Chapter 6 exercise answers: Constructions and Tilings

Open NCERT p. 140One way to think about it

1. Draw XY, say 10 units long. Open the compass to more than half of XY, say 7 units, and keeping this opening draw arcs from X and from Y above XY. They meet at A.
2. Change the opening to a different length that is still more than half of XY, say 9 units, and draw arcs from X and from Y below XY. They meet at B.
3. Join AB. It still cuts XY at its midpoint and at a right angle.
4. Why: within each pair the SAME opening was used from X and from Y, so AX = AY (7 units each) and BX = BY (9 units each). A and B are each equidistant from X and Y.
5. Every point that is equidistant from X and Y lies on the perpendicular bisector of XY, and two points fix a line. So AB is the perpendicular bisector, even though the two pairs used different radii.

In shortNo. The pair of arcs above XY and the pair below XY can have different radii. Only the two arcs within one pair (one from X, one from Y) must have the same radius.

Watch this explained “Loosening the recipe”, 7:20 into The perpendicular bisector, and the equidistance property that justifies it · हिंदी में देखें

Open NCERT p. 140One way to think about it

1. Draw XY, say 10 units long. With an opening of 7 units (more than half of XY), draw arcs from X and from Y above XY. They meet at A.
2. Change the opening to 9 units and again draw arcs from X and from Y above XY. They meet at C, higher up than A.
3. Join A and C and extend the line through XY. It cuts XY at its midpoint and at a right angle.
4. Why: AX = AY and CX = CY, so A and C are both equidistant from X and Y, and both lie on the perpendicular bisector of XY. Two points fix the line, so AC is the perpendicular bisector. It does not matter which side of XY the two points are on.
5. Take care: the second pair must use a different radius. With the same radius, the arcs above XY meet at the same point A again, and one point is not enough to draw the line.

In shortNo, it is not necessary. Both pairs of arcs can be drawn on the same side of XY, as long as the two pairs use different radii, because all we need is two different points that are each equidistant from X and Y.

Watch this explained “Loosening the recipe”, 7:20 into The perpendicular bisector, and the equidistance property that justifies it · हिंदी में देखें

Open NCERT p. 140One way to think about it

1. Draw XY. From X, draw an arc of one radius r1; from Y, draw an arc of a different radius r2 (still large enough for the two arcs to cross), and mark the crossing P.
2. By the way an arc is drawn, PX = r1 and PY = r2. Since r1 is not equal to r2, P is not equidistant from X and Y, so P does not lie on the perpendicular bisector of XY.
3. Now redraw both arcs with the SAME radius r from X and from Y, and mark the new crossing Q.
4. This time QX = r = QY, so Q is equidistant from X and Y and does lie on the perpendicular bisector.

In shortYes — within one pair, the two arcs (from X and from Y) must use the same radius, or their crossing point won't be equidistant from X and Y at all.

Watch this explained “One length, four times over”, 0:40 into The perpendicular bisector, and the equidistance property that justifies it · हिंदी में देखें

Open NCERT p. 140One way to think about it

1. Draw a segment AC and construct its perpendicular bisector. It cuts AC at its midpoint O, so the two lines are perpendicular at O.
2. Keep the compass at OA and mark B and D on the bisector, one on each side of O. Now OA = OB = OC = OD, on four rays at right angles.
3. Petal on OA: construct the perpendicular bisector of OA and call its midpoint M. Open the compass to MO, put its point on M, and mark P and Q on this bisector, one on each side of OA (MP = MQ = MO).
4. With centre P and radius PO draw the arc from O to A. With centre Q and radius QO draw the arc from O to A. These two arcs bound one petal.
5. Why the arcs end exactly at A: P and Q lie on the perpendicular bisector of OA, so PO = PA and QO = QA. Two circles meet in at most two points, so the two arcs meet only at O and A.
6. Repeat on OB, OC and OD, then rub out the straight construction lines.
7. Other choices also work: taking MP larger (the same on both sides and for all four petals) gives thinner petals, and smaller gives fatter ones.

In shortFour petals meet at O, each bounded by two arcs whose centres lie on the perpendicular bisector of that petal's segment (OA, OB, OC or OD), equally far on either side. This is one way; other choices of these centres give equally valid versions of the design.

Watch this explained “Why not just measure?”, 8:05 into The perpendicular bisector, and the equidistance property that justifies it · हिंदी में देखें

Open NCERT p. 142One way to think about it

1. The two loops at the rope's ends are excluded before folding, so the marked midpoint splits the WORKING length (the part between the pegs) into two truly equal halves.
2. Pulling that midpoint taut above XY to A stretches both halves fully, so AX = AY (each equal to half the working rope).
3. Pulling the same midpoint taut below XY to B gives BX = BY, the same half-length again — so AX = AY = BX = BY, one length four times over.
4. In triangles AXB and AYB: AX = AY, BX = BY, and AB is shared, so the triangles are congruent (SSS); this makes angle XAB = angle YAB.
5. In triangles AOX and AOY (O is where AB crosses XY): AX = AY, AO is shared, and the included angles are equal from the last step, so these triangles are congruent too (SAS); hence OX = OY and the two angles at O are equal.
6. X, O, Y lie on a straight line, so the two equal angles at O add to 180°, making each one 90°.

In shortAB is the perpendicular bisector because AX = AY = BX = BY forces, through two pairs of congruent triangles, both OX = OY and a right angle at O.

Watch this explained “Why it has to work”, 5:41 into A stretched rope as compass and straightedge: the Śulba-Sūtra constructions · हिंदी में देखें

Open NCERT p. 142One way to think about it

1. Peg the given point O on the line.
2. Tie a rope to the peg at O, pull it tight along the line to one side and mark X. With the same length, pull it tight to the other side and mark Y. Now OX = OY, so O is the midpoint of XY.
3. Fix pegs at X and Y. Take a rope with a loop at each end whose working part (leaving out the loops) is longer than XY. Fold the working part in half and mark its middle, then fasten the loops to X and Y.
4. Pull the middle mark away from the line until both halves are tight, and peg that point A. The two halves are equal, so AX = AY and A lies on the perpendicular bisector of XY.
5. O also lies on that bisector, because OX = OY. Stretch a rope tight from O through A: OA is perpendicular to the line at O. Only one pull is needed, because O is already one of the two points of the line.

In shortYes, there are several ways; here is one. Use the rope to mark X and Y at equal distances on either side of O, fasten a folded rope's ends at X and Y and pull its midpoint tight to a point A; the rope stretched from O through A makes a 90 degree angle with the line at O.

Watch this explained “Two more rope problems”, 7:29 into A stretched rope as compass and straightedge: the Śulba-Sūtra constructions · हिंदी में देखें

Open NCERT p. 144One way to think about it

1. Draw four angles of different sizes, for example a narrow acute angle, a wider acute angle, a right angle (constructed as before) and an obtuse angle. Call the corner of each one O.
2. For one angle: with centre O and any radius, draw an arc cutting the two arms at A and B, so OA = OB.
3. With centres A and B and one equal radius (more than half of AB; it need not be the first radius), draw two arcs that cross inside the angle at C.
4. Draw the ray OC. It is the bisector: OA = OB, AC = BC and OC is common, so triangles OAC and OBC are congruent (SSS), and angle AOC = angle BOC.
5. Repeat for the other three angles. You can check with a protractor that the two halves are equal each time.

In shortAny four different angles will do. For each one, mark OA = OB on the arms, draw equal arcs from A and B meeting at C, and draw ray OC; it bisects the angle whatever its size.

Watch this explained “Three steps, and 45 at last”, 5:56 into Bisecting an angle, and halving 90° to get 45° · हिंदी में देखें

Open NCERT p. 144One way to think about it

1. Draw two perpendicular lines through a point O (construct a 90 degree angle at O and extend both arms).
2. Bisect two neighbouring right angles and extend both bisectors through O. Now there are 8 rays from O, each 45 degrees from the next (360 ÷ 8 = 45).
3. With one compass opening, mark a point on each ray at the same distance from O: P1, P2, ..., P8. Each petal runs from O to one of these points.
4. Petal on OP1: construct the perpendicular bisector of OP1 and call its midpoint M. Open the compass to OP1, put its point on M, and mark C and C' on this bisector, one on each side of OP1.
5. With centre C and radius CO draw the arc from O to P1; with centre C' and radius C'O draw the arc from O to P1. C and C' lie on the perpendicular bisector of OP1, so CO = CP1 and C'O = C'P1: both arcs end exactly at O and P1. This is one petal.
6. Do the same on the other 7 rays, then rub out the construction lines.
7. Taking MC equal to OP1 gives thin petals like those in Fig. 6.5 (about a quarter as wide as they are long); a shorter MC gives fatter petals.

In shortThe 8 petals lie along 8 rays 45 degrees apart, made by bisecting right angles. Each petal is bounded by two equal arcs whose centres are on the perpendicular bisector of the petal's segment, one on each side at the same distance; other choices of that distance give valid variations.

Watch this explained “Two designs”, 8:36 into Bisecting an angle, and halving 90° to get 45° · हिंदी में देखें

Open NCERT p. 144One way to think about it

1. Draw angle XOY and mark A on OY and B on OX with OA = OB, as in Step 1.
2. With centres A and B and one equal radius larger than OA, draw arcs on the far side of O, outside the angle. They cross at C, beyond O.
3. Draw the line through C and O and extend it past O into angle XOY. Compare it with the bisector drawn the usual way: it is the same line.
4. Why: C is equidistant from A and B (equal arcs), and so is O (OA = OB). So C and O both lie on the perpendicular bisector of AB. The usual crossing point inside the angle is also equidistant from A and B, so it lies on this same line, and that line bisects angle XOY.
5. Take care: the ray from O towards C points away from angle XOY and bisects the vertically opposite angle. The part of the line inside angle XOY is the bisector of angle XOY.

In shortYes. The line OC is still the angle bisector; it is the same line as before. But the ray from O that bisects angle XOY is the one on the opposite side of O from C.

Watch this explained “The arcs on the far side”, 6:41 into Bisecting an angle, and halving 90° to get 45° · हिंदी में देखें

Open NCERT p. 144One way to think about it

1. Bisecting 90 degrees again and again gives 45, 22.5, 11.25, 5.625 degrees, and so on.
2. Bisecting 60 degrees (from an equilateral triangle) again and again gives 30, 15, 7.5 degrees, and so on.
3. Copying constructed angles side by side (adding) or one inside another (subtracting) gives more, for example 45 + 30 = 75, 90 + 45 = 135, 60 + 45 = 105, 45 + 22.5 = 67.5 and 90 − 60 = 30 degrees.
4. Every angle made this way is a whole number times 30 degrees, divided by 2 some number of times: 90 = 3 × 30 and 60 = 2 × 30, and halving, adding and subtracting keep this form.
5. If 65.5 degrees had this form, then 65.5 = 30 × c ÷ 2^n for whole numbers c and n, so 131 × 2^n = 60 × c. The right side is a multiple of 3, but 131 × 2^n is not, because neither 131 nor 2 is divisible by 3. So 65.5 degrees cannot be made this way. (In fact no ruler-and-compass construction gives exactly 65.5 degrees.)
6. You can only get close: 60 + 5.625 (90 halved four times) = 65.625 degrees, which is 1/8 of a degree too big.

In shortMany angles can be constructed, such as 45, 22.5, 30, 15, 7.5, 75, 105, 135 and 67.5 degrees. But 65.5 degrees cannot be constructed exactly; it can only be approached, for example by 65.625 degrees.

Watch this explained “What halving can reach”, 7:38 into Bisecting an angle, and halving 90° to get 45° · हिंदी में देखें

Open NCERT p. 144One way to think about it

1. Peg the corner O. Tie a rope at O, stretch it tight along one arm and mark A. With the same length, stretch it along the other arm and mark B. Now OA = OB.
2. Peg A and B. Take a rope with a loop at each end whose working part (leaving out the loops) is longer than AB. Fold the working part in half and mark its middle.
3. Fasten the loops at A and B. Pull the middle mark away from O until both halves are tight, and peg that point C. The halves are equal, so CA = CB.
4. Stretch a rope tight from O through C. This line bisects the angle: OA = OB, CA = CB and OC is common, so triangles OAC and OBC are congruent (SSS) and angle AOC = angle BOC.

In shortMark equal rope lengths OA = OB along the two arms, fasten a folded rope at A and B and pull its midpoint tight to a point C (so CA = CB), then stretch the rope from O through C: that line is the angle bisector.

Watch this explained “Two more rope problems”, 7:29 into A stretched rope as compass and straightedge: the Śulba-Sūtra constructions · हिंदी में देखें

Open NCERT p. 145One way to think about it

1. Draw the given square and find the midpoint of each of its 4 sides.
2. With centre at the midpoint of one side and radius equal to half that side's length, draw an arc from the near corner of the square curving in towards the centre.
3. With centre at the midpoint of the adjacent side and the same radius (half a side), draw the second arc bounding the same petal, from the same corner towards the centre.
4. Repeat at all 4 corners (using the 4 side-midpoints as centres) to get all 4 petals meeting at the centre of the square.

In shortEach petal is bounded by two arcs centred on the midpoints of two adjacent sides, with radius exactly half a side — that radius is what makes the petals meet exactly at the centre without a gap and without spilling outside the square.

Watch this explained “Two designs”, 8:36 into Bisecting an angle, and halving 90° to get 45° · हिंदी में देखें

Open NCERT p. 147One way to think about it

1. Draw angle XOY of any size and orientation. With centre O and any radius, draw an arc cutting both arms, at B (on OX) and C (on OY).
2. Draw a new ray from a new point O′ (the corner for the copy). With the SAME radius as before, centre O′, draw an arc cutting the new ray at B′.
3. Open the compass to the distance BC (between the two original arc-marks). With centre B′ and this radius, draw an arc cutting the first new arc at C′.
4. Draw ray O′C′: since O′B′ = OB, O′C′ = OC, and B′C′ = BC, triangle O′B′C′ is congruent to OBC (SSS), so angle X′O′Y′ exactly copies angle XOY.
5. Repeat for at least three more angles of different sizes and orientations.

In shortEach copy is exact because converting the angle into the single distance BC — and carrying only that distance with the compass — forces the new triangle to be congruent to the old one by SSS.

Watch this explained “Turn the angle into a length”, 1:51 into Copying an angle, and why triangle congruence proves it works · हिंदी में देखें

Open NCERT p. 147One way to think about it

1. Choose an arm length and an angle for the unit. Draw the first arm V1T1 slanting up to the right. At V1 draw the chosen angle so that the second arm goes up to the left, and mark T0 on it with V1T0 = V1T1 (compass).
2. With centre V1 and radius V1T1, draw the arc from T0 to T1. This is the first (shaded) unit.
3. At T1, copy the angle T0V1T1 (equal arcs at both corners, then carry the one distance across) so that one arm is T1V1 and the new arm points down to the right, on the other side of V1T1 from T0. Mark V2 on it with T1V2 = V1T1.
4. With centre T1 and the same radius, draw the arc from V1 to V2, bulging downwards. This is the second unit: the first unit turned upside down, sharing the arm V1T1.
5. At V2, copy the angle again with one arm V2T1 and the new arm going up to the right; mark T2 with V2T2 = V1T1 and draw the arc centred V2 from T1 to T2. Keep going, alternating up and down, and shade the upward units.
6. The copies are exact because every unit has two arms of one compass opening, the same angle (copied using SSS), and an arc of that same radius. The lower corners V1, V2, ... come out on one straight line, and so do the upper corners T0, T1, T2, ...

In shortFig. 6.6 is one sector-shaped unit copied along a line and turned upside down every other time; each new unit shares an arm with the one before, its angle is copied with ruler and compass, and its arms and arc all use the same compass opening.

Watch this explained “A band that repeats”, 0:00 into Copying an angle, and why triangle congruence proves it works · हिंदी में देखें

Open NCERT p. 148One way to think about it

1. Draw a line m and a transversal line l crossing it at A, making some angle with m.
2. Choose a point B on l where the parallel line should pass, and copy the angle at A onto B (equal radius from A and B on l, then the one transferred distance), so the new ray through B makes the same corresponding angle with l as m does at A.
3. Draw the new line through B both ways: it is parallel to m, because equal corresponding angles is exactly the test for two lines being parallel.
4. Repeat with different lines and points in at least 4 different orientations.

In shortCopying the corresponding angle at the new point with a compass and straight edge is enough — equal corresponding angles is exactly what makes the new line parallel to the old one.

Watch this explained “The parallel, built by copying”, 6:56 into Copying an angle, and why triangle congruence proves it works · हिंदी में देखें

Open NCERT p. 148One way to think about it

1. Draw two perpendicular lines through a centre O, then bisect two neighbouring right angles and extend both bisectors through O. This gives 8 rays, 45 degrees apart.
2. With one compass opening, mark A, B, C, D, E, F, G, H on the 8 rays, going round clockwise from the top, all at the same distance from O.
3. Keep the same opening. With centres A and B draw arcs; they cross at O and at a second point T. Join AT and BT.
4. OA = AT = TB = BO, so OATB is a rhombus, and so AT is parallel to OB and BT is parallel to OA; these are the parallel pairs of edges in the figure. (T can also be found with the parallel-line construction: through A draw a line parallel to OB, and through B a line parallel to OA.)
5. Repeat for each pair of neighbouring points: B and C give U, C and D give V, D and E give W, E and F give X, F and G give Y, G and H give Z, and H and A give S.
6. Join O to each outer point S, T, ..., Z, and shade one of the two triangles in every rhombus, the same one each time (O-A-T, O-B-U, O-C-V, and so on), as in the figure.

In shortThe figure is 8 rhombuses around O, each with a 45 degree angle at O and all sides equal to OA. The 8 inner points come from bisecting right angles, and each outer point is where equal arcs from two neighbouring inner points cross again.

Watch this explained “Two things to build”, 8:18 into Copying an angle, and why triangle congruence proves it works · हिंदी में देखें

Open NCERT p. 151One way to think about it

1. Draw the support lines of Fig. 6.11: a base AB, its perpendicular bisector, and a point P on it; join PA and PB. P is on the perpendicular bisector, so PA = PB.
2. Mark the midpoint M of PA and the midpoint N of PB (perpendicular bisector construction).
3. Left side, lower half AM: draw an arc from A to M that bulges outwards, away from the inside of the arch. The simplest choice is a semicircle, with its centre at the midpoint of AM and radius a quarter of AP.
4. Left side, upper half MP: draw an arc with the same radius from M to P that bulges inwards. The two arcs join smoothly at M and make an S-shaped curve, as in 'Wavy Wave'.
5. Right side: do the same, bulging outwards on BN and inwards on NP. The two S-curves meet in a point at P. Rub out the support lines.
6. Different arches: keep the support lines and choose a larger radius (it can never be less than a quarter of AP). For the arc on AM, put the compass point on the perpendicular bisector of AM, on the inner side, at the chosen radius from A (it is then the same distance from M), and draw the shorter arc from A to M. For MP, do the same on the outer side of MP with the same radius. A larger radius gives a flatter S and a slimmer arch.

In shortEach side of the pointed arch is an S-shaped curve made of two arcs of the same radius on the two halves of a support line, the lower one bulging out and the upper one bulging in. Changing that radius (a quarter of the support line or more) gives different arches.

Watch this explained “One opening, struck from both ends”, 4:03 into Constructing 60° from an equilateral triangle, and the arches built on it · हिंदी में देखें

Open NCERT p. 151One way to think about it

1. There are many correct answers: any arch built from support lines and arcs is fine. Here is one design, a pointed arch on an opening AB.
2. Draw the opening AB, say 6 cm, with a pillar going down from A and from B if you like.
3. With centre A and radius AB, draw an arc upwards from B. With centre B and the same radius, draw an arc upwards from A. They cross at P, above the middle of AB.
4. The arch is the arc from B to P (centre A) and the arc from A to P (centre B). AP = BP = AB, so triangle ABP is equilateral. The top is a point, and the arch rises about 5.2 cm on a 6 cm opening, higher than a semicircle would (3 cm).
5. Other designs: the trefoil arch of this chapter with a different angle at A and D or different arm lengths; the S-curve pointed arch of Fig. 6.11 with a different radius; or one arch repeated side by side to make a row of arches.

In shortMany designs are possible. One example: on an opening AB, draw arcs of radius AB centred at A and at B; they meet at P, and the arcs AP and BP form a pointed arch about 0.87 × AB high (5.2 cm for a 6 cm opening).

Watch this explained “One opening, struck from both ends”, 4:03 into Constructing 60° from an equilateral triangle, and the arches built on it · हिंदी में देखें

Open NCERT p. 154One way to think about it

(a) An Inflexed Arc

1. Draw two parallel upright sides (pillars) and mark their tops B and C at the same height. Find the midpoint M of BC with the perpendicular bisector.
2. On the perpendicular bisector of BC, mark P above M with MP = MB (compass). P will be the point of the arch.
3. At C, extend the pillar upwards (it is perpendicular to BC) and mark Q on it with CQ = CM. Do the same at B to get Q' with BQ' = BM.
4. MCQP has right angles at M and C and MC = CQ = MP, so it is a square and QP = QC. In the same way Q'P = Q'B.
5. With centre Q and radius QC, draw the arc from C up to P. With centre Q' and radius Q'B, draw the arc from B up to P. Each is a quarter circle; the two arcs sag inwards and meet in a sharp point at P. Rub out the construction lines.

In shortTwo quarter circles, centred level with the top point P and directly above the pillar tops, with radius half the width BC, meet in a point at P.

(b) it can also be constructed using only a compass!

1. Draw a circle with centre O and radius r. This is the middle of the flower.
2. Keep the same opening. Put the compass point anywhere on the circle and mark P1; from P1 mark P2 on the circle, and so on. The sixth step lands back on P1, giving 6 equally spaced points P1 to P6.
3. With the same opening r, lightly draw circles centred at P1 and at P2. Besides O, they cross at one more point Q, outside the middle circle (OQ is about 1.73r).
4. The petal between P1 and P2 is bounded by the arc of the circle centred P2 from P1 out to Q, and the arc of the circle centred P1 from P2 out to Q. Ink only these two arcs.
5. Do the same for each pair of neighbouring points (P2 and P3, ..., P6 and P1). Every line is an arc made with the one opening r, so no ruler is needed.

In shortSix petals round a circle of radius r; each petal is bounded by two arcs of radius r centred at two neighbouring points of the six equally spaced points on the circle, so the whole figure needs only a compass.

(c)

1. Draw a circle with centre O and radius r.
2. Keeping the opening r, step the compass round the circle, marking 6 points; the sixth step lands back on the first.
3. Join the 6 points in order with a ruler. Each side equals r, because each of the 6 triangles with a vertex at O has three sides equal to r, so this is a regular hexagon inside the circle.

In shortStep the radius r round the circle 6 times and join the marks: the regular hexagon's side equals the circle's radius.

(d)

1. Lightly draw a circle with centre O and radius r and step off 6 points P1 to P6 on it with the same opening, as in (c).
2. Construct the perpendicular bisector of P1P2 to find its midpoint M, and set the compass to P1M, which is half of r.
3. With this radius, draw a circle centred at each of P1 to P6, then rub out the first circle.
4. Neighbouring centres are r apart, which is half of r plus half of r, so each circle just touches its two neighbours (at the midpoints of the hexagon's sides) and the six make a closed ring.

In shortSix circles of radius half of r, centred at the six corners of a regular hexagon of side r, each touching its two neighbours.

(e)

1. Choose a length a. Draw a circle of radius 2a and step the opening 2a round it to get a regular hexagon of side 2a, as in (c).
2. Join the three pairs of opposite corners. These lines pass through the centre and cut the hexagon into 6 equilateral triangles of side 2a.
3. Mark the midpoint of every side of these 6 triangles (compass opening a from a corner). In each triangle join the three midpoints; these joins line up into straight lines across the hexagon, which is now a grid of 24 small equilateral triangles of side a.
4. In each small triangle, find its centre: find the midpoints of two of its sides (perpendicular bisector) and join each to the opposite corner; the two lines cross at the centre.
5. Join the centre to the three corners of that small triangle (not to the midpoints). Do this in all 24 small triangles to complete the pattern.

In shortA regular hexagon of side 2a, cut by a triangular grid into 24 equilateral triangles of side a, with the centre of each small triangle joined to its three corners.

Watch this explained “Build one, and a star on it”, 8:21 into Regular hexagons, and why the angles round a point must total 360° · हिंदी में देखें

Open NCERT p. 155One way to think about it

1. What you notice: a bright white triangle, pointing upwards, seems to lie on top of three black discs and of a black-outlined triangle pointing downwards. But no line of the white triangle is drawn.
2. How it happens: each disc has a 60 degree wedge cut out, and the straight edges of the three wedges point exactly at one another; the pieces of the outlined triangle stop exactly on those same lines. The brain explains all these gaps most simply as a white triangle covering them, so it sees edges that are not there. (This figure is known as the Kanizsa triangle.)
3. To recreate it: lightly draw a six-pointed star, that is, two equal equilateral triangles with the same centre, one pointing up and one pointing down (step the radius round a circle to get 6 points and join alternate points).
4. At each corner of the upward triangle, draw a small circle (radius about one eighth of a side) and shade it, except for the 60 degree wedge that lies inside the upward triangle.
5. Of the downward triangle, ink only the two short sides of each of its three points, the parts outside the upward triangle, stopping just before the upward triangle's sides. Rub out all the construction lines.

In shortYou see a white upward-pointing triangle that is not drawn at all. It appears because the cut-out wedges of the discs and the broken corners of the outlined triangle all line up along the edges of one triangle, and the eye fills in those edges. To recreate it, start from a six-pointed star: put discs with 60 degree wedges at the corners of one triangle and ink only the points of the other triangle.

Open NCERT p. 155One way to think about it

1. Draw a circle with centre O and step its radius round it 6 times to get the corners A, B, C, D, E, F of a regular hexagon, with A at the top. Join them in order.
2. Join alternate corners: A to C, C to E and E to A, and B to D, D to F and F to B. This gives two equilateral triangles that cross to make a six-pointed star whose points are the corners of the hexagon.
3. Ink the outline of the star, from each corner of the hexagon to the nearest crossing points, and rub out the small hexagon in the middle where the two triangles overlap.
4. The angles (the hint): each angle of the regular hexagon is 120 degrees. Triangle ABC has AB = BC and angle B = 120 degrees, so its other two angles are 30 degrees each. So at A the 120 degrees is split into 30 + 60 + 30: every point of the star is 60 degrees, and each gap between the star and the hexagon is a triangle with angles 30, 30 and 120 degrees.

In shortDraw a regular hexagon, join every other corner to get two equilateral triangles, and keep only their star-shaped outline. Each point of the star is 60 degrees, and each gap between the star and the hexagon is a triangle with angles 30, 30 and 120 degrees.

Watch this explained “Build one, and a star on it”, 8:21 into Regular hexagons, and why the angles round a point must total 360° · हिंदी में देखें

Open NCERT p. 155One way to think about it

1. Draw line l and mark point P outside it.
2. With centre P, open the compass wide enough to reach past line l, and swing an arc that cuts l at two points; call them X and Y.
3. Since X and Y are both on one arc centred at P, PX = PY, so P is equidistant from X and Y.
4. Construct the perpendicular bisector of XY in the usual way (equal-radius arcs from X and from Y, above and below, then join the crossings).
5. This perpendicular bisector passes through P (because P is equidistant from X and Y) and is perpendicular to l — it is the required perpendicular.

In shortStrike one arc from P to cut l at two points X and Y, then construct the perpendicular bisector of XY in the usual way — it passes through P because P is equidistant from X and Y.

Watch this explained “A point off the line”, 6:47 into Constructing a 90° angle at a chosen point on a line · हिंदी में देखें

Open NCERT p. 156One way to think about it

1. The seven pieces: 2 large triangles, 1 medium triangle, 2 small triangles, 1 small square and 1 parallelogram. Every figure must use all seven, with no gaps and no overlaps. You may slide a piece, turn it, or flip it over (flipping only makes a difference for the parallelogram); you may not cut or stretch a piece.
2. Measure sizes in 'small squares' (the square piece). Each large triangle covers 2 small squares, the medium triangle, the square and the parallelogram cover 1 each, and each small triangle covers half. So the whole set covers 2 + 2 + 1 + 1 + 1 + 1/2 + 1/2 = 8 small squares, whatever figure you make.
3. General way in: place the two large triangles first (together they are half of the figure), then the medium triangle, the square and the parallelogram, and keep the two small triangles for the corners that are left. If the figure will not close up, move a large triangle rather than forcing a small piece.
4. Worked example, the C-shaped figure. It is 3 small squares wide and 4 tall, with a 2 by 2 gap cut out of its right side, so the bar and the two arms are 1 small square thick. Its area is 3 × 4 − 2 × 2 = 8 small squares, which matches the set.
5. Put one large triangle in the top-left corner of the C, with its right angle in the corner: one short side runs 2 squares along the top edge, the other 2 squares down the left edge. Put the other large triangle the same way in the bottom-left corner. Their tips meet at the middle of the left edge.
6. Between the two large triangles and the inner edge of the bar there is a triangular gap whose long side is the inner edge (2 squares long). The medium triangle fits it exactly, long side against the inner edge.
7. Top arm: put the square at the right-hand end of the arm. The slanted gap left between the square and the large triangle is filled by one small triangle.
8. Bottom arm: lay the parallelogram along the slanted edge of the lower large triangle, touching the bottom edge. The corner left at the top right of the arm's end is filled by the last small triangle.
9. Check: all 7 pieces used once, no gap, no overlap, and 2 + 2 + 1 + 1/2 + 1 + 1 + 1/2 = 8 small squares, the area of the C. The other figures are built the same way; most of them can be made in more than one way.

In shortEach figure is made from all seven tangram pieces, used once each, by sliding, turning and (for the parallelogram) flipping them so they cover the outline with no gaps and no overlaps. The pieces always cover 8 small squares, so every figure has that area. There can be more than one arrangement; one arrangement of the C-shaped figure: large triangles in its top-left and bottom-left corners, the medium triangle along the inside of the bar, the square and 1 small triangle in the top arm, and the parallelogram and 1 small triangle in the bottom arm.

Watch this explained “A way in”, 7:19 into Tangrams: rearranging pieces without changing the area · हिंदी में देखें

Open NCERT p. 160Checked by computer

1. Number the rows 1 to 4 from the top and the columns 1 to 4 from the left. The region has columns 1 and 2 in rows 1 and 2, and all four columns in rows 3 and 4: 2 + 2 + 4 + 4 = 12 squares. The tile covers 3 squares in an L shape (a 2 × 2 block with one square missing).
2. 12 ÷ 3 = 4, so exactly 4 tiles are needed. Counting does not rule it out, so try to place them.
3. Tile 1: row 1 column 1, row 1 column 2 and row 2 column 1.
4. Tile 2: row 2 column 2, row 3 column 2 and row 3 column 3.
5. Tile 3: row 3 column 1, row 4 column 1 and row 4 column 2.
6. Tile 4: row 3 column 4, row 4 column 3 and row 4 column 4.
7. Each tile is an L shape (3 squares of a 2 × 2 block), no square is used twice, and all 12 squares are covered. (This is in fact the only way to do it.)

AnswerYes. The 12 squares can be covered exactly by 4 L-shaped tiles of 3 squares each.

Watch this explained “It is not about twos”, 6:57 into What it takes to cover a region with no gaps and no overlaps · हिंदी में देखें

Open NCERT p. 160Checked by computer

1. Read the region off the figure: it is the union of two 7x7 squares of unit cells, one shifted one row down and one column right from the other, giving 62 cells in all (7 + 6x8 + 7).
2. 62 is even, so a plain count of cells does not rule the tiling out on its own.
3. Colour the region like a chessboard, by (row + column): this gives 32 cells of one colour and 30 of the other.
4. Every domino, upright or sideways, covers two cells that touch along an edge, and two touching cells always have different colours — so every domino covers exactly one cell of each colour.
5. 31 dominoes would need exactly 31 cells of each colour, but the region has 32 of one colour and only 30 of the other, so no arrangement of dominoes can cover it.

AnswerNo — a checkerboard colouring of the region gives 32 cells of one colour and 30 of the other, and a domino always takes one of each, so the tiling is impossible even though the total (62) is even.

Watch this explained “The big one”, 6:11 into What it takes to cover a region with no gaps and no overlaps · हिंदी में देखें