Constructions & Tilings
— every in-text & exercise question, solved.
Straightedge-and-compass constructions, angle bisection, regular hexagons, arches and plane tiling — worked step-by-step with labelled diagrams, so you can follow the exact same lines a compass would draw.
In-Text Questions
Every guided question ( ? ) that appears alongside the reading — the small reasoning checks that build up each construction.
6.1 — The “Eyes” Construction & Perpendicular Bisector
pg 136–139- Draw the support line XY.
- Open the compass to any radius. With X as centre, draw an arc above XY, then with the same radius and Y as centre, draw a second arc above XY. They meet at point A.
- Keeping the same or a different (but again equal) radius, repeat the process below XY to get point B.
AB passes exactly through the midpoint O of XY, and it crosses XY at a 90° angle. In other words, AB is perpendicular to XY and bisects it — this is why line AB is called the perpendicular bisector of XY.
Yes — always. This is true regardless of the length of XY or the radius chosen, and it is proved rigorously using triangle congruence in the next question.
We are given AX = AY = BX = BY. Let O be the point where AB meets XY.
- Show ΔABX ≅ ΔABY — by SSS: AX = AY, BX = BY, and AB is common. This gives ∠XAB = ∠YAB, i.e. ∠XAO = ∠YAO.
- Show ΔAOX ≅ ΔAOY — by SAS: AX = AY (given), ∠XAO = ∠YAO (from step 1), and AO is common.
- By CPCT, OX = OY (so O is the midpoint) and ∠AOX = ∠AOY.
- Since ∠AOX and ∠AOY together form a straight angle on line XY, ∠AOX + ∠AOY = 180°. Combined with ∠AOX = ∠AOY, each equals 90°.
Choose a different pair of points C and D such that CX = CY = DX = DY (using a different radius than the one used for A and B). Drawing the eye’s arcs from C and D instead of A and B produces an eye of a different curvature/shape — flatter or more pointed, depending on the radius chosen.
Yes. C and D are each equidistant from X and Y. We already know that any point equidistant from X and Y lies on the unique perpendicular bisector of XY — and since XY has only one perpendicular bisector (line AB), both C and D must lie on line AB.
Let P be any point with PX = PY, and let O be the already-established midpoint of XY (so OX = OY).
- In ΔPXO and ΔPYO: PX = PY (given), OX = OY (O is midpoint), and PO is common.
- By SSS, ΔPXO ≅ ΔPYO, so ∠POX = ∠POY (CPCT).
- Since ∠POX and ∠POY form a straight angle, each must equal 90°. So PO ⊥ XY.
- Open the compass to a radius more than half of XY.
- With X as centre, draw arcs both above and below XY.
- Keeping the same radius, with Y as centre, draw arcs above and below XY — cutting the previous arcs at A (above) and B (below).
- Draw line AB with the ruler. AB is the required perpendicular bisector, crossing XY at its midpoint.
Construction of a 90° Angle at a Given Point
pg 141Yes, and it needs only one pair of arcs, because O is already a known point on the perpendicular bisector we’re about to draw.
- Draw the given line l and mark point O on it.
- With O as centre, and any convenient radius, draw arcs cutting line l at X and Y on either side of O — this makes O automatically the midpoint of XY.
- With the same (or any equal) radius, draw arcs from X and from Y that meet at a point A (on one side of the line).
- Draw OA. Since O is already the midpoint of XY, joining O to any single point A that is equidistant from X and Y gives the perpendicular bisector — so OA ⊥ l, i.e. ∠AOY = 90°.
Angle Bisection for a Design
pg 143–144The figure needs 8 rays from a common centre, each pair of adjacent rays equally spaced. Since a full turn is 360°, and we need it divided into 8 equal parts:
- Construct a 90° angle at the centre (as shown above) — this gives 4 rays, 90° apart.
- Bisect each of the four 90° angles (method below) to get 8 rays, 45° apart.
- On each pair of adjacent rays, draw a symmetric “petal” using two intersecting arcs — exactly like the eye construction.
Construct a 90° angle first (method shown earlier), then bisect it using the angle-bisection method (below). Half of 90° is 45°.
- With O as centre and any radius, draw an arc cutting the two arms of ∠XOY at points A and B, so OA = OB.
- With A and B as centres (same radius for both, sufficiently long), draw two arcs that intersect at point C.
- Draw ray OC. OC bisects ∠XOY.
- Given ∠A, and a ray starting at point X where the copy is needed.
- With A as centre, any radius, draw an arc cutting both arms of ∠A at B and C.
- With the same radius, and X as centre, draw a similar arc cutting the given ray at Z.
- Measure the distance BC with the compass (without changing its width).
- With Z as centre and this width (BC), draw an arc cutting the previous arc (from X) at Y.
- Draw ray XY. Then ∠YXZ = ∠BAC.
Construction of a Line Parallel to a Given Line
pg 147- Draw any transversal line l that cuts m at point A, forming an angle a between l and m.
- Choose the point B on l through which the parallel line must pass.
- Copy angle a at point B (using the angle-copy method above), on the same side, so it becomes the corresponding angle.
- Draw the line through B along the new arm. This line n is parallel to m.
Arch Designs — Trefoil & Pointed Arch
pg 149–150- Draw the base line AD.
- At A and at D, construct equal angles opening upward (using the angle-copy method, so ∠BAD = ∠CDA automatically).
- On these two new arms, mark points B and C with the same compass width, so AB = CD.
Yes. This uses the same idea as the “Wavy Wave” pattern from Grade 6: each curved side of the arch is an arc whose centre is the far (bottom) endpoint of the opposite support segment, with radius equal to the segment’s own length.
Regular Hexagons
pg 151–153Yes to both. A regular hexagon can be split into 6 equilateral triangles that share a common central vertex O.
- Each angle of an equilateral triangle is 60°. Placed edge-to-edge around a point, six of them give a total angle of 6 \times 60° = 360° — exactly filling the space around O, with no gaps and no overlaps.
- Since all six triangles are congruent (same side length s), every outer side of the resulting hexagon is equal to s.
- Each hexagon vertex angle is made of two adjacent 60° triangle-angles: 60° + 60° = 120° — so all interior angles are equal too.
All angles around a point add up to 360°:
310° + \text{gap} = 360°
\text{gap} = 50°
Since the actual gap angle ∠AOI is 50°, a 70° angle is too big — it will not fit into the gap; a triangle with a 70° vertex angle would overlap into the neighbouring shapes by 20°.
The hexagon is made of six 60° equilateral triangles: ΔOAB, ΔOBC, ΔOCD, ΔODE, ΔOEF, ΔOFA, all meeting at centre O.
Since ∠AOD = 180°, A, O and D lie on one straight line. The exact same reasoning (adding three consecutive 60° triangle-angles) shows \angle BOE = 180° and \angle COF = 180° — so BOE and COF are straight lines too.
- Draw ray AX.
- With A as centre, any radius, draw an arc cutting AX at B.
- With the same radius, and B as centre, draw an arc cutting the first arc at C.
- Join AC. Then \angle CAX = 60°.
To build the full hexagon: at each new vertex, construct a fresh 120° interior angle (or copy the 60°/120° angle using the angle-copy method) and mark off the same side-length — repeating this six times brings you back to the start, since 6 \times 60° = 360° of exterior turning.
Yes. In the construction, AB = AC (both are the same radius drawn from A) and BC = AB (the arc from B was drawn with the same radius, so it also equals the radius). Hence AB = BC = CA — triangle ABC is equilateral, so all its interior angles, including ∠CAX = ∠CAB, are exactly 60°.
- Construct a 60° angle (equilateral-triangle method above).
- Bisect it (angle-bisection method) to get \dfrac{60°}{2}=30°.
- Bisect the 30° angle again to get \dfrac{30°}{2}=15°.
The 6-pointed star (hexagram) is built from a regular hexagon: mark six points around a centre, each 60° apart (using the 60°-angle construction repeatedly, or by stepping the compass — same radius — six times around a circle). Joining alternate points forms two overlapping equilateral triangles, whose overlap creates the inner hexagon and whose “points” stick out as six small triangles.
6.2 — Tiling: Reasoning Questions
pg 157–1604×6 grid (24 squares): Yes — simply place vertical 2×1 tiles down each of the 6 columns (4 rows = 2 tiles per column).
4×7 grid (28 squares): Yes — tile the first 6 columns as above with vertical dominoes; for the last column (4 rows, 1 column), stack 2 more vertical dominoes. Since the number of rows (4) is even, this works for any number of columns.
5×7 grid (35 squares): Not possible. Each 2×1 tile always covers exactly 2 unit squares, so n tiles cover 2n squares — always an even number. But 5 \times 7 = 35 is odd, so it can never equal 2n for a whole number n.
(a) Both even — Tileable. Since the number of rows m is even, every column of height m can be exactly covered by m/2 vertical dominoes stacked on top of each other. Repeat for all n columns.
(b) One even, one odd — Tileable. Say m (rows) is even and n (columns) is odd. The same column-by-column strategy still works: each of the n columns (whatever n is) gets covered by m/2 vertical dominoes, since only the row-count needs to be even for this method.
(c) Both odd — Not tileable. Total squares = m \times n = \text{odd}\times\text{odd} = \text{odd}. Since every tile covers 2 squares, the total covered area by any number of tiles is always even. An odd total area can never be fully covered.
Having an even number of squares is necessary but not sufficient. It depends on which square is removed / which shape results — some even-area regions genuinely cannot be tiled. We test this properly using a checkerboard colouring argument (next question).
Colour and tileability go together in both directions — if either version is tileable, so is the other, because the colouring doesn’t change the shape, only labels each square. A valid tiling of one is automatically a valid tiling of the other.
Is Fig. 6.14’s black-and-white region tileable? Every 2×1 domino, once placed on a checkerboard-coloured grid, covers exactly one black and one white square. So any tileable region must have an equal number of black and white squares.
A 5×3 grid has 15 squares. In a standard checkerboard colouring where both dimensions are odd, the four corners share the same colour — giving 8 squares of that colour and 7 of the other.
- If you remove a square of the majority colour (8→7), you’re left with 7 and 7 — equal counts, so the region may be tileable (needs to be checked further).
- If you remove a square of the minority colour (7→6), you’re left with 8 and 6 — unequal — so it is guaranteed non-tileable, exactly as with Fig. 6.13.
Squares clearly tile the plane. Equilateral triangles also work — six of them meet at every vertex: 6 \times 60° = 360°. Regular hexagons work too — three meet at every vertex: 3 \times 120° = 360°.
“By introducing colours and making the problem more complicated, it becomes easier to tackle.” — that’s the spirit of this whole chapter.
Exercise Questions — Figure It Out
All the numbered “Figure it Out” practice sets, in the order they appear across the chapter.
Figure It Out — Perpendicular Bisector
pg 140No, it is not necessary. The radius used above XY (to locate A) and the radius used below XY (to locate B) can be different from each other.
Geometrically, it still works even if both points are found on the same side of XY, provided the two radii used are different enough to give two distinguishable points, both equidistant from X and Y.
Yes, this is essential. To locate point A such that AX = AY, the arc drawn from X (say radius r_1) and the arc drawn from Y (radius r_2) must intersect at a point where both distances match.
- Construct two perpendicular lines through a common centre O (using the 90°-angle construction) — this gives 4 rays, 90° apart.
- Treat each pair of adjacent rays as a support “XY” line for one petal.
- On each pair, draw two symmetric intersecting arcs between the ray-tips — exactly like the eye construction — to form each petal.
Figure It Out — Śulba-Sūtra Rope Method
pg 142The rope is folded exactly in half to find its midpoint, so the two halves are always equal in length — call this length k.
- When the loops are fixed at X and Y and the midpoint is pulled taut above XY to point A: since both halves of the rope are equal, AX = AY = k.
- Similarly, pulling the midpoint taut below XY to point B gives BX = BY = k.
- Fix a peg at O on the line. Using the rope as a fixed measuring length, mark two points X and Y on the line, on either side of O, at equal rope-lengths — so O becomes the midpoint of XY.
- Using the same folded-rope method as before, find point A equidistant from X and Y by pulling the rope’s midpoint taut above the line.
- Join O to A — OA is perpendicular to the line at O.
Figure It Out — Angle Bisection
pg 144Draw any four angles of different sizes and orientations. For each one, apply the standard bisection method:
- From the vertex O, draw an arc cutting both arms at equal distance — points A and B.
- From A and B (equal radius), draw two arcs meeting at C.
- Draw OC — this is the bisector, since ΔOAC ≅ ΔOBC (SSS) gives ∠AOC = ∠BOC.
As shown earlier: build 4 perpendicular rays (90° apart), bisect each right angle to get 8 rays at 45° spacing, then draw a symmetric petal (two intersecting arcs) between every adjacent pair of rays.
Yes — the resulting line is still the true bisector line, just extended in the opposite direction through O.
Starting from the two angles we can build directly — 90° (perpendicular method) and 60° (equilateral triangle) — and combining them with addition, subtraction and repeated bisection, we can reach a large family of angles, all multiples of 7.5°:
From 60°: 30°, 15°, 7.5° …
By combining (e.g. 90° − 60° = 30°, 90° + 60° = 150°, (90°+60°)/2 = 75°): 15°, 30°, 45°, 60°, 75°, 90°, 105°, 120°, 135°, 150°, 165°, 180° — and all of their repeated bisections.
65.5° cannot be constructed this way. It is not a multiple of 7.5° (65.5 \div 7.5 is not a whole number), so it can never be reached by bisecting or combining 90° and 60° with a ruler and compass.
- Using the rope as a measuring tool, mark points A and B on the two arms of the angle, at equal rope-length from the vertex O, so OA=OB.
- Take a second rope, fold it exactly in half, and fasten its two ends (loops) at A and B.
- Pull the midpoint of this second rope taut (away from O) to a point C — since the rope is folded exactly in half, AC = BC.
- Join O to C. By the same SSS congruence logic (ΔOAC ≅ ΔOBC), OC is the angle bisector.
- Draw the given square (side length s).
- With each corner of the square as centre, and radius equal to the full side length s, draw a quarter-circle arc joining the two adjacent corners.
- Repeat this from all 4 corners. The four overlapping quarter-circle arcs intersect each other at the square’s centre and along its sides, carving out the 4-petal shape.
Figure It Out — Copying Angles
pg 147For each angle drawn, apply the angle-copy method: draw an arc from the vertex cutting both arms (radius r); draw an equal arc from the new vertex; transfer the chord-length between the two arm-intersections onto the new arc using the compass; join the new vertex to that transferred point.
This pattern repeats a single “fan” unit (two arms with an angle between them, same arm length) in two orientations along a line. Construct one fan unit; then use the angle-copy method to reproduce the same angle (flipped) at the next point along the base line, keeping arm lengths equal throughout with the compass, so every copy is identical.
Figure It Out — Parallel Lines
pg 148For each pair: draw a line m, a transversal l through it at some point A (any angle), pick a second point B on l, and copy the angle at A onto B (corresponding position) to fix the direction of the new line n through B. Then m \parallel n. Repeat with different transversal angles/orientations for the remaining 3 pairs.
- Divide the full 360° around a centre into 12 equal 30° sectors — this can be done by first getting 60° angles (equilateral triangle method) and then bisecting each once (60° ÷ 2 = 30°).
- Draw 12 rays from the centre, 30° apart, alternating a “long” outer point and a “short” inner vertex along the star’s edge.
- Join consecutive outer/inner points in order (S→B→T→C… as shown), then shade alternate triangular regions to complete the star pattern.
Figure It Out — Pointed Arch
pg 151- Draw two equal-length line segments meeting at a top apex point, spread symmetrically (like an inverted “V”).
- From the bottom end of the right segment, draw an arc of radius = segment length, curving from the apex down to the base — this forms the left curved side of the arch.
- Symmetrically, from the bottom end of the left segment, draw the matching arc for the right side.
- Try radii slightly larger or smaller than the segment length itself — this changes how “pointed” or “rounded” the arch looks, while keeping it symmetric.
Open-ended creative task. A reliable recipe: start with two equal, symmetric support segments (as above) or an equal-angle base (as in the trefoil arch); then build the curved outline using arcs centred at the segment endpoints or midpoints, keeping the left and right halves mirror-symmetric by using matching radii on both sides.
Figure It Out — Related Constructions
pg 154–155(a) Inflexed arc: Draw two equal vertical support segments; connect their tops with an arc that curves inward (concave), centred at a point below the segments, so the curve dips toward the centre instead of bulging outward.
(b) Compass-only flower: Draw a circle of radius r centred at O. Choose any point A on it. With the same radius r and centre A, draw a new circle (its edge automatically passes through O). Repeat, stepping around the original circle — each new circle centred where the previous one crosses it — until 6 outer circles surround the centre one, forming a 6-petal flower, drawn with compass only, no ruler needed.
(c) Hexagon in a circle: Draw a circle of radius r. Starting at any point on it, step the same radius r around the circle six times (each new arc centred at the previous point) to mark 6 vertices; join them in order.
(d) Six circles in a ring: Same stepping method as (b), but only draw the 6 outer circles (omit the central one) — each centred on the main circle’s edge, radius equal to the main circle’s radius.
(e) Hexagon with star-triangle fill: Construct the regular hexagon (60°/120° method), then draw all three main diagonals through the centre and join alternate vertices to overlay the triangular star pattern inside it.
This is a classic Kanizsa triangle illusion — although no actual triangle is drawn anywhere in the figure, our eyes clearly “see” a bright white triangle floating in front of the other shapes.
To recreate: Place three circles (each with a wedge cut out, facing the centre) at the corners of an imaginary triangle, plus three bent (“<“) line-corners positioned between them to match the triangle’s other edges.
First construct a regular hexagon (each interior angle = 120°, using the 60°/120° method). Then construct an equilateral triangle (all 60° angles) pointing outward from every alternate side of the hexagon, using the angle-copy method to keep each triangle identical. The combination of the hexagon’s 120° angles and the star-points’ 60° angles produces the full star pattern.
- With P as centre, draw a single arc of any convenient radius that cuts line l at two points X and Y.
- Since both X and Y lie on the same arc from P, PX = PY automatically — so P is equidistant from X and Y.
- P must therefore lie on the perpendicular bisector of XY. Construct that perpendicular bisector using the standard method (equal-radius arcs from X and Y meeting at a second point, say Q).
- Draw line PQ — this passes through P and is perpendicular to l.
Figure It Out — Tangram
pg 155–156Tangram figures are solved by spatial trial, not calculation — but the same method works for every shape:
- Study the outline and mentally break it into triangular/square/parallelogram chunks matching the 7 standard pieces: 2 large right-triangles, 1 medium right-triangle, 2 small right-triangles, 1 square, 1 parallelogram.
- Start by placing the two large triangles first — they cover the most area and usually form the “body” or largest region of the figure.
- Use the square and medium triangle for the next-largest recognisable block (e.g. a head, a wing, a torso).
- Fit the two small triangles into pointed extremities (a beak, an ear, a foot, a corner).
- The parallelogram is the trickiest piece — remember it can be flipped over (mirrored), not just rotated, which is often the key to fitting tails, arms, or slanted limbs.
- Match edge lengths carefully — every tangram piece’s edges are built from the same unit length or its √2 multiple, so aligned edges always fit exactly if the outline has been read correctly.
Figure It Out — Tiling Regions
pg 160- Count the total unit squares in the region.
- Since each tile covers 2 squares, check whether the total is even — this is the first necessary test.
- If even, try to actually place tiles working from a corner inward, or apply a checkerboard-colour count to confirm equal black/white coverage before concluding “yes”.
Apply the same two-step test used throughout this section:
- Parity check: count the squares in the region — the total must be even.
- Colour check: colour the region like a checkerboard and count black vs white squares — they must be exactly equal, since every 2×1 tile always covers one of each colour.
- If both checks pass, construct an explicit tiling (e.g. row-by-row with horizontal tiles, or column-by-column with vertical tiles) to confirm it is genuinely possible.