NCERT Solutions for Class 7 Maths Chapter 14 Constructions and Tilings (PDF) – 2027

From X and from Y, draw arcs of the same radius above XY and below XY. The upper arcs meet at A and the lower arcs meet at B. Then AX = AY = BX = BY.

To draw the eye: with centre A draw the arc from X to Y (the lower arc); with centre B draw the arc from X to Y (the upper arc).

AB cuts XY at its midpoint, and the angle between them is 90°.

Yes, always. Only AX = AY = BX = BY is used, and that holds for any length of XY and any radius for which the arcs meet.

ΔAOX and ΔAOY.

Keep the same XY, but take a different pair of centres C and D on the perpendicular bisector of XY, with CX = CY = DX = DY. Draw the lower arc from C and the upper arc from D.

Yes. C and D must lie on the line AB.

Let P be any point with PX = PY. We show P is on the perpendicular bisector AB.

1. Open the compass to some fixed radius, more than half of XY. With centre X draw a long arc above XY; with the same radius and centre Y draw another long arc above XY. Call their crossing point A.
2. With the same radius, draw the matching pair of arcs below XY. Call their crossing point B.
3. Join A and B with the ruler. AB is the perpendicular bisector of XY.

No, it is not necessary. The upper pair of arcs may use one radius and the lower pair a different radius. AB is still the perpendicular bisector.

Both pairs may be drawn on the same side — provided the two pairs use different radii, so that they cross at two different points.

Yes — within one pair the two radii must be equal. This is the one thing you cannot change.

The design is four equal eyes (petals) meeting at one point O.

1. Construct the perpendicular bisector of a segment; call the crossing point O. This gives two perpendicular lines.
2. With centre O, cut the four half-lines at P, Q, R, S, so OP = OQ = OR = OS.
3. On the perpendicular bisector of OP mark C and D, one on each side, equally far from OP. Draw the arcs from O to P with centres C and D: one petal.
4. Repeat for OQ, OR and OS with the same compass settings; rub out the supporting lines.

Yes. First make O the midpoint of a segment XY; then the perpendicular bisector of XY passes through O.

1. Extend the line on both sides of O.
2. With centre O and any radius, cut the line at X and Y. Now OX = OY.
3. With a larger radius, draw arcs from X and from Y that meet above the line at A.
4. Join AO. Then ∠AOX = ∠AOY = 90°.

Extend the line on both sides of O. Set the compass to any convenient opening, put the point at O, and cut the line on both sides — at X and at Y.

No — one pair is enough.

Because the rope gives exactly the four equal distances we need.

One method: make the given point a midpoint (the same idea as on paper). Let O be the given point on the line.

1. With a short rope held at O, mark equal lengths on both sides of O along the line: pegs at X and Y, so OX = OY.
2. Tie a longer rope to the pegs at X and Y, fold it to find its midpoint, and pull the midpoint taut to one side. Mark it A.
3. Stretch a rope from A to O. ∠AOX = 90°, since AX = AY and OX = OY.

Draw the eight supporting lines first, then draw one petal on each.

1. Draw a line through a point O and construct the perpendicular at O. This gives 4 rays at 90°.
2. Bisect each of the four right angles. Now there are 8 rays, 45° apart.
3. With centre O and one radius, cut all eight rays at the same distance from O.
4. On each of these eight segments draw an eye-shaped petal (two arcs with centres on its perpendicular bisector), keeping the compass settings the same.
5. Trace the petals and rub out the supporting lines.

45°.

Construct a 90° angle, then bisect it.

1. At O construct the perpendicular OA, so ∠AOX = 90°.
2. With centre O and any radius, cut OA at P and OX at Q. So OP = OQ.
3. With the same (or any equal) radius, draw arcs from P and from Q crossing at R.
4. Join OR. Then ∠ROX = 45°.

Make the three pairs of sides equal, and SSS does the rest.

Use the same three steps for every angle — the method does not care how big the angle is.

1. Draw ∠AOB (try one acute, one right, one obtuse and one very wide obtuse angle).
2. With centre O and one radius, cut both arms at P and Q, so OP = OQ.
3. With one equal radius, draw arcs from P and from Q meeting at R.
4. Join OR. It is the bisector.

Eight rays 45° apart, then eight identical petals.

1. Mark O on a line and construct the perpendicular at O (4 rays, 90° apart).
2. Bisect each right angle (8 rays, 45° apart).
3. With centre O and one radius, cut all eight rays at P₁, P₂, …, P₈.
4. For OP₁: construct its perpendicular bisector, mark two centres C and D on it at equal distances from OP₁, and draw the arcs from O to P₁ with centres C and D. That is one petal.
5. Repeat with the same compass settings on the other seven rays; rub out the supporting lines.

Yes. The line through C and O still bisects ∠AOB: the part of the line CO extended beyond O cuts ∠AOB into two equal angles.

Starting from 90° and 60°, repeated bisection gives many angles, and adding or subtracting them gives more. But 65.5° cannot be constructed this way.

Use the rope as both compass and ruler, and copy the compass steps.

1. With a rope of fixed length held at O, mark P and Q on the two arms, so OP = OQ.
2. With a second rope of fixed length, sweep arcs from P and from Q; peg their crossing point R, so PR = QR.
3. Stretch a rope from O through R. OR bisects ∠AOB (ΔOPR ≅ ΔOQR by SSS).

The four petals are largest when each arc is a semicircle drawn on a side of the square as diameter.

1. Construct the square ABCD.
2. Construct the perpendicular bisectors of the sides. They give the midpoints of the four sides and meet at the centre O.
3. With centre at the midpoint of AB and radius half of AB, draw the semicircle inside the square. It passes through A, O and B.
4. Repeat from the other three midpoints with the same radius. The overlaps of the semicircles are the four petals.

The figure is one fan-shaped unit (two equal arms, closed by an arc drawn from the point where the arms meet), repeated with its point alternately down and up. Neighbouring units share an arm.

To construct it, copy the unit exactly: the arm length with the compass, and the angle between the arms by copying the angle (next question). Full steps: Figure it Out Q2, page 147.

Turn the angle into a triangle, then rebuild the same triangle at the new place.

1. Let the angle be at A. Draw a ray from a new point X.
2. With centre A and any radius, draw an arc cutting the arms at B and C.
3. With the same radius and centre X, draw an arc cutting the ray at Z.
4. Set the compass to BC. With centre Z, cut the new arc at Y, so ZY = BC.
5. Join XY. Then ∠YXZ = ∠BAC.

Draw four angles freely (one narrow, one nearly a right angle, one wide, one opening downwards) and copy each by the arc-and-chord method:

1. Draw a ray from a new point X, pointing any way you like.
2. Draw an arc of radius r from the vertex A, cutting the arms at B and C, and an arc of the same radius r from X, cutting the ray at Z.
3. Set the compass to BC and, from Z, cut the second arc at Y.
4. Join XY. Then ∠YXZ = ∠BAC.

Build one fan-shaped unit, then copy it, turning every second copy upside down. Neighbouring units share an arm.

1. From a point V₁ draw two upward arms V₁P and V₁Q of equal length l with the chosen angle between them; with centre V₁ draw the arc PQ.
2. At Q copy the angle ∠PV₁Q on the other side of QV₁ and cut QV₂ = l; with centre Q draw the arc V₁V₂.
3. At V₂ copy the angle again, using V₂Q as one arm, cut V₂R = l, and draw the arc QR with centre V₂.
4. Continue, alternating up and down.

Draw a transversal and copy the corresponding angle at the new point.

1. Let m be the given line. Draw a line l crossing m at A, and choose B on l.
2. With one radius draw an arc from A cutting m at C and l at D, and an arc of the same radius from B cutting l at E.
3. Set the compass to CD and, from E, cut the arc at F.
4. Join B and F and extend it. This line n is parallel to m.

Repeat the copy-the-corresponding-angle construction four times, choosing a different slant for the given line each time.

The figure is eight congruent rhombi around the centre O (for example OHSA), each with a 45° angle at O and its opposite corner at a tip.

1. Through O draw a line and its perpendicular; bisect the four right angles: 8 rays, 45° apart.
2. With one compass opening, cut the rays at A, B, C, D, E, F, G, H (A on the upward ray).
3. Through H draw the line parallel to OA, and through A the line parallel to OH. They meet at the tip S.
4. Repeat for each pair of neighbouring rays to get T, U, …, Z. Join O to each tip and shade alternate triangles.

They were first drawn full size on a flat surface such as paper, a board or the stone itself, using a straight edge and a compass (or a peg and a cord for a large arch). The stones were then cut to follow the drawn arcs.

First draw the support lines: a base AD, and two equal slanting lines AB and DC with ∠BAD = ∠CDA (how to construct them is the next question). Then draw three arcs on this frame:

1. a small arc from A to B,
2. a matching small arc (same radius) from D to C,
3. a large arc from B to C, with its centre on the perpendicular bisector of BC.

One length carried by the compass, one angle carried by the copy-an-angle construction.

With the frame ABCD ready, draw three arcs, always changing the two side arcs together.

1. Left lobe: take a centre on the perpendicular bisector of AB and draw the arc from A to B.
2. Right lobe: take the matching centre on the perpendicular bisector of CD (same radius) and draw the arc from D to C.
3. Top lobe: take a centre on the perpendicular bisector of BC and draw the arc from B to C.
4. If the arch looks too flat or too round, change the radii (the two side radii equally) and redraw.

The supporting lines are two line segments of equal length, PT and QT, meeting at the top point T. On each of them we draw a wavy wave (as in Grade 6): two arcs, one bulging outwards and one bulging inwards.

Yes. The midpoint M splits PT into two equal halves, PM and MT, and the arch is made of one arc on each half.

1. Construct the perpendicular bisector of PM and take a centre on it inside the arch; draw the arc from P to M. It bulges outwards.
2. Construct the perpendicular bisector of MT and take a centre on it outside the arch, at the same distance from MT; draw the arc from M to T. It bulges inwards.
3. Do the same on QT with the midpoint N, using the same radius.

Keep the support lines PT and QT (with midpoints M and N) fixed. On each half (PM, MT, QN, NT) draw an arc, bulging outwards on the lower half and inwards on the upper half, all with the same radius. Then repeat with a different radius.

start from a symmetric frame, then choose the arcs.

1. Draw the base AD, construct equal angles at A and D, and cut equal lengths AB = DC.
2. Choose the lobes: one big arc gives a round arch, three arcs give a trefoil, five give a cinquefoil.
3. Choose the curve: arcs bulging outwards look full and round; outward-then-inward arcs give a pointed arch.
4. For every arc on the left, draw a matching arc on the right with the same radius. Ink the outline and rub out the support lines.

Making the sides equal is easy: set the compass to one length and step it round, turning a little each time. But the angles then usually come out unequal, so the figure is not regular. A regular hexagon needs every angle to be 120°, which we can construct from 60°. A regular pentagon needs ideas from later classes.

Yes. Joining the centre O to the six corners splits a regular hexagon into six equilateral triangles, and an equilateral triangle can be constructed with ruler and compass. The next questions show why this works, by starting from six equilateral triangles.

The three long diagonals AD, BE and CF pass through one point O and cut the hexagon into six equilateral triangles. The book checks this by working the other way round: it starts with six equilateral triangles and shows that they fit around a point and make a regular hexagon (next question).

Yes to both.

The gap angle is 50°, so the 70° angle does not fit — it is 20° too wide.

Because the three 60° angles on one side of each line add up to a straight angle.

1. Mark a point O. With the compass set at 4 cm, draw a circle with centre O.
2. Mark any point A on the circle.
3. Without changing the compass opening, put the point at A and cut the circle at B. From B cut at C, from C at D, from D at E, from E at F.
4. F should land exactly 4 cm from A. Join A–B–C–D–E–F–A with the ruler.

Construct a 60° angle at a point on a line — then the angle on the other side of it is automatically 120°.

Build an equilateral triangle. Its angles are 60°, so the corner you make is the angle you want.

1. Suppose the 60° angle is needed at A on the segment AX. With centre A and any radius r, draw an arc cutting AX at B.
2. Without changing the radius, put the compass point at B and cut the first arc at C.
3. Join A to C. Then ∠CAX = 60°.

Yes — ΔABC is equilateral, and that is exactly why the angle is 60°.

Same method as the 4 cm hexagon, with the compass set to 5 cm.

1. Draw a circle of radius 5 cm with centre O.
2. Mark a point A on it.
3. Keeping the compass at 5 cm, step round the circle: A → B → C → D → E → F. The sixth step returns to A.
4. Join the six points in order.

Another way: draw AB = 5 cm, construct a 120° angle at B (60° on one side of the line gives 120° on the other) and cut BC = 5 cm; repeat at C, D, E and F until the figure closes.

Bisect 60° to get 30°, then bisect 30° to get 15°.

1. Construct ∠CAX = 60° with the equilateral-triangle method.
2. With centre A cut both arms at P and Q (AP = AQ). With one equal radius cut arcs from P and Q meeting at R. Join AR — then ∠RAX = 30°.
3. Bisect ∠RAX the same way: cut the two arms of the 30° angle, cross two equal arcs, join. That ray makes 15° with AX.

Yes, there is a hexagon: the six inner points G, H, I, J, K, L form a small regular hexagon, and the six tips A, B, C, D, E, F form a bigger one. One way to construct the star:

1. Draw a circle with centre O. Step its radius round the circle to mark A, B, C, D, E, F (the corners of a regular hexagon).
2. Join A, C and E to get one equilateral triangle, and B, D and F to get another.
3. The two triangles cross at G, H, I, J, K, L. Ink the outline of the star.

Yes, all six are equilateral. Take ΔAGH. A, B, C, D, E, F are the corners of a regular hexagon (each angle 120°, all sides equal).

(a) Draw two equal vertical sides with tops P and Q, and a point T on the perpendicular bisector of PQ. On PT and QT draw the pointed arch of page 150: an arc bulging outwards up to the midpoint, then one bulging inwards up to T.

(b) Compass only. Draw a circle of radius r; step r round it to mark six points. For neighbouring points P, Q, draw arcs of radius r outwards from Q (centre P) and from P (centre Q); they meet at a petal tip.

(c) Draw a circle, step its radius round it six times and join the six points.

(d) Mark six points as in (c). With each as centre, draw a circle of radius half of r; neighbouring circles just touch.

(e) Divide a regular hexagon of side 2 units into 24 equilateral triangles of side 1 unit. In each small triangle, join its centre (where two angle bisectors meet) to its three corners.

You see a bright white triangle sitting on top of the figure — but there is no triangle drawn at all. Only three notched circles and three bent corners are on the page.

To recreate it: construct an equilateral triangle lightly. At each corner draw a small filled circle and cut out the 60° wedge that faces into the triangle. Draw the three corners (V-shapes) of an upside-down triangle between them, then rub out the construction lines.

Draw a regular hexagon, then draw the two equilateral triangles that join its alternate vertices. The star’s six points land exactly on the hexagon’s six corners.

1. Draw a circle of radius r, and step r around it to mark A, B, C, D, E, F. Join them — the regular hexagon.
2. Join A–C–E–A. Join B–D–F–B. Two equilateral triangles.
3. Ink the hexagon and the star; rub out the parts of the triangles inside the small hexagon.

Make P equidistant from two points of l. Then P is on their perpendicular bisector, which is the perpendicular you want.

1. With centre P and a radius big enough, draw an arc cutting l at X and Y. Now PX = PY, so P is on the perpendicular bisector of XY.
2. With equal radii, draw arcs from X and from Y on the other side of l, meeting at Q.
3. Join P and Q. PQ is perpendicular to l and passes through P.

Each figure uses all seven pieces exactly once, with no gaps and no overlaps. One way to make each figure is shown below.

Yes. It takes 12 tiles, and there are many ways to do it.

An easy tiling. The grid has 4 rows, an even number. So fill every column from top to bottom with 2 vertical tiles: 2 tiles per column × 6 columns = 12 tiles. No gaps, no overlaps.

Yes. It takes 14 tiles.

No — a 5 × 7 grid can never be tiled with 2 × 1 tiles.

Yes, always. Fill each column with vertical tiles.

Yes. Tile along the even direction.

No — never.

Examples. 3 × 3 = 9, 5 × 7 = 35, 7 × 9 = 63 — all odd, all impossible.

Yes. With the top-right corner square removed, 14 squares are left and 7 tiles cover them.

A tiling. The top row now has only 2 squares — cover them with one horizontal tile. What is left is a full 4 × 3 block (rows 2 to 5). Each of its 3 columns is 4 squares tall, so it takes 2 vertical tiles. That is 6 more tiles.

No. This is the complete 5 × 3 grid, with nothing removed.

It cannot be tiled — even though it has an even number of squares.

Counting alone cannot settle this, because the count passes the test. Something finer is needed: the black-and-white colouring of the next question.

Colour the grid like a chessboard and count the two colours.

Another square: in the full 5 × 3 grid (corners white) there are 8 white and 7 black squares. Removing any black square leaves 8 and 6, so the region cannot be tiled: for example the middle square of the bottom row, the centre square, or an end square of the 2nd or 4th row.

Yes. Take the tiling of the plain grid and just look at it in colour.

Yes. Rub the colours out and the same arrangement is a tiling of the plain grid.

No. The colours do not balance.

Remove any square of the minority colour — for example the middle square of the bottom row.

The 7 black squares are: the middle square of the top row (Fig. 6.13), the two end squares of the 2nd row, the centre square, the two end squares of the 4th row, and the middle square of the bottom row. Removing any one of them makes the grid non-tileable.

Yes, it is possible. Four L-shaped tiles cover the region exactly.

No — this one is impossible, even though the count of squares is even.

A square is the easiest one. Lay squares edge to edge in rows and columns and they cover the whole plane, with no gap and no overlap — like floor tiles in a room.

Yes: equilateral triangles and regular hexagons. Together with squares, these are the only regular polygons that can tile the plane.

Yes. For example:

• Floors and walls: square tiles, hexagonal paving blocks, brick walls.
• Designs: repeating patterns on cloth, jaali screens and rangoli.
• Nature: the front face of a beehive and some wasp nests are tiled with hexagonal cells, used to keep eggs, larvae and pupae safe and to store food.