Any point equidistant from X and Y lies on the perpendicular bisector of XY — two such points give the bisector.
90° at a point: make the point the midpoint of a segment and draw its perpendicular bisector.
Angle bisector and copying an angle both use SSS congruence.
60° comes from an equilateral triangle; bisecting gives 30°, 15° …; a regular hexagon is six equilateral triangles.
Tiling: no gaps, no overlaps. A 2 × 1 tile covers one black and one white square of a chessboard colouring.
๐ Textbook Page 137
๐ Page 137Eyes
Construct an eye whose upper and lower arcs are symmetrical about the line XY.
1Draw XY. With the same radius (more than half of XY), draw arcs from X and from Y above and below XY.
2The arcs meet at A (above) and B (below): AX = AY = BX = BY.
3With centre A and radius AX draw the lower arc from X to Y; with centre B and radius BX draw the upper arc.
4Add a circle for the eyeball at the middle of XY.
Arcs of the SAME radius from X and Y meet at A and B; arcs centred at A and B make the eye.
✅ Answer: See the construction
๐ Page 137Where does AB meet XY?
In Fig. 6.1, join A and B. Where does AB meet XY, and at what angle?
1Measure: AB passes through the midpoint O of XY.
2∠AOX = 90°.
AB bisects XY at right angles.
✅ Answer: At the midpoint of XY, at 90° — AB is the perpendicular bisector
๐ Textbook Page 138
๐ Page 138Why always?
Will the line joining the two meeting points always be the perpendicular bisector of XY, whatever the radius? Which triangles must be congruent?
1AX = AY, BX = BY, AB common → △ABX ≅ △ABY (SSS) → ∠XAO = ∠YAO.
2Then in △AOX and △AOY: AX = AY, ∠XAO = ∠YAO, AO common → congruent (SAS).
3So OX = OY and ∠AOX = ∠AOY. They add up to 180°, so each is 90°.
AX = AY = BX = BY. △ABX ≅ △ABY (SSS), then △AOX ≅ △AOY (SAS) → OX = OY and ∠AOX = 90°.
✅ Answer: Yes — by congruence △AOX ≅ △AOY
๐ Textbook Page 139
๐ Page 139Eyes of other shapes
If CX = CY = DX = DY for two other points C and D, will C and D lie on AB? Justify: any point equidistant from X and Y lies on the perpendicular bisector.
1Joining any two such points (one above, one below XY) gives the perpendicular bisector of XY.
2XY has only one perpendicular bisector, the line AB.
3So C and D lie on AB. In the same way any point P with PX = PY, joined with B, gives the same line — so P is on it.
✅ Answer: Yes — they lie on the perpendicular bisector AB
๐ Textbook Page 140
๐ Page 140Figure it Out · Q1
When constructing the perpendicular bisector, must the arcs above and below XY have the same radius?
1No. The point above is equidistant from X and Y, and so is the point below, even with a different radius.
2Both lie on the perpendicular bisector, and two points fix a line.
✅ Answer: No — only each pair of arcs needs one common radius
๐ Page 140Figure it Out · Q2
Can both pairs of arcs be drawn on the same side of XY?
1Yes: draw one pair with radius r₁ and another pair with a different radius r₂, both above XY.
2Each meeting point is equidistant from X and Y, so both lie on the bisector; join them and extend.
3The radii must be different, otherwise you get the same point twice.
✅ Answer: Yes — with two different radii
๐ Page 140Figure it Out · Q3
In one pair of intersecting arcs, must both arcs have the same radius?
1Yes. If the arc from X has radius 5 and the one from Y has radius 4, the meeting point is 5 from X and 4 from Y.
2It is not equidistant from X and Y, so it is NOT on the perpendicular bisector.
✅ Answer: Yes — otherwise the point is not equidistant from X and Y
๐ Page 140Figure it Out · Q4
Recreate this four-petal design using only a ruler and compass.
1Draw a line and construct its perpendicular at O.
2With the compass mark four equal lengths OP from O along the lines.
3For each segment OP, construct an ‘eye’: the perpendicular bisector of OP gives two centres at equal distances; draw the two arcs from O to P.
4Trace the petals in colour.
✅ Answer: See the construction
๐ Textbook Page 141
๐ Page 14190° at a point
Draw a line and mark a point O on it. Construct a 90° angle at O.
1With centre O, cut arcs on both sides: X and Y with OX = OY.
2With a bigger radius, draw arcs from X and from Y that meet at A.
3Join OA. ∠AOY = 90° (O is already on the bisector, so one pair of arcs is enough).
Mark X and Y with OX = OY, then cut arcs of equal radius from X and Y at A. OA ⊥ the line.
✅ Answer: OA ⊥ the line at O
๐ Textbook Page 142
๐ Page 142Figure it Out · Q1
Justify why AB in the rope construction (Fig. 6.4) is the perpendicular bisector of XY.
1The rope is folded at its midpoint, so its two halves are equal: XA = YA and XB = YB.
2A and B are both equidistant from X and Y.
3Any such points lie on the perpendicular bisector, so AB is the perpendicular bisector.
The rope’s two halves are equal: XA = YA and XB = YB, so A and B are equidistant from X and Y.
✅ Answer: A and B are equidistant from X and Y
๐ Page 142Figure it Out · Q2
Suggest other ways to make a 90° angle at a point on a line using a rope.
1Method 1: with the rope as a compass mark X and Y equally on both sides of O, then use the folded-rope method to find a point A with XA = YA; OA ⊥ line.
2Method 2 (ลulba-Sลซtra triangle): tie knots to divide a rope into 12 equal parts and make a triangle with sides 3, 4 and 5 parts — the angle between the 3 and 4 sides is 90°.
✅ Answer: e.g. equal marks + folded rope, or a 3-4-5 knotted rope triangle
๐ Textbook Page 143
๐ Page 143Angle between adjacent lines
In the 8-petal figure, what is the angle between two adjacent supporting lines? How can we get it with ruler and compass?
18 equal angles fill 360°: 360 ÷ 8 = 45°.
2Construct 90° and bisect it to get 45°.
Supporting lines at 45° and the petals.
✅ Answer: 45°
๐ Textbook Page 144
๐ Page 144Angle bisection
Explain the steps to bisect any angle XOY and why they work.
1Mark A on OY and B on OX with OA = OB (one arc from O).
2With the same radius (long enough), draw arcs from A and B meeting at C.
3Join OC.
4△OAC ≅ △OBC by SSS (OA = OB, AC = BC, OC common), so ∠AOC = ∠BOC.
OA = OB, AC = BC → △OAC ≅ △OBC (SSS), so OC bisects ∠XOY.
✅ Answer: OC bisects ∠XOY
๐ Page 144Figure it Out · Q1–Q2
Construct 4 different angles and their bisectors; construct the 8-petalled figure.
1Use the steps of angle bisection for each angle.
28 petals: draw two perpendicular lines, bisect the four right angles to get 8 lines at 45°, mark equal lengths, and draw an ‘eye’ (petal) on each.
✅ Answer: See the construction steps
๐ Page 144Figure it Out · Q3
If the arcs of equal radius from A and B are drawn on the other side (beyond O), is OC still the angle bisector?
1C is still equidistant from A and B, and so is O (OA = OB).
2So O and C both lie on the perpendicular bisector of AB, which is the bisector of the angle in isosceles △OAB.
3Extend CO beyond O: it bisects ∠XOY. (C must not coincide with O.)
Even with C on the other side of O, OA = OB and AC = BC, so C is equidistant from A and B and lies on the bisector line.
✅ Answer: Yes — the line CO (extended) bisects the angle
๐ Page 144Figure it Out · Q4
Which other angles can be constructed using angle bisection? Can you construct 65.5°?
2All these are made from 15° halved again and again (15, 7.5, 3.75 …).
365.5° = 131 ÷ 2, and 131 is not a multiple of 15 — no sum of such pieces gives 65.5° exactly.
✅ Answer: Many (45°, 30°, 15°, 22.5°, 67.5° …); 65.5° cannot be constructed exactly this way
๐ Page 144Figure it Out · Q5
Give a method to bisect an angle using a rope.
1Fix a peg at O. Using the rope as a compass, mark A and B on the two arms with OA = OB.
2Tie the rope’s ends at A and B and pull its midpoint out between the arms until both halves are tight: that point is C (AC = BC).
3OC is the bisector.
✅ Answer: Equal lengths OA = OB, then folded rope from A and B
๐ Textbook Page 145
๐ Page 145Figure it Out · Q6
Construct this figure: four petals in a square, each as large as possible.Each petal: arcs of radius half the side, centred at the two side-midpoints next to that corner.
1Draw the square and mark the midpoint of each side (perpendicular bisectors) and the centre O.
2Each petal joins O to a corner.
3For the top-right petal, draw arcs of radius half the side centred at the top midpoint and at the right midpoint, from O to the corner.
4Repeat for the other three corners. Using half the side as radius makes the petals touch the square — the largest possible.
✅ Answer: Arcs of radius half the side, centred at the side midpoints
๐ Page 145Repeating units
Construct Fig. 6.6 made of fans in two orientations.Copies of one fan (same radius, same angle) in two orientations.
1All fans need the same radius and the same angle (here 90°).
2Draw a line; at equal steps construct the fans with apex on the line (arms at 45° on each side of the vertical, using bisection).
3Upside-down fans: their apex is where two neighbouring arcs meet; their arms lie along the arms of the first fans; draw the arc below with the same radius.
✅ Answer: Copy the same angle and radius every time
๐ Textbook Page 146
๐ Page 146Copying an angle
Draw an angle A and copy it at X using only ruler and compass. Why does it work?
1Draw an arc with centre A cutting the arms at B and C.
2Draw a ray from X; with the same radius draw an arc from X cutting it at Z.
3Take BC in the compass; from Z cut the arc at Y. Join XY.
4AB = XY, AC = XZ, BC = YZ → △ABC ≅ △XYZ (SSS) → ∠X = ∠A.
Same radius from A and X; YZ = BC (compass). △ABC ≅ △XYZ (SSS) → ∠X = ∠A.
✅ Answer: ∠YXZ = ∠BAC
๐ Textbook Page 148
๐ Page 148Parallel line
Construct a line through B parallel to a line m, using a transversal l.
1Draw a transversal l through B meeting m at A.
2Draw arcs of equal radius from A (cutting m at C and l at D) and from B (cutting l at F, and the line to be drawn at E).
3Transfer CD from F to fix E, i.e. copy the angle at A to B. Join BE: line n.
4Corresponding angles are equal, so n ∥ m.
Copy the angle at A to B (corresponding angles equal) → n ∥ m.
✅ Answer: n ∥ m
๐ Page 148Figure it Out · Q2
Construct the 8-pointed star figure (points S, T, U, V, W, X, Y, Z).My drawing: 16 rays at 22.5° from the centre — get them by bisecting 90° twice.
1Draw 8 lines through the centre at 45° (90° bisected) — they give the inner points A–H at equal distance.
2Bisect each 45° to get rays at 22.5° for the outer points S–Z; mark them at a larger equal distance.
3Join each outer point to the centre and to its two neighbouring inner points; shade alternate triangles.
4Use the parallel-line method to check that opposite edges are parallel.
✅ Answer: Use 90° → 45° → 22.5° bisections
๐ Textbook Page 149
๐ Page 149Trefoil arch
Construct the trefoil arch: support lines AB = CD with ∠BAD = ∠CDA.
1Draw AD. Construct equal angles at A and D (copy the angle).
2Mark B and C with AB = CD (compass).
3Find the midpoints of AB, BC and CD (perpendicular bisectors) and draw semicircles outward on each.
4Add the side pillars from A and D.
Equal angles at A and D, AB = CD; semicircles on AB, BC and CD make the trefoil arch.
✅ Answer: See the construction
๐ Textbook Page 151
๐ Page 151Pointed arch
Use two equal support lines to construct a pointed arch.
1One way: draw AD; with centre D and radius AD draw an arc from A upwards; with centre A, the same radius, draw an arc from D.
2The arcs meet at the top point (an equilateral triangle!).
3Different radii (bigger than half AD) give taller or flatter arches.
Arcs centred at D and A with radius AD meet at the top point.
✅ Answer: See the construction
๐ Textbook Page 152
๐ Page 152Regular hexagon from triangles
Can six congruent equilateral triangles fit around a point to make a regular hexagon? What are its angles?
1Each triangle has a 60° angle at the centre; 6 × 60° = 360°, so they fit exactly with no gap.
2All sides are equal (sides of the triangles).
3Each hexagon angle = 60° + 60° = 120°.
Six equilateral triangles round O.
✅ Answer: Yes — a regular hexagon with angles 120°
๐ Page 152Angles round a point
Angles 40°, 60°, 50°, 30°, 40°, 90° are placed round a point. Will a 70° angle fit in the gap?
140 + 60 + 50 + 30 + 40 + 90 = 310°
2Gap = 360 − 310 = 50°
370° > 50° → it does not fit.
40 + 60 + 50 + 30 + 40 + 90 = 310°, so the gap is 360 − 310 = 50°.
✅ Answer: No — the gap is only 50°
๐ Page 152Straight lines in the hexagon
In the regular hexagon ABCDEF with centre O (six equilateral triangles), why are AOD, BOE and COF straight lines?
2So A, O, D lie on a straight line; similarly for BOE and COF.
✅ Answer: Three 60° angles make 180°
๐ Textbook Page 153
๐ Page 153Hexagon of side 4 cm / 5 cm
Construct a regular hexagon of side 4 cm (and of side 5 cm).
1Draw a circle of radius 4 cm.
2Keeping the compass at 4 cm, start at any point on the circle and step off arcs round the circle: you get 6 points.
3Join them in order. (Each side equals the radius, as the six triangles are equilateral.)
4For 5 cm, use radius 5 cm.
Step the radius round the circle 6 times.
✅ Answer: See the construction
๐ Page 15360° angle
Construct a 60° angle at A on AX. Why is it 60°?
1Arc from A cuts AX at B.
2Same radius, arc from B cuts the first arc at C.
3AB = AC = BC (all equal to the radius) → △ABC is equilateral → ∠CAX = 60°.
AB = AC = BC (same radius) → △ABC is equilateral → ∠CAX = 60°.
✅ Answer: ∠CAX = 60°
๐ Textbook Page 154
๐ Page 15430° and 15°
How will you construct 30° and 15° angles?
1Construct 60°, then bisect it → 30°.
2Bisect 30° → 15°.
✅ Answer: Bisect 60° once (30°) and twice (15°)
๐ Page 1546-pointed star
Construct the 6-pointed star. Are the six point-triangles equilateral?
1Construct a regular hexagon (step the radius round a circle).
2Join alternate vertices: two big equilateral triangles ACE and BDF.
3Each small point-triangle has a 60° angle at its tip (angle of a big triangle) and 60° base angles (they are angles of the other big triangle, vertically opposite / corresponding).
4All three angles 60° → equilateral.
✅ Answer: Yes — all their angles are 60°
๐ Page 154Figure it Out · Q1 (a)
Construct an inflexed arc.Quarter circles centred at the red points meet in a point at the top.
1Draw two equal vertical pillars a distance 2w apart.
2Mark a point straight above each pillar-top at a height equal to half the gap (w). These are the two centres.
3Draw quarter circles of radius w: they curve up and meet at a point above the middle.
✅ Answer: See the construction
๐ Textbook Page 155
๐ Page 155Figure it Out · Q1 (b)–(e)
Construct the flower, the hexagon in a circle, the ring of 6 circles and the hexagon star pattern.
1(c) Hexagon in a circle: step the radius 6 times round the circle and join.
2(d) Ring of circles: mark the 6 hexagon points on a circle and draw equal circles centred at them (radius = half the side so they touch).
3(b) Flower: a centre circle; at its 6 hexagon points draw circles/arcs for petals — this can be done with a compass alone, as every point comes from stepping the radius.
4(e) Divide a hexagon into equilateral triangles using lines through the hexagon points, then join the points in a star pattern.
(b)(c)(d)
✅ Answer: See the hints
๐ Page 155Figure it Out · Q2
Optical illusion: what do you notice in this figure? How does it happen?
1We ‘see’ a white triangle on top, pointing up, and an outlined triangle below it — but neither is drawn!
2The cut-outs of the three discs and the broken lines line up, so the brain fills in the missing edges.
3To recreate: mark the vertices of an equilateral triangle, draw discs there with 60° wedges cut out, and draw the V-shaped corners of a second (inverted) triangle.
✅ Answer: A triangle that is not drawn appears (an illusion)
๐ Page 155Figure it Out · Q3
Construct the star inside a hexagon.
1Construct a regular hexagon.
2Join alternate vertices to make two equilateral triangles — the star.
3Angles: each hexagon angle 120° is split into 30° + 60° + 30°.
✅ Answer: Hexagon + two triangles through alternate vertices
๐ Page 155Figure it Out · Q4
Construct a perpendicular from a point P outside a line l.
1With centre P, draw an arc cutting l at X and Y (PX = PY).
2P is equidistant from X and Y, so it is on the perpendicular bisector of XY.
3From X and Y draw equal arcs on the other side of l meeting at Q. Join PQ: PQ ⊥ l.
✅ Answer: PQ ⊥ l
๐ Textbook Page 156
๐ Page 156Tangram figures
How can the 7 tangram pieces be rearranged to form each figure on page 156?The 7 tangram pieces: 2 large, 1 medium and 2 small triangles, a square and a parallelogram.
1Start with the 2 large triangles — they usually form the main body.
2The square and the parallelogram make legs, necks or tails; the small triangles make ears, beaks, feet.
3All 7 pieces must be used without overlapping. The total area is always the same (the whole square).
✅ Answer: Puzzle — use the hints and try!
๐ Textbook Page 157
๐ Page 1574 × 6 and 4 × 7 grids
Can 4 × 6 and 4 × 7 grids be tiled with 2 × 1 tiles? What about 5 × 7?A 4 × 6 grid (4 rows, 6 columns)
14 × 6: yes — e.g. vertical tiles in every column (4 rows is even).
24 × 7: yes — again vertical tiles in each of the 7 columns.
35 × 7 = 35 squares. Each tile covers 2, so the tiles always cover an even number of squares — 35 is odd → impossible.
4 × 6 grid tiled with vertical 2 × 1 tiles
✅ Answer: 4 × 6 yes, 4 × 7 yes, 5 × 7 no
๐ Textbook Page 158
๐ Page 158m × n grids
Is an m × n grid tileable with 2 × 1 tiles when (a) both are even (b) one is even (c) both are odd?
1(a) Yes: fill every column with vertical tiles.
2(b) Yes: if m (rows) is even, use vertical tiles in each column; if n is even, use horizontal tiles in each row.
3(c) No: m × n is odd, but tiles always cover an even number of squares.
✅ Answer: (a) Yes (b) Yes (c) No
๐ Page 1585 × 3 with a corner removed
A 5 × 3 grid with its top-right square removed: is it tileable?
114 squares → 7 tiles needed.
2Yes — one tiling is shown.
✅ Answer: Yes
๐ Page 1585 × 3 with a middle square removed
Is this region (second-row middle square removed) tileable?
1Yes — one tiling is shown.
✅ Answer: Yes
๐ Textbook Page 159
๐ Page 159Fig. 6.13
A 5 × 3 grid with the top middle square removed: is it tileable? Find another square whose removal makes it non-tileable.
1Colour like a chessboard. Each 2 × 1 tile always covers one black and one white square.
2The region has 8 squares of one colour and 6 of the other → impossible.
3Removing any square of the colour that the 4 corners do NOT have gives the same problem: e.g. the first square of the 2nd row, or the middle square of the 3rd row.
✅ Answer: Not tileable; e.g. also removing the left square of row 2 makes it non-tileable
๐ Page 159Black-and-white grid
If the plain grid is tileable, is the black-and-white grid tileable? And the other way round?
1Neighbouring squares always have different colours, so any 2 × 1 tile on the grid automatically covers one black and one white square.
2So every plain tiling is a black-and-white tiling, and the other way round.
✅ Answer: Yes, both ways
๐ Textbook Page 160
๐ Page 160Figure it Out · Q1
Can this region be tiled with the L-shaped tile of 3 squares?
112 squares → 4 tiles needed.
2Yes — one tiling is shown (found by a computer search).
✅ Answer: Yes
๐ Page 160Figure it Out · Q2
An 8 × 8 grid with two opposite corners removed: can it be tiled with 2 × 1 tiles?
1Colour it like a chessboard. Opposite corners have the SAME colour.
2After removing them: 30 squares of that colour and 32 of the other.
3Each tile covers one of each, so 31 tiles would need 31 + 31 → impossible.
✅ Answer: No
๐ Textbook Page 161
๐ Page 161Tiling the plane
Which regular polygons can tile the whole plane on their own?
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