Class 7 Maths Ganita Prakash Chapter 14 Solutions (NCERT 2026–27) – Constructions and Tilings

These Class 7 Maths Ganita Prakash Chapter 14 solutions cover Constructions and Tilings, which is Chapter 6 of Ganita Prakash Part II (the site lists it as Chapter 14 in continuous numbering). Every Figure it Out task, Math Talk and Try This is answered step by step — with each ruler-and-compass construction described in words and verified — so you can build the perpendicular bisector, angle bisector, angle copy, parallel lines and a regular hexagon, and reason confidently about tiling.

Class: 7 Subject: Mathematics Book: Ganita Prakash (Part II) Chapter: 14 (Part II, Ch. 6) Exercises: Six “Figure it Out” sets + Math Talk & Try This Session: 2026–27
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Chapter 14 Overview

Chapter 14 of Ganita Prakash, Constructions and Tilings (Part II, Chapter 6), teaches accurate geometric constructions using only an unmarked ruler and a compass, then explores how shapes cover a region or the whole plane. Starting from the playful ‘Eyes’ design, it derives the perpendicular bisector and proves it works using triangle congruence, then extends the idea to build a 90° angle, bisect any angle (giving 45°, 30°, 15°), copy an angle, draw parallel lines, and construct a 60° angle and a regular hexagon. It also visits the ancient Śulba-Sūtras rope methods, arch and star designs, tangrams, and finally tiling — deciding when an m × n grid can be covered by 2 × 1 tiles using a clever black-and-white colouring argument. The Class 7 Maths Ganita Prakash Chapter 14 solutions below work through every Figure it Out, Math Talk and Try This task step by step.

Key Concepts & Definitions

Bisection: dividing a line segment (or any geometrical object) into two identical parts.

Perpendicular bisector: a line that bisects a given segment and is perpendicular to it. Key property — any point equidistant from the two endpoints of a segment lies on its perpendicular bisector.

Angle bisector: a ray that divides an angle into two equal angles. It can be drawn by building two congruent triangles (SSS) on the two arms.

Copying an angle: reproducing the exact measure of a given angle elsewhere by transferring an isosceles triangle of arcs (SSS congruence makes ∠A = ∠X).

Congruence conditions used: SSS (three sides equal) and SAS (two sides and the included angle equal) — the proofs in this chapter rely on these.

Regular polygon: a polygon with all sides equal and all angles equal; a regular hexagon splits into 6 congruent equilateral triangles.

Tiling: covering a region using a set of shapes without gaps or overlaps.

Śulba-Sūtras: ancient Indian geometric texts (Vedā&ndotbelow;ga) that give rope-and-peg methods to construct perpendiculars and perpendicular bisectors.

Important Constructions & Facts (Chapter 14)

Perpendicular bisector of XY: with one fixed radius, draw arcs from X and from Y above XY (meeting at A) and below XY (meeting at B); the line AB is the perpendicular bisector.

90° at a point O on a line: mark X and Y equidistant from O, then draw the perpendicular bisector of XY — it passes through O at 90° (only one extra arc point is needed since O is already on it).

Angle bisector of ∠AOB: mark A, B with OA = OB; cut equal arcs from A and B meeting at C; OC bisects the angle (ΔOBC ≅ ΔOAC by SSS).

Special angles: 90° → bisect → 45°; 60° → bisect → 30° → bisect → 15°.

60° angle: draw an arc from A; with the same radius cut it from B to get C; ∠CAX = 60° (ΔABC is equilateral).

Regular hexagon: six congruent equilateral triangles meet at the centre (6 × 60° = 360°); each interior angle is 60° + 60° = 120°.

Tiling rule (2 × 1 tiles): each tile covers 2 unit squares, so a region must have an even number of unit squares; the colouring test then needs equal black and white squares.

Figure it Out — Perpendicular Bisector (Page 140)

Questions are reproduced verbatim from the NCERT Ganita Prakash textbook; the worked solutions are original and verified.

1. When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore this through construction, and then justify your answer. [Hint 1: Any point that is of the same distance from X and Y lies on the perpendicular bisector. Hint 2: We can draw the whole line if any two of its points are known.]

SOLUTION No, it is not necessary. Use one radius (say r₁) for the two arcs that meet above XY at point A, and a different radius (say r₂) for the two arcs that meet below XY at point B. Point A is equidistant from X and Y (both arcs had radius r₁), so AX = AY → A lies on the perpendicular bisector. Point B has BX = BY (both arcs had radius r₂), so B also lies on the perpendicular bisector. By Hint 2, two points fix a unique line. Since both A and B lie on the perpendicular bisector, the line AB is the perpendicular bisector — even though the two pairs used different radii. ∴ the radius above and below need not be equal; only the two arcs that meet at the same point must share a radius.

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

SOLUTION Yes, both pairs can be on the same side. Draw a first pair of equal arcs (radius r₁) from X and Y meeting above XY at A, and a second pair of equal arcs of a different radius (r₂) from X and Y also meeting above XY at a different point, say A′. Both A and A′ are equidistant from X and Y, so both lie on the perpendicular bisector. Joining A and A′ (and extending the line) gives the full perpendicular bisector. ∴ we do not need one pair above and one below; any two distinct equidistant points determine the line, even if both lie on the same side. (The above-and-below method is just the most convenient.)

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

SOLUTION Yes — for a single intersection point it is necessary. Suppose we draw an arc from X with radius r₁ and an arc from Y with a different radius r₂, and they meet at a point P. Then PX = r₁ and PY = r₂. If r₁ ≠ r₂, then PX ≠ PY, so P is not equidistant from X and Y and does not lie on the perpendicular bisector. ∴ the two arcs that intersect at one point must use the same radius, so that the meeting point is equidistant from X and Y. (Question 1 only allowed different radii for different meeting points, not for the two arcs sharing one point.)

4. Recreate this design using only a ruler and compass —

SOLUTION (method) This is a construction (drawing) task; here is the verified ruler-and-compass method, since the design is built from perpendicular bisectors and arcs. Step 1. Draw a base line segment and construct its perpendicular bisector by the standard method (equal arcs from each endpoint above and below, joined). Step 2. Mark the point where the bisector meets the segment as the centre; from here and from the equidistant points, draw arcs of equal radius to form the symmetric loops/petals of the design. Step 3. Keep the compass radius unchanged for matching arcs so the two halves are congruent and the figure is symmetric about the supporting line. Step 4. After completing the design, trace its boundary with a coloured pencil (using the ruler/compass) so the final figure stands out from the faint supporting lines and arcs.

Figure it Out — Śulba Rope Method (Page 142)

1. Justify why AB in Fig. 6.4 is the perpendicular bisector.

SOLUTION In the rope method, the rope is folded in half and its midpoint marked; the two loop-ends are then fixed to the pegs at X and Y. When the midpoint is pulled taut above XY to point A, the two halves of the rope are equal, so AX = AY. Similarly, pulling the midpoint taut below XY to point B keeps the two halves equal, so BX = BY. Thus both A and B are equidistant from X and Y, so each lies on the perpendicular bisector of XY. Two points determine the line, so AB is the perpendicular bisector of XY — exactly the property used in the compass construction.

2. Can you think of different methods to construct a 90° angle at a given point on a line using a rope? (Math Talk)

SOLUTION Method A (perpendicular bisector idea). On the line, fix pegs at X and Y so that the given point O is exactly the midpoint of XY (fold a rope of length XY to find its midpoint and place O there). Then build the perpendicular bisector of XY with the rope as above; it passes through O and is perpendicular to the line, giving a 90° angle at O. Method B (3–4–5 right-angle rope). Use a rope knotted into 12 equal parts. Peg it as a triangle with sides 3, 4 and 5 units, placing the corner of the 3-unit and 4-unit ropes at O along the line. Because 32 + 42 = 52, the angle at O is exactly 90° — a classic Śulba-style construction.

Figure it Out — Angle Bisection (Page 144)

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

SOLUTION (method) Take four different angles, for example 40°, 70°, 110° and 150°. For each angle ∠AOB: 1. With centre O and any radius, cut the two arms at A and B (so OA = OB). 2. With the same sufficiently long radius, draw arcs from A and from B that meet at C. 3. Join OC. Then OC is the bisector, because ΔOBC ≅ ΔOAC (OB = OA, BC = AC, OC common → SSS), so ∠BOC = ∠AOC. Check: the bisector of 40° gives two 20° angles, of 70° gives two 35°, of 110° gives two 55°, and of 150° gives two 75° angles.

2. Construct the 8-petalled figure shown in Fig. 6.5.

SOLUTION (method) The 8 petals need 8 equal supporting lines through a centre, so consecutive lines make 360° ÷ 8 = 45°. 1. Draw a line through the centre O. 2. Construct a 90° angle at O (perpendicular-bisector method), giving 4 lines at 90° apart. 3. Bisect each 90° angle to get 45° lines — now there are 8 supporting lines, each 45° apart. 4. Between every pair of adjacent lines, draw two equal arcs (same compass radius) to form a petal; repeat around all 8 sectors. Equal radii keep all 8 petals congruent.

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

SOLUTION It will not bisect the original angle ∠AOB. If the equal arcs from A and B meet at a point C on the far side of O, then OC points roughly opposite to the angle’s interior. By the same SSS argument (OA = OB, AC = BC, OC common), ΔOAC ≅ ΔOBC, so ∠AOC = ∠BOC — meaning OC is equally inclined to OA and OB. Hence the line containing OC is the bisector of the vertically opposite angle, i.e. OC (extended) lies along the bisector of the angle on the other side of O. ∴ the ray OC drawn on the other side bisects the opposite angle, not ∠AOB itself; but its opposite ray is the genuine bisector of ∠AOB. So the line is still an angle-bisector line, just of the reflex/opposite angle.

4. What are the other angles that can be constructed using angle bisection? Can you construct 65.5° angle? (Math Talk)

SOLUTION Repeated bisection halves an angle each time. Starting from 90° we get 45°, 22.5°, 11.25°, …; from 60° we get 30°, 15°, 7.5°, …; combining and bisecting standard angles also gives 75°, 105°, 120°, 135°, 150°, etc. In general, by bisecting and adding constructible angles we can build any angle of the form (multiple of 15°) and their repeated halves — e.g. 7.5°, 3.75°, and so on. 65.5°: this is not obtainable by bisection of 60°/90°, since 65.5° is not a sum/half of the standard constructible angles (15°, 30°, 45°, … and their halves). So 65.5° cannot be constructed by simple ruler-and-compass angle bisection here.

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

SOLUTION (method) 1. From the vertex O, use the rope as a compass to mark A on one arm and B on the other so that OA = OB (same rope length for both). 2. Stretch the rope between A and B and fold it to find/mark its midpoint, or use the rope to draw equal arcs from A and from B that meet at a point C inside the angle. 3. Stretch the rope from O through C to draw the line OC. Since OA = OB and AC = BC, ΔOAC ≅ ΔOBC (SSS), so OC bisects ∠AOB.

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

SOLUTION (method) 1. Draw the square and find its centre by drawing both diagonals (they cross at the centre) or by constructing the perpendicular bisectors of two sides. 2. The largest petals fit when their arcs use the side of the square as a diameter / chord. Set the compass to half the side length and, taking the midpoints of the sides as centres, draw arcs that curve inward to meet near the centre — this makes each petal as large as the square allows without crossing outside. 3. Keep the same radius for every arc so all four petals are congruent and the design is symmetric.

Figure it Out — Copying an Angle (Page 147)

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

SOLUTION (method) Draw four arbitrary angles at vertex A in different tilts. To copy each angle to a new vertex X on a ray XZ: 1. From A, draw an arc cutting both arms at B and C (this gives the isosceles triangle ΔABC). 2. With the same radius, draw an arc from X cutting the ray at Z. 3. Open the compass to the chord length BC and, from Z, cut the new arc at Y so that YZ = BC. 4. Join XY. By SSS, ΔABC ≅ ΔXYZ, so ∠A = ∠X — the angle is copied exactly, with no measuring.

2. Construct the Fig. 6.6.

SOLUTION (method) Fig. 6.6 is a single ‘unit’ (two equal arms with a fixed angle between them) repeated in two orientations. To make exact copies: 1. Draw the first unit with chosen arm length (set with the compass) and a chosen angle between the arms. 2. Using the copy-an-angle method above, reproduce the same angle at the next position, and transfer the same arm length with the compass. 3. Flip the orientation for the alternate units (copy the angle on the opposite side). Equal arms + equal copied angles guarantee identical units, so the repeating pattern lines up perfectly.

Figure it Out — Parallel Lines (Page 148)

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

SOLUTION (method) For each pair, draw a line m and a transversal l meeting m at A. Choose a point B on l where the parallel line should pass. 1. At A, draw an arc cutting m at C and l at D (the angle between m and l). 2. With the same radius, draw an arc from B cutting l at E. 3. Transfer the chord length CD onto this arc from E to get F (EF = CD). 4. Draw the line through B and F, call it n. Then n makes the same corresponding angle with l as m does, so by the corresponding-angles test m ∥ n. Repeat with four different tilts of l and m.

2. Construct the following figure. (the design lettered S T A B U V W X Y Z … F G H, built from parallel lines)

SOLUTION (method) The figure is made of several sets of parallel lines crossing each other to form the bordered design. 1. Draw the outer square/frame (e.g. STWX) and its main diagonals or guidelines. 2. Using the copy-an-angle parallel-line method, draw each inner line parallel to a chosen side so that pairs like SX ∥ TW and ST ∥ XW are exactly parallel. 3. Repeat for the inner octagon-like ring (points A–H), keeping equal spacing by transferring equal compass lengths. The intersections of these parallel families create the lettered vertices of the design.

Figure it Out — Pointed Arch (Page 151)

1. Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs.

SOLUTION (method) The supporting lines are two equal line segments leaning towards each other (like ‘Wavy Wave’). Mark the midpoint of each segment. 1. Take the left segment’s lower endpoint as a centre and a radius equal to the full segment length; draw an arc that rises up to the top meeting point — this is the right-hand curve of the arch. 2. Take the right segment’s lower endpoint as a centre with the same radius; draw the matching arc for the left-hand curve. The two arcs meet at a point at the top, forming the pointed (ogive) arch. 3. To make different arches, change the compass radius (a larger radius gives a flatter, wider arch; a smaller radius gives a taller, sharper point).

2. Make your own arch designs.

SOLUTION (method) Combine the techniques above to invent new arches. For a trefoil arch, construct equal angles at the two base points A and D, mark B and C with AB = CD, then draw three equal arcs between them. For a multi-lobed arch, add more equidistant centres along the springing line and draw equal arcs. Keep symmetry by using equal radii on the left and right and by placing the centres symmetrically about the perpendicular bisector of the base — then adjust the radii until the arch looks aesthetically pleasing.

Figure it Out — Designs & Perpendicular through P (Page 154)

1. Construct the following figures: (a) An Inflexed Arc (b) The fun part about this figure is that it can also be constructed using only a compass! Can you do it? (c)   (d)   (e)

SOLUTION (method) (a) Inflexed arc: draw a base segment and its perpendicular bisector; using points on the bisector as centres, draw arcs that curve inward (concave) instead of outward to get the ‘inflexed’ (inward-bending) arc shape. (b) Compass-only figure: yes — mark a centre O and draw a circle; without changing the radius, step the compass around the circle. The radius divides the circle into exactly 6 equal arcs (since each chord equals the radius and subtends 60°), giving the 6-point flower/petal design using only the compass. (c), (d), (e): each is built from circles/arcs of equal radius drawn from centres marked on a base line and its perpendicular bisector; keep one radius throughout so the overlapping arcs form the symmetric pattern shown.

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

SOLUTION What you notice: the lines or circles look bent, tilted or unequal even though they are actually straight, parallel and equal. This is a true optical illusion — the eye is misled by the surrounding pattern. Why it happens: the closely spaced background arcs/lines create misleading angle and length cues, so the brain misjudges direction and size. Measuring with a ruler/compass shows the lines really are straight/parallel and the lengths equal. Recreate: first draw the true straight/parallel lines accurately with a ruler, then add the slanted background strokes — the illusion appears once the background is added.

3. Construct this figure. [Hint: Find the angles in this figure.]

SOLUTION (method) Following the hint, first work out the angles the figure needs (for symmetric star/flower designs these are usually 360° shared equally, e.g. 60° for 6 directions or 45° for 8 directions). 1. Construct a 60° angle (equilateral-triangle method) or a 45° angle (bisect 90°) as required, and repeat around the centre to lay down the supporting lines at equal angles. 2. Draw equal arcs between the lines using one compass radius to complete the petals/points symmetrically.

4. Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P. [Hint: Find a line segment on l whose perpendicular bisector passes through P.] (Try This)

SOLUTION (method) 1. With P as centre and a suitable radius, draw an arc that cuts the line l at two points, say X and Y. Then PX = PY (both are the same radius), so P is equidistant from X and Y. 2. By the key property, P lies on the perpendicular bisector of XY. Construct that perpendicular bisector: from X and Y draw equal arcs that meet at a point Q on the other side of l. 3. Draw the line PQ. Since both P and Q are equidistant from X and Y, PQ is the perpendicular bisector of XY, hence PQ ⊥ l and passes through P — the required perpendicular from the external point P.

Figure it Out — Tangram (Page 156)

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

SOLUTION (method) A tangram has 7 pieces (2 large triangles, 1 medium triangle, 2 small triangles, 1 square and 1 parallelogram) cut from one square. Every figure must use all 7 pieces, flat, without overlaps or gaps, and pieces may be rotated or flipped. Strategy: place the two large triangles first — they usually form the biggest part (body/base) of the silhouette. Then fit the square and parallelogram into the straight-edged gaps, and finally slot the medium and two small triangles into the remaining corners. Because every tangram outline has the same total area as the original square, if a placement leaves a gap or an overlap, swap a triangle for the parallelogram (or flip a piece) until all 7 pieces fit exactly. (Cut out cardboard pieces and try — each given outline can be filled using all seven pieces.)

Figure it Out — Are the Tilings Possible? (Page 158)

1. Are the following tilings possible? (Region to be tiled / Tile shown)

SOLUTION Method — count and colour. Each 2 × 1 tile covers exactly 2 unit squares, so the region must contain an even number of unit squares. Next colour the region like a chessboard (black/white): each tile always covers one black and one white square, so the region must have equal numbers of black and white squares. If the region passes both tests (even total and equal black/white), a tiling is possible — show it by actually placing tiles. If the counts of black and white squares differ, no tiling is possible, no matter how the tiles are arranged. Apply the test to the given region: count its squares; if odd → impossible. If even, chessboard-colour it; if black ≠ white → impossible; if black = white → tile it directly to confirm it is possible.

2. Are the following tilings possible? (Region to be tiled / Tile shown)

SOLUTION Use the same two-step check. For an L-shaped or non-rectangular tile, the ‘equal-area’ rule still applies: the region’s area must be a multiple of the tile’s area (number of unit squares). Then look at fit at the corners and edges — sometimes the area divides evenly but a corner cell cannot be covered without overlapping. If you can place copies of the tile (rotating as allowed) to cover every cell once, the tiling is possible; if a forced gap or overlap always appears (shown clearly by the colouring/parity argument), it is not possible.

Math Talk & Try This — Answered

These are the in-text Math Talk and Try This tasks in the chapter; the determinate ones are answered, the open ones are guided.

Try This — Regular pentagon & hexagon How do we construct a regular pentagon (5-sided figure) and a regular hexagon (6-sided figure)? To begin with, try to construct a pentagon and hexagon with equal sidelengths. Answer. A regular hexagon is within reach: it is made of 6 congruent equilateral triangles meeting at the centre (6 × 60° = 360°), so construct a 60° angle (equilateral-triangle method) and repeat. The regular pentagon needs deeper triangle/pentagon theory studied in later years, so a perfect one can only be approximated for now — start by making the 5 sides equal with a compass.
Math Talk — Will the 70° angle fit the gap? Will the 70° angle fit into the gap? What is the gap angle ∠AOI? (Given 40° + 60° + 50° + 30° + 40° + 90° + gap angle = 360°.) Answer. Add the known angles: 40 + 60 + 50 + 30 + 40 + 90 = 310°. So the gap angle ∠AOI = 360° − 310° = 50°. Since the spare angle is 70° but the gap is only 50°, the 70° angle does NOT fit — it would overlap by 20°. (Angles fit around a point only when they add up to exactly 360°.)
Why AOD, BOE, COF are straight lines (Fig. 6.12) In Fig. 6.12 can you explain why AOD, BOE and COF are straight lines? Answer. At the centre O, the six equilateral triangles each contribute a 60° angle. For a pair of opposite triangles, the two angles on the same side of a line through O add to 60° + 60° + 60° = 180° on each side, i.e. the angle AOD is a straight angle (180°). So A, O, D are collinear; the same holds for B, O, E and C, O, F. These are the three ‘diagonals’ joining opposite vertices of the hexagon.
Math Talk — Copy an angle Draw an angle. Create a copy of this angle using only a ruler and compass. Answer. Draw an arc from the vertex cutting both arms (isosceles ΔABC). Draw an equal-radius arc from the new vertex X cutting its ray at Z. Transfer the chord BC onto that arc from Z to get Y (YZ = BC). Join XY. By SSS, ΔABC ≅ ΔXYZ, so ∠X = ∠A — an exact copy with no protractor.
6-Pointed Star — are the six triangles equilateral? Are the six triangles forming the 6 points of the star — ΔAGH, ΔBHI, ΔCIJ, ΔDJK, ΔELK, ΔFLG — equilateral? Why? [Hint: Find the angles.] Answer. Yes, all six are equilateral. The star is a regular hexagon (GHIJKL) with an equilateral triangle built on each side. Each side of the inner hexagon is equal, and the points are formed so that every angle of each small triangle is 60°. With all three angles 60° (and equal sides), each ΔAGH, ΔBHI, … is equilateral — which is why the star has 6-fold rotational symmetry.
Math Talk — Tile a 4 × 6 grid with 2 × 1 tiles? Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles? (We may rotate a tile.) Answer. Yes. A 4 × 6 grid has 24 unit squares = 12 tiles. The simplest way: fill each of the 6 columns with two vertical 2 × 1 tiles (4 rows = 2 tiles per column). It works because the number of rows (4) is even. Many other arrangements are also possible.
Math Talk — 4 × 7 and 5 × 7 grids Can a 4 × 7 grid be tiled using 2 × 1 tiles? What about a 5 × 7 grid? Answer. 4 × 7: Yes — 28 squares = 14 tiles; fill each of the 7 columns with two vertical tiles (4 rows are even). 5 × 7: No — it has 35 unit squares, an odd number. Each tile covers exactly 2 squares, and tiles can only cover an even total, so 35 squares can never be tiled by 2 × 1 tiles.
Math Talk — m × n grids (even/odd cases) Is an m × n grid tileable with 2 × 1 tiles if both m and n are even? If one is even and the other odd? If both are odd? Answer. Both even: Yes — cover each column with vertical tiles (even rows). One even, one odd: Yes — if the number of rows is even use vertical tiles in each column; if the number of columns is even use horizontal tiles in each row. Both odd: No — m × n is then odd (odd × odd = odd), and an odd number of squares can never be split into 2 × 1 tiles.
Math Talk — 5 × 3 grid with one square removed Here is a 5 × 3 grid with a unit square removed. Now it has an even number of unit squares. Is it tileable with 2 × 1 tiles? Answer. Having an even number of squares (14) is necessary but not enough. Chessboard-colour the grid: a full 5 × 3 grid has 8 squares of one colour and 7 of the other. Removing a square of the majority colour leaves 7 black + 7 white → tileable. Removing a square of the minority colour leaves 8 and 6 → not tileable, because every tile needs one of each colour.
Math Talk — Fig. 6.13 region; black-and-white argument Were you able to tile Fig. 6.13? How can we be sure that this is not tileable? If the plain grid is tileable, is the black-and-white grid tileable, and vice versa? Answer. The region of Fig. 6.14 has 8 white and 6 black squares. Every 2 × 1 tile covers exactly one black and one white square, so any tileable region must have equal black and white squares. Since 8 ≠ 6, this region can never be tiled. And yes — the plain grid is tileable exactly when the black-and-white grid is tileable; the colouring just makes the obstruction easy to see.
Math Talk — Another non-tileable removal Use this idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable. Answer. In a chessboard-coloured 5 × 3 grid the corners share the majority colour (8 of one colour, 7 of the other). Removing any square of the minority colour (one of the 7) leaves 8 vs 6 squares, which can never be tiled. For example, removing a square that is the same colour as the centre square (a minority-colour cell) makes the grid non-tileable.

Common Mistakes to Avoid

Watch out for these

  • Using different radii for the two arcs that meet at one point — that point will not be equidistant from X and Y, so it will not lie on the perpendicular bisector.
  • Changing the compass setting between the two arms when copying an angle — OA must equal OB, and the transferred chord must equal BC, or SSS congruence fails.
  • Confusing the supporting (construction) lines with the final figure — only the boundary should be inked; arcs and centres are guides.
  • Thinking an even number of squares guarantees a tiling — you also need equal black and white squares (e.g. removing the wrong square from a 5 × 3 grid).
  • Forgetting that a 2 × 1 tile may be rotated (vertical or horizontal) — both orientations are allowed.
  • Believing every angle is constructible — angles like 65.5° cannot be built from 60°/90° by bisection.

Practice MCQs & Assertion–Reason

1. The perpendicular bisector of a segment XY is the set of all points that are:

(a) closer to X    (b) closer to Y    (c) equidistant from X and Y    (d) on the segment XY

2. To construct a 45° angle with ruler and compass, you first construct a 90° angle and then:

(a) copy it    (b) bisect it    (c) double it    (d) add 60°

3. A 60° angle is constructed by drawing:

(a) a square    (b) a perpendicular bisector    (c) an equilateral triangle    (d) two parallel lines

4. The angle-bisection and angle-copy constructions are justified using:

(a) area formulas    (b) the SSS congruence condition    (c) the Pythagoras rule only    (d) measurement with a protractor

5. Each interior angle of a regular hexagon is:

(a) 60°    (b) 90°    (c) 108°    (d) 120°

6. A regular hexagon can be split into how many congruent equilateral triangles?

(a) 3    (b) 4    (c) 6    (d) 8

7. A 5 × 7 grid cannot be tiled by 2 × 1 tiles because it has:

(a) 35 (an odd number of) unit squares    (b) too few rows    (c) curved edges    (d) 36 unit squares

8. In a chessboard colouring, every 2 × 1 tile always covers:

(a) two black squares    (b) two white squares    (c) one black and one white square    (d) no square

9. “Covering a region using shapes without gaps or overlaps” is called:

(a) bisection    (b) tiling    (c) congruence    (d) symmetry

10. The ancient Indian texts giving rope methods for perpendiculars are the:

(a) Vedā&ndotbelow;gas only    (b) Purā&ndotbelow;as    (c) Śulba-Sūtras    (d) Siddhāntas

Answer key: 1-(c), 2-(b), 3-(c), 4-(b), 5-(d), 6-(c), 7-(a), 8-(c), 9-(b), 10-(c).

For each Assertion–Reason question, choose: (A) Both Assertion and Reason are true and the Reason is the correct explanation of the Assertion; (B) Both are true but the Reason is not the correct explanation; (C) Assertion is true but Reason is false; (D) Assertion is false but Reason is true.

A-R 1. Assertion: Any point equidistant from X and Y lies on the perpendicular bisector of XY.

Reason: Such a point forms two congruent triangles with X and Y, giving equal angles at the foot of the perpendicular.

A-R 2. Assertion: When copying an angle, the line OC bisects the angle.

Reason: Copying an angle uses the SSS congruence condition to reproduce the same angle measure.

A-R 3. Assertion: A 60° angle can be constructed using only a ruler and a compass.

Reason: Constructing an equilateral triangle creates an angle of 60° at each vertex.

A-R 4. Assertion: A 5 × 7 grid cannot be tiled with 2 × 1 tiles.

Reason: It has 35 unit squares, and each tile covers exactly 2 squares, so an odd total cannot be tiled.

A-R 5. Assertion: An m × n grid with an even number of unit squares can always be tiled by 2 × 1 tiles.

Reason: Every 2 × 1 tile covers one black and one white square in a chessboard colouring.

Answer key: 1-(A), 2-(D), 3-(A), 4-(A), 5-(D).

Quick Revision Summary

  • Bisection splits a segment (or object) into two identical parts; the perpendicular bisector is perpendicular and passes through the midpoint.
  • Any point equidistant from the endpoints of a segment lies on its perpendicular bisector — this is the basis of the ruler-and-compass construction.
  • A 90° angle is built by the perpendicular-bisector method; bisecting angles gives 45°, 30°, 15°, etc.
  • Angles can be bisected and copied using SSS/SAS triangle congruence; copying an angle lets us draw parallel lines.
  • A 60° angle comes from an equilateral triangle; six equilateral triangles (6 × 60° = 360°) make a regular hexagon (interior angle 120°).
  • Tiling covers a region without gaps or overlaps; 2 × 1 tiles need an even number of squares and equal black/white squares.
  • The black-and-white (chessboard) colouring is a quick way to prove some regions cannot be tiled.

How to score full marks in this chapter

For every construction, write the steps in order and state the congruence (SSS/SAS) that makes it correct — examiners give marks for the reason, not just the drawing. Keep arcs faint and ink only the final figure. For special angles, show how you reached them (90° → 45°, 60° → 30° → 15°). In tiling questions, always do the two-step check: first count the squares (must be even), then chessboard-colour and compare black vs white — this single argument settles most ‘is it tileable?’ problems.

Frequently Asked Questions

What is Class 7 Maths Ganita Prakash Chapter 14 about?

Chapter 14, Constructions and Tilings (Ganita Prakash Part II, Chapter 6), teaches ruler-and-compass constructions — the perpendicular bisector, 90° and 60° angles, angle bisection, copying an angle, parallel lines and a regular hexagon — along with arch and star designs, the Śulba-Sūtra rope methods, tangrams, and tiling regions and the plane.

How many Figure it Out exercises are in Chapter 14?

There are several “Figure it Out” sets across the chapter (on the perpendicular bisector, the Śulba rope method, angle bisection, copying an angle, parallel lines, the pointed arch, designs and the perpendicular through an external point, the tangram, and tiling), plus many Math Talk and Try This tasks — all solved on this page.

How do you construct a perpendicular bisector with a compass?

With one fixed radius, draw arcs from each endpoint of the segment above the line meeting at a point A, and below the line meeting at a point B. Both A and B are equidistant from the endpoints, so the line through A and B is the perpendicular bisector.

When can a region be tiled with 2 × 1 tiles?

The region must have an even number of unit squares (each tile covers 2), and when chessboard-coloured it must have equal numbers of black and white squares (each tile covers one of each). If either test fails, the region cannot be tiled.

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