or rope?

To draw a perpendicular, a line meeting another at a right angle, at a given point O on a line, first make O the midpoint of a segment. This adapts the perpendicular-bisector construction without measuring an angle.

1. Extend the given line on both sides of O if necessary.
2. With centre O and a fixed compass opening, mark X and Y on opposite sides, so OX = OY.
3. Draw equal-radius arcs from X and Y that intersect at a point A away from the line.
4. Join O to A. Both points lie on the perpendicular bisector of XY, so OA meets the original line at 90°.

Worked example 2. Construct a 90° angle at a point O on a given line.

Answer: Mark equally distant points X and Y on opposite sides of O. Draw equal-radius intersecting arcs from X and Y and name their intersection A. Join OA. Only one pair of intersecting arcs is needed because O is already known to lie on the perpendicular bisector.

How does a rope preserve equal distances?

The Śulba-Sūtras are geometric texts of the Vedic period concerning the construction of fire altars for rituals. They are the earliest known Indian texts containing these construction methods. Their constructions use a rope, which can draw arcs or stretch into a straight line.

Fix pegs at X and Y. Take a sufficiently long rope with loops at its ends. Excluding the portions used for the loops, fold the rope to mark its midpoint. Fasten its ends to the two pegs.

Pull the midpoint above XY until both halves are fully stretched, and mark A. Repeat below XY to mark B. Equal rope halves give AX = AY and BX = BY. Therefore AB is the perpendicular bisector. Keeping the rope fully stretched is essential to this reasoning.

How is an angle bisected to make an eight-petalled design?

An angle bisector divides an angle into two equal angles. In the eight-petalled design, eight equal angles meet at the centre. A complete turn is 360°, so each angle between neighbouring supporting directions is 45°.

Property: Equal-sided triangles justify angle bisection

Let ∠XOY be the given angle, with vertex O and arms OX and OY. An arm is one of the rays forming an angle; a ray starts at a point and continues in one direction. Mark B on OX and A on OY so OA = OB.

1. Use one compass opening to mark A and B at equal distances from O.
2. From A and B, draw arcs with the same sufficiently long radius so that they meet inside the angle.
3. Name this intersection C and draw the ray OC.
4. Use OA = OB, AC = BC and the common side OC to establish that ΔOAC and ΔOBC are congruent by SSS.

The corresponding angles ∠AOC and ∠BOC are equal. Thus OC bisects the original angle. The first pair of equal lengths comes from the arc centred at O; the second comes from the equal-radius arcs centred at A and B.

Worked example 3. Find the equal angle between neighbouring supporting directions of an eight-petalled design, and construct it.

Answer: Divide the complete angle equally: 360° ÷ 8 = 45°. Construct a 90° angle using a perpendicular, then bisect it. Each resulting angle is 90° ÷ 2 = 45°.

Eight narrow petals meet at a common centre. The supporting drawing has equally spaced directions around that centre, including horizontal, vertical and sloping directions.

See Fig. 6.5 in your NCERT textbook

The construction combines two ideas: equal angles organise the supporting directions, and equal-radius arcs shape the petals. The supporting lines explain the design's arrangement even when they are omitted from its finished boundary.

How can we copy an angle and construct a parallel line?

Copying an angle reproduces its opening without measuring it in degrees. For a repeating unit to be an exact copy, both its arm lengths and the angle between its arms must match. A compass transfers lengths and supplies the equal sides needed for congruence.

How is the angle transferred?

1. Let A be the original angle's vertex. Draw an arc centred at A, meeting its arms at B and C.
2. Draw a starting ray from a new point X. With the same radius, draw an arc centred at X, meeting that ray at Z.
3. Set the compass opening equal to BC. From Z, cut the new arc at Y, so YZ = BC.
4. Join XY. The triangles ABC and XYZ have three equal corresponding sides, so they are congruent by SSS.

Here AB matches XY, AC matches XZ, and BC matches YZ. Hence the copied angle ∠YXZ equals ∠BAC. The new angle can face a different direction while keeping the same size.

How do corresponding angles give parallel lines?

Parallel lines lie in the same plane and do not meet when extended. A plane is a flat surface extending in all directions. A transversal is a line crossing other lines. Corresponding angles occupy matching positions at its intersections with them.

Call the given line m and a crossing line l. Let A be their intersection, and choose B on l away from m. Copy the angle between m and l at B in the corresponding position. Extend the new arm into line n.

Equal corresponding angles give m ∥ n, where ∥ means “is parallel to”. This uses the angle-copying construction directly: equal-radius arcs and a transferred distance reproduce the required opening.

Line l crosses m at A. Equal-radius arcs are drawn at A and B. The distance CD is transferred from F to locate E; the final line through B and E is n, parallel to m.

See Figs. 6.8, 6.9 and 6.10 in your NCERT textbook

How do supporting lines help construct arches?

An arch is a curved outline spanning an opening. An arch design can first be drawn on a plane surface such as paper or stone. Supporting lines organise its equal lengths and angles before the curved boundary is constructed.

What gives a trefoil arch its symmetry?

A trefoil arch has a three-part curved outline. For its supporting figure, name the base endpoints A and D, and the two upper supporting points B and C. The required equalities are AB = CD and ∠BAD = ∠CDA.

Begin with AD. Construct equal angles at A and D, with their arms directed towards the upper part of the design. Mark B and C on these arms at equal distances from A and D. The compass transfers these equal lengths.

Use the supporting figure to construct the arcs. If required, adjust their radii to make the arch look more aesthetically pleasing. The construction therefore combines exact symmetry conditions with choices about the final curved appearance.

What supports a pointed arch?

A pointed arch has a pointed top. Its supporting figure consists of two equal-length line segments meeting at the top. Mark their midpoints to guide the curved construction. Changing the radii of the arcs produces different arch designs.

Two sloping segments meet at an upper point. Each segment has a marked midpoint. These straight supports guide the pointed curved outline shown above them.

See Fig. 6.11 in your NCERT textbook

These designs apply familiar tools in a new setting. Copying an angle creates equal openings on the two sides; transferring a length creates equal supports; and bisecting a segment locates its midpoint. Identify these relationships before drawing the decorative boundary.

How can we construct 60°, 120°, 30° and 15° angles?

An equilateral triangle has three equal sides and three angles of 60°. It provides a direct way to construct 60° using a ruler and compass. Equal-radius arcs establish the equal side lengths without measuring the angle.

How does the equilateral-triangle method work?

1. Choose point A on a segment AX, where X indicates the direction of the starting arm.
2. With centre A, draw an arc meeting AX at B.
3. Without changing the compass opening, draw an arc centred at B to meet the first arc at C.
4. Join AC. The lengths AB, AC and BC are equal, so the triangle ABC is equilateral and ∠CAX is 60°.

Worked example 4. Construct a 60° angle at A on AX.

Answer: Mark B on AX with an arc centred at A. With centre B and the same radius, cut the first arc at C. Join AC. Since AB = AC = BC, the triangle is equilateral and ∠CAX = 60°.

To obtain 120°, extend the starting line in the opposite direction. The neighbouring angle on the straight line is 180° − 60° = 120°, where − means subtraction. These two angles together form the straight angle.

What does repeated bisection produce?

Worked example 5. Construct 30° and 15° angles using angle bisection.

Answer: First construct 60° from an equilateral triangle. Bisect this angle to obtain 60° ÷ 2 = 30°. Bisect one of the resulting 30° angles to obtain 30° ÷ 2 = 15°. Each bisection uses equal-distance points and equal-radius intersecting arcs.

The sequence matters: 30° comes from halving 60°, and 15° comes from halving 30°. At each stage, keep the construction arcs until the new ray is located. Their intersections provide the geometric reason that the two smaller angles are equal.

Why do six equilateral triangles form a regular hexagon?

A polygon is a closed figure made of straight sides. A regular polygon has equal sides and equal angles. A hexagon has six sides, so a regular hexagon must satisfy both these equality conditions.

Place six congruent equilateral triangles with one vertex of each at the same point O. Each contributes a 60° angle at O. Their angles total 360°, allowing them to fit around O without gaps or overlaps.

Property: A complete angle checks the fit around a point

Angles whose measures add to 360° can be arranged at one point to cover the region around it without overlapping. In the hexagon construction, the equal outer sides come from congruent triangles. Each outer interior angle combines two 60° angles.

Worked example 6. Find the angles formed by six congruent equilateral triangles arranged around one centre.

Answer: Their central angles total 6 × 60° = 360°, where × means multiplication. Each interior angle of the resulting hexagon is 60° + 60° = 120°, where + means addition. The equal outer sides and equal interior angles make the hexagon regular.

The six outer vertices are A, B, C, D, E and F in order around O. Segments joining opposite vertices divide the figure into six triangles meeting at O.

See Fig. 6.12 in your NCERT textbook

For a regular hexagon of side 4 cm, where cm means centimetres, construct the six congruent equilateral triangles with sides of that length. Joining their outer edges gives the required boundary. Three successive 60° angles form 180°, explaining the straight lines through opposite vertices.

Worked example 7. Angles of 40°, 60°, 50°, 30°, 40° and 90° surround a point, leaving one gap. Will a 70° angle fit exactly?

Answer: The given angles total 310°. The gap is 360° − 310° = 50°. A 70° angle is larger than this gap, so it will not fit without overlapping an occupied part.

The same hexagon relationships support a six-pointed star. Its construction asks us to examine the hexagon within the design and the angles of the triangles forming its six points.

What are tangrams and what counts as tiling?

A tangram is a puzzle using seven pieces obtained by dividing a square. Tangrams originated in China. Rearranging the pieces produces different figures, allowing the same collection of shapes to form different outlines.

The pieces can be cut from cardboard using the square's division as a guide. Keep their shapes unchanged while trying arrangements. The challenge is to place the pieces together to match a target outline, rather than redraw or resize the pieces.

What conditions must a tiling satisfy?

Definition: Tiling means covering a region with a set of shapes without leaving gaps and without overlaps.

A gap is an uncovered part of the intended region. An overlap occurs when pieces cover the same part of the region. A valid tiling must avoid both: complete coverage is insufficient if some pieces lie over one another.

A tangram arrangement directs attention to how boundaries fit. Rectangular-grid problems apply the same idea to repeated copies of a tile. In both settings, the pieces must cover exactly the intended region.

How should a tiling attempt be checked?

Inspect the outer boundary and then the joins between pieces. A matching outer outline does not by itself establish that every interior part is covered correctly. Check that neighbouring pieces meet along their boundaries, with neither an empty space nor a shared covered area.

Two kinds of reasoning help: constructing an arrangement shows that a tiling is possible; finding an obstruction shows that it is impossible. Counting squares and colouring them provide such obstructions for the rectangular tiles considered next.

Which rectangular grids can be covered with two-square tiles?

A unit square is a square with side length one chosen unit. A rectangular grid arranges unit squares in rows and columns. The notation 4 × 6 describes four rows and six columns; here × separates the two grid dimensions.

A 2 × 1 tile is a rectangle covering two neighbouring unit squares. It may be rotated and used vertically or horizontally. Rotation changes its direction without changing its shape or the number of squares it covers.

Worked example 8. Can a 4 × 6 grid be tiled using 2 × 1 tiles?

Answer: Yes. Place vertical tiles in each column, pairing the four rows. Each column needs two tiles, so the six columns need 6 × 2 = 12 tiles. They cover all 24 unit squares without gaps or overlaps.

How does an even dimension provide a strategy?

For an m × n grid, let m be the number of rows and n the number of columns. An even number is divisible by two; an odd number leaves one unpaired when grouped in twos. If m is even, tile each column vertically.

If n is even, tile each row horizontally. Thus a complete rectangular grid can be tiled when at least one dimension is even. If both dimensions are odd, the number of unit squares is odd and cannot be covered in pairs.

Worked example 9. Compare tiling a 4 × 7 grid and a 5 × 7 grid using 2 × 1 tiles.

Answer: The 4 × 7 grid can be tiled vertically because its four rows can be paired. The 5 × 7 grid has 35 squares. Every tile covers exactly two squares, so complete coverage would require an even number of squares. Hence the second grid cannot be tiled.

Why can an even-sized region still be impossible to tile?

An even number of squares is necessary for tiling with 2 × 1 tiles, but it does not settle every problem. A region with a square removed is no longer a complete rectangle, so the simple row-pairing or column-pairing strategy may fail.

How does black-and-white colouring help?

Colour the grid alternately black and white, so squares sharing an edge have different colours. A 2 × 1 tile then covers exactly one black square and one white square, whether placed horizontally or vertically.

This creates a stronger counting test. A tileable region must have equal numbers of black and white squares. Adding another tile covers one of each colour, so no arrangement of these tiles can correct an imbalance between the two colour counts.

A five-row, three-column region has its top-middle square removed. The second drawing colours the remaining squares alternately black and white and shows horizontal and vertical tiles, each containing both colours.

See Figs. 6.13 and 6.14 in your NCERT textbook

Worked example 10. A region has 8 white squares and 6 black squares under alternating colouring. Each allowed 2 × 1 tile covers one square of each colour. Can the region be tiled?

Answer: No. The total of 14 squares is even, but the colour counts differ. Covering the eight white squares would require eight tiles and therefore eight black squares. Only six black squares are available, so complete tiling is impossible.

Colouring does not change which squares touch or how a tile fits. Any tiling of the plain grid can be coloured to match the grid; removing the colours from a coloured tiling gives a plain tiling. Thus the obstruction also applies to the original uncoloured region.

Use the tests in order: first count all the squares, then examine the colour counts if the total is even. Equal colour counts pass this particular test; the argument alone does not supply a tiling arrangement.

How can shapes tile the whole plane?

Tiling an entire plane extends the covering beyond a fixed boundary. Copies of squares can be placed in continuing rows and columns. Equilateral triangles and regular hexagons can also tile the plane without gaps or overlaps.

Which regular polygons are illustrated?

A plane can also be tiled using more than one shape or using polygons that are not regular. The Dutch artist Escher explored mathematical themes including tiling and created tilings with animal shapes. A tile need not resemble a familiar regular polygon.

The hexagon construction connects local angle-fitting to a larger pattern. Checking how shapes meet around a point is useful, but a plane tiling must continue across the whole surface without introducing gaps or overlaps elsewhere.

Where are tilings seen outside a drawing?

Tilings are often used in buildings and designs. They also occur in nature. The front face of bee hives and some wasp nests are tiled using hexagonal cells. These cells protect eggs, larvae (an immature insect stage) and pupae (a later developing stage), and also store food.

Because the region is tiled, no space is wasted between the cells. Scientists still wonder how bees and wasps are able to make hexagonal cells.

Tiling remains an active area of geometry. The same questions run through the topic: which shapes fit, what relationships make them fit, and what evidence proves that a proposed covering is possible or impossible?

Glossary

• Arc — A part of a circle drawn at a fixed distance from its centre.
• Radius — The distance from a circle's centre to any point on the circle.
• Supporting line — A line guiding a construction that need not appear in the finished design.
• Bisection — Division of a line segment or geometrical quantity into two identical parts.
• Midpoint — The point on a line segment that divides it into two equal lengths.
• Perpendicular bisector — A line passing through a segment's midpoint and meeting it at a right angle.
• Congruent triangles — Triangles of the same shape and size, with equal corresponding sides and angles.
• Angle bisector — A ray dividing a given angle into two angles of equal measure.
• Transversal — A line that crosses other lines and forms angles at the intersections.
• Regular polygon — A closed straight-sided figure whose sides and interior angles are all equal.
• Equilateral triangle — A triangle with three equal sides and three angles of 60° each.
• Tangram — A puzzle using seven pieces obtained by dividing a square and rearranging them.
• Tiling — Covering a region with a set of shapes without any gaps or overlaps.
• Unit square — A square with each side equal to one chosen unit of length.

Common errors and misconceptions

• Misconception: Any line through a segment's midpoint is its perpendicular bisector. Correct: It must also meet the segment at a right angle.
• Misconception: The two arcs in an intersecting pair can use unrelated radii. Correct: Equal radii guarantee that their intersection is equally distant from the endpoints.
• Misconception: A right angle at a known midpoint requires intersections on both sides. Correct: The midpoint already provides one point on the perpendicular bisector.
• Misconception: Copying an angle requires a measurement in degrees. Correct: Equal-radius arcs and a transferred length establish congruent triangles and equal angles.
• Misconception: Equal sides alone establish that a polygon is regular. Correct: A regular polygon must have equal angles as well as equal sides.
• Misconception: Every region containing an even number of unit squares can be tiled with 2 × 1 tiles. Correct: Unequal black-and-white counts can make such a region impossible to tile.
• Misconception: Rotating a 2 × 1 tile avoids the colouring obstruction. Correct: A horizontal or vertical tile covers one square of each colour.

Exam-style questions with model answers

Q1. What two conditions make a line the perpendicular bisector of a segment? [2 marks]

1. The line passes through the segment's midpoint, dividing its length into two equal parts.
2. It meets the segment at a right angle of 90°.

Q2. A design has eight equal angles meeting around its centre. Find each angle and explain how to construct it from a right angle. [2 marks]

1. A complete angle is 360°, so each of the eight equal angles is 360° ÷ 8 = 45°.
2. Construct a right angle of 90° and bisect it using equal-radius intersecting arcs to obtain 45°.

Q3. Point O lies on a given straight line. Describe a ruler-and-compass construction of a perpendicular through O, explaining why only one pair of intersecting arcs is needed. [4 marks]

1. With centre O and a fixed compass opening, mark X and Y on opposite sides of O along the given line.
2. These points satisfy OX = OY, so O is the midpoint of segment XY.
3. Draw equal-radius arcs from X and Y to meet at A away from the line. Join OA to obtain the perpendicular bisector of XY.
4. Only one pair is needed because O already lies on the perpendicular bisector. A supplies the second point determining that line.

Q4. Six congruent equilateral triangles, each with angles of 60°, are placed with one vertex at a common centre. Explain why their outer boundary is a regular hexagon. [3 marks]

1. The angles at the centre total 6 × 60° = 360°, so the six triangles fit around that point without a gap or overlap.
2. The outer boundary has six equal sides because these are sides of congruent equilateral triangles.
3. Each interior angle of the boundary is formed by two triangle angles: 60° + 60° = 120°. Equal sides and equal angles make the hexagon regular.

Q5. Around a point, adjacent non-overlapping angles measure 40°, 60°, 50°, 30°, 40° and 90°, leaving one gap. Find the gap and decide whether a 70° angle fits it exactly. [3 marks]

1. Add the occupied angles: 40° + 60° + 50° + 30° + 40° + 90° = 310°. This is the part of the complete angle already covered.
2. The full angle around a point is 360°, so the remaining gap measures 360° − 310° = 50°.
3. A 70° angle is larger than the 50° gap. It cannot fit exactly and would overlap an already occupied part of the region.

Q6. A region has 8 white and 6 black unit squares in an alternating colouring. Every allowed 2 × 1 tile covers one square of each colour. Explain why the region cannot be tiled, although its total square count is even. [5 marks]

1. The region contains 8 + 6 = 14 unit squares. Its total is even, so counting all the squares alone does not rule out a tiling.
2. Each allowed tile covers exactly one black square and one white square. Therefore every placed tile uses equal numbers of the two colours.
3. Any complete tiling by these tiles must consequently cover equal numbers of black and white squares throughout the region.
4. Here the counts are unequal: there are eight white squares but only six black squares. Covering all eight white squares would also require eight black squares.
5. The required black squares are unavailable, so the region cannot be tiled. An even total is necessary but is insufficient for this region.

Q7. Let X and Y be endpoints of a segment. Points A and B are on opposite sides with AX = AY = BX = BY. Their joining line meets XY at O between A and B. Prove that AB is the perpendicular bisector of XY using triangle congruence. [6 marks]

1. Compare triangles ABX and ABY. The given equalities supply AX = AY and BX = BY, while AB is their common third side.
2. The triangles are congruent by SSS, the side-side-side condition. Hence the corresponding angles XAB and YAB are equal.
3. Since O lies on AB between A and B, these equal angles give ∠XAO = ∠YAO for the smaller triangles.
4. Triangles AOX and AOY have AX = AY, common side AO and equal included angles. They are congruent by SAS, the side-angle-side condition.
5. Their corresponding sides give OX = OY. Their corresponding angles at O are also equal and together form the straight angle of 180°.
6. Each angle at O is therefore 90°. Thus AB passes through the midpoint of XY and is perpendicular to it, establishing the required perpendicular bisector.

Q8. Compare a complete rectangular grid of 4 rows and 7 columns with one of 5 rows and 7 columns. Decide whether each can be tiled by rotatable 2 × 1 tiles, and justify your conclusions. [3 marks]

1. The grid with four rows can be tiled by placing vertical tiles in each column. The rows pair up without leaving an uncovered square.
2. The grid with five rows and seven columns contains 5 × 7 = 35 unit squares, which is an odd number.
3. Each tile covers exactly two unit squares, so any complete tiling covers an even number. Therefore the second grid cannot be tiled, regardless of tile orientation.

Key takeaways

• A perpendicular bisector divides a segment equally and meets it at a right angle; both conditions are required.
• Equal-radius intersecting arcs locate points equally distant from two endpoints, providing a compass construction of their perpendicular bisector.
• Triangle congruence explains why angle bisection and angle copying work, rather than relying on a figure's appearance.
• Constructing a 60° angle uses an equilateral triangle; successive bisections then produce 30° and 15° angles.
• Six congruent equilateral triangles fit around a point and form a regular hexagon with interior angles of 120°.
• A complete rectangular grid can be tiled by two-square tiles when at least one of its dimensions is even.
• Alternating colouring can prove impossibility because each two-square tile covers exactly one black square and one white square.
• Squares, equilateral triangles and regular hexagons tile the plane; tilings can also use other shapes or combinations of shapes.

Test yourself

What is the difference between a midpoint and a perpendicular bisector?

A midpoint is a point dividing a segment equally. Its perpendicular bisector is a line through that point meeting the segment at 90°.

Why must intersecting arcs from two endpoints use equal radii in the standard construction?

Equal radii make their intersection equally distant from both endpoints, placing it on the perpendicular bisector.

Which equal lengths justify bisecting ∠AOB with OC?

Choose OA = OB and AC = BC. Together with common side OC, these give congruent triangles OAC and OBC by SSS.

How is a 15° angle obtained from a constructed 60° angle?

Bisect the 60° angle to obtain 30°, then bisect the 30° angle to obtain 15°.

Why do six 60° angles fit around one point?

Their sum is 360°, exactly the complete angle around a point, so they leave no angular gap.

How many pieces are used in a tangram, and where did tangrams originate?

A tangram uses seven pieces obtained by dividing a square. Tangram puzzles originated in China.

Why can rotating a two-square tile not fix unequal colour counts?

Under alternating colouring, both horizontal and vertical placements cover one black square and one white square.

Name three regular polygons whose copies can tile the plane.

Squares, equilateral triangles and regular hexagons can each tile the plane without gaps or overlaps.