NCERT Solutions for Class 7 Mathematics Ganita Prakash II Chapter 6
NCERT solutions for Class 7 Maths Chapter 6: Constructions and tilings – complete answers and explanations
Are you looking for step-by-step NCERT solutions for Class 7 Maths Chapter 6: Constructions and tilings from Ganit Prakash II? This chapter is a fascinating blend of geometric construction and logical reasoning — two skills that lie at the heart of Maths at this level. Students learn how to draw precise geometric shapes using only a ruler and compass, explore the beauty of angle bisection, arch designs, regular hexagons, and discover the mathematical world of tiling patterns. Download the worksheet and practice alongside these solutions for better clarity and learning. You can also book a free trial now to get expert guidance and improve writing and comprehension skills. This blog provides complete NCERT solutions for Class 7 Maths, strictly based on the worksheet, to help students check their work, understand each answer clearly, and build confidence for their exams.
What this NCERT chapter covers?
1. Perpendicular bisector: Students learn what a perpendicular bisector is, how to construct it using a ruler and compass, and how to justify the construction through congruence.
2. Construction of 90° angles: The chapter shows how the perpendicular bisector method can be extended to draw a 90° angle at any point on a line.
3. Construction methods from the Śulba-Sūtras: Students explore the ancient Indian rope-based method from the Kātyāyana-Śulbasūtra for constructing perpendicular bisectors, connecting geometry to cultural history.
4. Angle bisection: A general method to bisect any angle using a compass is developed, with justification through the SSS congruence condition.
5. Copying an angle: Students learn how to create an exact copy of any given angle using only a ruler and compass.
6. Constructing parallel lines: The angle-copying method is applied to draw a line parallel to a given line.
7. Arch designs: The chapter explores how trefoil arches and pointed arches can be constructed geometrically using support lines, angle constructions, and arcs.
8. Regular hexagons: Students construct regular hexagons using equilateral triangles and the 60° angle method, and investigate related shapes like the 6-pointed star.
9. Tiling: The second half of the chapter introduces the concept of tiling — covering a region without gaps or overlaps — using shapes like tangram pieces, 2 × 1 tiles, and explores tiling the entire plane with squares, equilateral triangles, and regular hexagons.
10. The black-and-white colouring argument: Students use a checkerboard colouring method to prove whether a region can or cannot be tiled with 2 × 1 tiles.
How to use these NCERT solutions?
1. Attempt every question on your own first before checking the answers here. This is the most important habit for building strong Maths skills at the Class 7 level.
2. Use these NCERT solutions to verify your answers section by section, in the exact order they appear in the worksheet.
3. For construction-based questions, compare your steps carefully — focus on which points are drawn, which radius is used, and what the final result should look like.
4. Parents and teachers can use these solutions to explain each step to students, especially the justification-type questions which require reasoning rather than just drawing.
5. For tiling questions, check whether you have used the correct logic (odd/even number of squares, checkerboard colouring argument) rather than just trying to fit tiles visually.
6. All answers follow the exact order of the worksheet, making it easy to locate any specific question quickly.
Important tips and tricks for students
1. Always use a sharp pencil and a well-functioning compass for construction questions — even a small slip can change the answer.
2. In perpendicular bisector questions, remember that any two points equidistant from X and Y lie on the perpendicular bisector. This single property explains almost every variation in these questions.
3. For angle bisection and angle copying, the key is to use the same radius consistently. Different radii will produce incorrect results.
4. When constructing a 60° angle, remember that you are essentially forming one side of an equilateral triangle — the method is simple and elegant.
5. For tiling questions, always start by counting the total number of unit squares. If it is odd, a 2 × 1 tiling is immediately impossible. If it is even, use the checkerboard colouring argument to confirm.
6. Do not skip the justification steps in "Figure it Out" questions. Class 7 NCERT expects students to explain their reasoning using congruence, angle properties, and logical arguments — not just draw the figure.
7. In arch design questions, support lines are the key — always set them up correctly before drawing any arcs.
NCERT solutions – complete answer key
6.1 Figure it Out (Perpendicular Bisector)
Question 1: When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY?
Answer:
Yes, it is not necessary to have the same radius above and below XY.
Any two points equidistant from X and Y will lie on the perpendicular bisector.
Hence, the line joining them will still be the perpendicular bisector.
Question 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?
Answer:
Yes, arcs can be constructed on the same side of XY.
As long as the points obtained are equidistant from X and Y, joining them gives the perpendicular bisector.
Question 3: While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them?
Answer:
No, it is necessary to use the same radius.
If different radii are used, the points will not be equidistant from X and Y, so the line will not be the perpendicular bisector.
Question 4: Recreate this design using only a ruler and compass.
Answer:
Explanation:
Draw the design step-by-step using a ruler and compass by constructing equal arcs and lines symmetrically as shown in the figure.
Figure it Out (Śulba-Sūtras / Rope Method)
Question 1: Justify why AB in Fig. 6.4 is the perpendicular bisector.
Answer:
In the rope method, the rope is folded in half so that its midpoint is marked.
The two ends of the rope are fixed at X and Y. The length from one end to the midpoint equals the length from the other end to the midpoint — both equal half the rope's length (excluding loop parts).
So when the midpoint is pulled to position A (above XY), we have AX = AY.
Similarly, when pulled to position B (below XY), we have BX = BY.
Since both A and B are equidistant from X and Y, they lie on the perpendicular bisector of XY.
Hence, line AB is the perpendicular bisector of XY.
Question 2: Can you think of different methods to construct a 90° angle at a given point on a line using a rope?
Answer:
Method 1: Mark point O on the line. Fix equal lengths from O to X and Y on the line. Pull the rope midpoint perpendicular to the line using the technique from Śulba-Sūtras. The vertical rope gives a 90° angle at O.
Method 2: Use a rope to form an isosceles triangle with its apex directly above O. The line from the apex to O will be perpendicular to the base, giving a 90° angle.
Method 3: Use the 3-4-5 right triangle property with the rope — mark lengths 3, 4, and 5 units, and form a triangle. The angle between the sides of length 3 and 4 will be 90°.
Figure it Out (Angle Bisection)
Question 1: Construct at least 4 different angles. Draw their bisectors.
Answer:
Explanation:
Step 1: Draw an angle ∠XOY of any measure (e.g., approximately 60°, 90°, 120°, 150°).
Step 2: With centre O, draw an arc cutting both arms at A and B (OA = OB).
Step 3: With equal radius from A and B, draw intersecting arcs. Mark the intersection as C.
Step 4: Draw ray OC — this is the angle bisector.
Repeat for at least 4 different angles in different orientations.
Question 2: Construct the 8-petalled figure shown in Fig. 6.5.
Answer:
Explanation:
Step 1: Draw a horizontal line through centre O.
Step 2: Construct a 90° angle at O (perpendicular bisector method) to get a vertical line.
Step 3: Bisect each 90° angle to get 45° angles. This gives 4 lines through O at 45° intervals — 8 rays in total (360° ÷ 8 = 45° each).
Step 4: On each adjacent pair of rays, use the eye-construction method — choose two points as centres on the angle bisector and draw arcs between the two rays to form a petal shape.
Step 5: Repeat for all 8 pairs of adjacent rays to complete the 8-petalled figure.
Question 3: In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, will the line OC still be an angle bisector?
Answer:
Yes, OC will still be the angle bisector.
If arcs of equal radius are drawn on the other side (outside the angle), they still intersect at a point C' such that C'A = C'B and OA = OB.
By SSS congruence, ∆OBC' ≅ ∆OAC', so ∠BOC' = ∠AOC'.
The ray OC' is the extension of OC (the angle bisector), so line OC still bisects the angle (it is the same line, just the point C is on the opposite side).
Question 4: What are the other angles that can be constructed using angle bisection? Can you construct a 65.5° angle?
Answer:
Using a ruler and compass, we can construct 90°, and then repeatedly bisect to get:
90° → 45° → 22.5° → 11.25° etc.
We can also construct 60°, and bisect to get:
60° → 30° → 15° → 7.5° etc.
Combining these, angles like 90° + 45° = 135°, 60° + 30° = 90°, etc. can be formed.
65.5° = 60° + 5.5° — since 5.5° cannot be obtained by bisecting constructible angles from standard constructions (it requires trisecting 16.5° which is not constructible by ruler and compass alone), 65.5° cannot be constructed using only ruler and compass.
Question 5: Come up with a method to construct the angle bisector using a rope.
Answer:
Step 1: Fix a peg at vertex O of angle ∠XOY.
Step 2: Using a rope of fixed length, mark equal distances OA and OB along the two arms of the angle (OA = OB).
Step 3: Fix pegs at A and B.
Step 4: Take a rope, fold it in half, and fasten each end to pegs A and B.
Step 5: Pull the midpoint of the rope tightly away from O. The midpoint C lies on the angle bisector.
Step 6: The line OC bisects angle ∠XOY.
Question 6: Construct the following figure (4-petalled design inside a square).
Answer:
Explanation:
Step 1: Draw a square PQRS using ruler and compass.
Step 2: Find the midpoints of all four sides using the perpendicular bisector method.
Step 3: Draw both diagonals of the square; they intersect at the centre O.
Step 4: Each petal is formed between two adjacent midpoints of sides, using arcs centred at the opposite vertices of the square.
Step 5: To maximise petal size, use the side length of the square as the radius for each arc.
Step 6: Draw four arcs, each from one vertex, forming one petal between adjacent midpoints. The four arcs together form the 4-petalled design inside the square.
Figure it Out (Copying Angles)
Question 1: Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.
Answer:
Step 1: Draw 4 different angles freely (without a protractor) — e.g., an acute angle, an obtuse angle, a reflex-like angle, and an angle in a different orientation (tilted).
Step 2: For each angle at vertex A:
(a) Draw an arc from A with any radius, cutting both arms at B and C.
(b) On a new ray from X, draw an arc of the same radius.
(c) Use compass to transfer length BC; mark Y.
(d) Draw ray XY. Angle at X equals angle at A.
Repeat for all 4 angles to get their copies.
Question 2: Construct Fig. 6.6.
Answer:
Explanation:
Fig. 6.6 shows a repeating unit (a sector/petal shape) repeated in two orientations.
Step 1: Draw the basic unit — a sector with two straight arms and an arc at the top.
Step 2: Copy this angle exactly using the angle copying method.
Step 3: Place the second unit so that its right arm coincides with the left arm of the first unit (shared arm), but oriented in the opposite/mirror direction.
Step 4: Repeat to build the chain of repeating units as shown.
Ensure all arm lengths are equal (use compass) and all angles are identical (use copied angle construction).
Figure it Out (Parallel Lines)
Question 1: Construct 4 pairs of parallel lines in different orientations.
Answer:
Explanation:
For each pair:
Step 1: Draw a line m in any orientation (horizontal, vertical, diagonal, etc.).
Step 2: Draw a transversal l cutting m at point A.
Step 3: Mark a point B on l.
Step 4: Copy the corresponding angle at B using the angle copying method.
Step 5: Extend the new line through B — this is the parallel line n || m.
Repeat for 4 different orientations.
Question 2: Construct the following figure (8-pointed star).
Answer:
Explanation:
Step 1: Draw a horizontal line and mark its centre O.
Step 2: Construct a vertical line through O (90° angle).
Step 3: Bisect each 90° angle to get 45° lines — giving 8 equally spaced directions (rays) from O.
Step 4: On each ray, mark equal lengths from O using a compass (say OA = OB = OC ... = OH for inner points, and equal extended lengths for outer points ST, UV, WX, YZ).
Step 5: Each spike of the star is formed by connecting adjacent inner and outer points with straight lines.
Step 6: Connect points as labelled in the figure to form the 8-pointed star, shading alternate triangular portions as shown.
Trefoil Arch
How did they make these arches?
Answer:
The arches are drawn on a plane surface (paper or stone) using geometric constructions.
The trefoil arch uses support lines with AB = CD and ∠BAD = ∠CDA for symmetry.
To construct: Draw base line AD. Construct equal angles at A and D. Mark B and C such that AB = CD. Use these support lines to draw arcs from suitable centres to form the arch.
Adjust the radii of the arcs for aesthetic appearance.
How would you construct the support lines for the trefoil arch?
Answer:
Step 1: Draw base line AD (horizontal).
Step 2: At A, construct an angle ∠BAD (e.g., 60° or as desired) using the angle bisection/copy method.
Step 3: At D, copy the same angle ∠CDA = ∠BAD on the same side.
Step 4: Mark B on the ray from A and C on the ray from D such that AB = CD (using compass).
Step 5: BC forms the top support line.
Use A, B, C, D as support points to draw the three arcs of the trefoil arch.
Use these support lines to construct an arch.
Answer:
Explanation:
After marking support points A, B, C, D:
Step 1: The central arc is drawn from the midpoint of BC as centre, with BC/2 as radius (or adjusted for appearance).
Step 2: The left arc is drawn from B as centre with a suitable radius.
Step 3: The right arc is drawn from C as centre with the same radius.
Step 4: Adjust radii so the arcs meet smoothly to form a pleasing trefoil arch.
Pointed Arch
How do we construct a pointed arch?
Answer:
The supporting lines for a pointed arch are two line segments of equal length arranged in a V-shape (like two arms of the 'Wavy Wave' from Grade 6).
Step 1: Draw two equal line segments meeting at a point (like an inverted V), forming the two arms of the arch.
Step 2: Mark the midpoints of both arms.
Step 3: Using the midpoints as reference, draw arcs from the far ends of each arm, using the arm length as radius.
Step 4: The two arcs cross at the top, forming the pointed tip of the arch.
If their midpoints are marked, will you be able to construct a pointed arch?
Answer:
Yes. The midpoints serve as the centres from which the arcs of the pointed arch are drawn.
From the midpoint of the left arm, draw an arc with radius equal to half the arm length to form the left curve.
From the midpoint of the right arm, draw an arc with the same radius to form the right curve.
The two arcs meet at the top, creating the pointed arch shape.
Figure it Out (Arch Designs)
Question 1: Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs.
Answer:
Explanation:
Step 1: Draw two equal line segments (support lines) forming an inverted V-shape. Mark the midpoints M1 and M2 of each arm.
Step 2: Using M1 as centre, draw an arc from the base of the left arm curving upward.
Step 3: Using M2 as centre, draw an arc from the base of the right arm curving upward.
Step 4: The two arcs meet at the top, forming the pointed arch.
To make different arches: change the radius of the arcs (a smaller radius gives a more sharply pointed arch; a larger radius gives a flatter one).
Question 2: Make your own arch designs.
Answer:
Explanation:
Students should create original arch designs by:
Step 1: Drawing support lines (horizontal base and slanted arms of varying angles).
Step 2: Using the perpendicular bisector and angle bisection techniques to ensure symmetry.
Step 3: Drawing arcs from suitable centres to form arches of different shapes — rounded, pointed, trefoil, or multi-lobed.
Step 4: Combining multiple arch units for a decorative design.
Regular Hexagons
How do we construct a regular hexagon?
Answer:
A regular hexagon can be constructed by arranging six congruent equilateral triangles around a common centre.
Each angle of an equilateral triangle is 60°. Six such angles at the centre add up to 6 × 60° = 360°, so the triangles fit exactly around the centre without gaps or overlaps.
Can six congruent equilateral triangles be placed together as in Fig. 6.12? If yes, will it result in a regular hexagon?
Answer:
Yes. When six congruent equilateral triangles are placed with all their vertices meeting at a common centre O (as in Fig. 6.12 with vertices A, B, C, D, E, F around O):
Each interior angle of the hexagon = 60° + 60° = 120° (angles of two adjacent equilateral triangles).
All sides are equal (each side is one side of an equilateral triangle).
Since all sides and all angles are equal, the resulting figure is a regular hexagon.
Consider the figure with angles 40°, 60°, 50°, 30°, 40°, 90° around a point. Will the 70° angle fit into the gap?
Answer:
Sum of given angles = 40° + 60° + 50° + 30° + 40° + 90° = 310°
Gap angle ∠AOI = 360° − 310° = 50°
Since the gap is 50° and the remaining angle is 70°, the 70° angle does NOT fit into the gap.
In Fig. 6.12, can you explain why AOD, BOE and COF are straight lines?
Answer:
In Fig. 6.12, the six equilateral triangles are arranged symmetrically around centre O.
O is the common vertex, and the triangles are congruent.
Taking triangle OAB and triangle ODE: since the six triangles together span 360° around O, and each triangle occupies 60°, the triangle directly opposite each triangle is separated by 3 × 60° = 180°.
A straight angle is 180°, so A, O, D are collinear → AOD is a straight line.
Similarly, B, O, E are collinear → BOE is a straight line.
And C, O, F are collinear → COF is a straight line.
Construct a regular hexagon with sidelength 4 cm using a ruler and compass.
Answer:
Step 1: Draw a circle of radius 4 cm with centre O.
Step 2: Mark a point A on the circle.
Step 3: With compass set to 4 cm (same radius), cut arcs on the circle successively from A to get B, C, D, E, F.
Step 4: Join all six points to form the regular hexagon ABCDEF.
How do we construct a 120° angle using a ruler and compass?
Answer:
First construct a 60° angle at a point.
The supplementary angle on the other side of the 60° angle = 180° − 60° = 120°.
So by constructing a 60° angle at a point on a line, we automatically get a 120° angle on the other side.
Construct a regular hexagon of sidelength 5 cm.
Answer:
Explanation:
Step 1: Draw a horizontal line. Mark point A. Using the 60° construction, construct angle of 60° at A.
Step 2: Mark point B on ray AX at distance 5 cm.
Step 3: Construct a 60° angle at B (pointing upward/inward). Mark C at 5 cm from B.
Step 4: Construct a 60° angle at C. Mark D at 5 cm from C.
Step 5: Continue constructing 60° angles and marking points E and F at 5 cm each.
Step 6: Join F back to A. The hexagon ABCDEF with sidelength 5 cm is complete.
Related Constructions
How will you construct 30° and 15° angles?
Answer:
30° angle:
Step 1: Construct a 60° angle.
Step 2: Bisect the 60° angle using the angle bisection method.
The resulting angle is 30°.
15° angle:
Step 1: Construct a 30° angle (as above).
Step 2: Bisect the 30° angle.
The resulting angle is 15°.
Construct the following 6-pointed star. Are the six outer triangles equilateral?
Answer:
Construction:
Step 1: Construct a regular hexagon GHIJKL (using the 60° angle method).
Step 2: On each side of the hexagon, construct an equilateral triangle pointing outward.
Step 3: The six outer points of the star (A, B, C, D, E, F) are the apex vertices of these equilateral triangles.
Step 4: Connect alternate vertices of the hexagon to form two overlapping equilateral triangles (the classic Star of David / 6-pointed star shape).
Are the six outer triangles equilateral?
Answer:
Yes. Each outer triangle (∆AGH, ∆BHI, ∆CIJ, ∆DJK, ∆ELK, ∆FLG) is equilateral.
Justification: Each interior angle of the regular hexagon is 120°. The angle at each outer vertex of the star = 180° − 120° = 60°. Since the sides are equal (sides of the hexagon) and the angle between them is 60°, each outer triangle is equilateral.
Figure it Out (Related Constructions)
Question 1(a): Construct an inflexed arch figure.
Answer:
Explanation:
Step 1: Draw two vertical parallel lines using parallel line construction.
Step 2: Draw a horizontal base connecting the two lines.
Step 3: At the top centre, construct a pointed arch (inflexed arc) — use two arcs whose centres are inside the arch (below the arc), so the arcs curve inward at the top.
Step 4: Connect the arcs smoothly to the vertical sides.
Question 1(b): Construct the 6-petalled flower figure (can also be constructed using only a compass).
Answer:
Explanation:
Step 1: Construct a regular hexagon with sidelength r.
Step 2: Use each vertex of the hexagon as a centre and draw a circle of radius r.
Step 3: The 6 circles overlap at the centre forming a 6-petalled flower pattern.
Note: This figure can also be constructed using only a compass — place the compass at any point on a circle and step off equal arcs around the circle using the same radius.
Question 1(c): Construct the regular octagon figure.
Answer:
Explanation:
Step 1: Draw a circle with centre O.
Step 2: Draw two perpendicular diameters (using perpendicular bisector construction).
Step 3: Bisect each 90° angle to get 45° lines — giving 4 diameters at 45° intervals.
Step 4: Mark the 8 points where these diameters meet the circle.
Step 5: Join consecutive points to form the regular octagon.
Question 1(d): Construct the 7-circle pattern.
Answer:
Explanation:
Step 1: Draw a central circle of radius r.
Step 2: Place the compass at any point on the circle and mark the next point at distance r around the circumference.
Step 3: Continue marking 6 points around the central circle (the radius r fits exactly 6 times around a circle of radius r).
Step 4: Draw circles of radius r centred at each of the 6 outer points.
Step 5: This creates the 6 outer circles around the central one — 7 circles total.
Question 1(e): Construct the Flower of Life / hexagonal star pattern.
Answer:
Explanation:
Step 1: Start with a regular hexagon constructed using 60° angles.
Step 2: Using the hexagon's vertices, construct additional equilateral triangles and hexagons, extending the pattern outward.
Step 3: Continue adding layers of equilateral triangles and hexagons to build the complex Flower of Life / hexagonal star pattern.
Step 4: Add inner construction lines connecting all intersection points to complete the detailed geometric design.
Question 2: Optical Illusion — Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.
Answer:
Explanation of the illusion: The figure shows a Kanizsa triangle — an optical illusion where the brain perceives a bright white triangle in the centre even though no triangle is actually drawn. Three Pac-Man-like shapes and three angles create the illusion of a complete triangle.
To recreate:
Step 1: Lightly mark three points as the vertices of an equilateral triangle.
Step 2: At each vertex, draw an incomplete circle (a disc with a wedge cut out, opening toward the centre of the triangle).
Step 3: At each side's midpoint, draw an angle (two short lines forming a corner pointing outward).
Step 4: The brain will complete the triangle. The illusion arises due to the principle of contour completion in human perception.
Question 3: Construct this figure (6-pointed star with angles).
Answer:
Explanation:
Step 1: Construct a regular hexagon ABCDEF using the 60° angle method.
Step 2: Extend alternate sides of the hexagon outward (or connect alternate vertices).
Step 3: The intersection of these extended lines forms the 6 outer points of the star.
Step 4: The angles at the star's tips are 60° (since the outer triangles are equilateral — hint: find angles as directed).
Step 5: Draw the complete star by joining all intersection points.
Question 4: Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P.
Answer:
Explanation:
Step 1: Draw line l and mark point P outside it.
Step 2: Using P as centre, draw arcs that cut line l at two points, say A and B.
Step 3: Now construct the perpendicular bisector of segment AB using the standard method (arcs from A and B, equal radius, above and below l).
Step 4: The perpendicular bisector of AB passes through P (since PA = PB by construction), and is perpendicular to l.
Step 5: The perpendicular from P to l is the line joining P to the foot of the perpendicular bisector on l.
6.2 Tiling
Figure it Out (Tangrams)
How can the tangram pieces be rearranged to form each of the following figures?
Answer:
To solve this problem, we prepare 10 sets of 7 tans obtained from a tangram-based shape.
The tangram arrangements (i) through (x) are shown in the figures with numbered pieces indicating how each of the 7 tans is placed to form the respective shape.
Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles?
Answer:
Yes. A 4 × 6 grid has 24 unit squares. Each 2 × 1 tile covers 2 unit squares. 24 ÷ 2 = 12 tiles needed.
Use 12 horizontal tiles, covering 2 columns per row.
Can a 4 × 7 grid be tiled using 2 × 1 tiles?
Answer:
Yes. A 4 × 7 grid has 28 unit squares (an even number). 28 ÷ 2 = 14 tiles.
Strategy: Tile the 4 × 6 portion with horizontal/vertical tiles (12 tiles), and tile the remaining 4 × 1 column with 2 vertical tiles.
Total = 14 tiles.
What about a 5 × 7 grid?
Answer:
No, a 5 × 7 grid cannot be tiled with 2 × 1 tiles.
The grid has 5 × 7 = 35 unit squares, which is an odd number.
Each 2 × 1 tile covers exactly 2 unit squares.
To tile the grid, we would need 35 ÷ 2 = 17.5 tiles, which is not a whole number.
Therefore, it is impossible to tile a 5 × 7 grid with 2 × 1 tiles.
Complete the justification:
Answer:
The 5 × 7 grid has 35 unit squares (odd number).
Each 2 × 1 tile covers exactly 2 unit squares.
If the grid were tileable, the number of unit squares covered = 2 × (number of tiles) = an even number.
But 35 is odd, which cannot equal an even number.
Therefore, a 5 × 7 grid cannot be tiled with 2 × 1 tiles.
Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even?
Answer:
Yes. If both m and n are even, the grid has m × n unit squares (an even × even = even number of squares).
One general strategy: Cover each column with vertical tiles. Since m (number of rows) is even, each column of m squares can be covered by m/2 vertical tiles placed one above the other.
This works for all n columns. Total tiles = (m/2) × n.
Is an m × n grid tileable with 2 × 1 tiles, if one of m or n is even and the other is odd?
Answer:
Yes. If at least one of m or n is even, then m × n is even.
Strategy: Suppose m is even and n is odd (or vice versa). Cover each column with m/2 vertical tiles (since m is even). This works for all n columns regardless of whether n is odd or even.
Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd?
Answer:
No. If both m and n are odd, then m × n = odd × odd = odd number of unit squares. Since each tile covers 2 squares, tiling requires an even number of squares. An odd total cannot be covered, so the grid is not tileable.
A 5 × 3 grid with a unit square removed — is it tileable with 2 × 1 tiles?
Answer:
A 5 × 3 grid has 15 unit squares. After removing 1, we have 14 squares (even number).
It depends on which square is removed.
If the removed square, when we colour the grid like a checkerboard (alternating black and white), results in an equal number of black and white squares remaining — then tiling is possible.
A 5 × 3 grid has 8 squares of one colour and 7 of the other.
Removing one square of the colour that has 8 leaves 7 black and 7 white — equal — so tiling may be possible.
Removing one square of the colour that has 7 leaves 8 and 6 — unequal — so tiling is impossible.
Is the following region (staircase-shaped region, Page 158) tileable with 2 × 1 tiles?
Answer:
Apply the black-and-white colouring argument.
Count the black squares and white squares in the region.
If they are equal, tiling may be possible; if unequal, tiling is impossible.
For the specific staircase region shown in the book: count the squares and check — if the counts are equal, try to find a tiling; if unequal, it is non-tileable.
What about the region in Fig. 6.13?
Answer:
The region in Fig. 6.13 (a 5 × 3 grid with two squares removed at the top to form an irregular shape) should be tested using the black-and-white colouring method.
Colour the region like a checkerboard. Count black and white squares.
If the counts differ, the region is non-tileable with 2 × 1 tiles.
If the plain grid is tileable, is the black-and-white-grid tileable?
Answer:
Yes to both.
If the plain grid is tileable with 2 × 1 tiles, we can colour the tiles black-and-white and place them on the corresponding coloured squares — the black-and-white grid is tileable.
Figure it Out (Are the following tilings possible?)
Question 1:
Answer:
The given tiles can be used for tiling the region.
The tiling shall use 4 tiles of the given shape.
Question 2:
Answer:
The black-and-white region of the given region is shown in the figure.
This region is to be tiled by tiles of the form shown in the figure.
Why NCERT solutions help students?
Class 7 Maths Chapter 6 Constructions and tilings is one of the most visually rich and reasoning-heavy chapters in the NCERT Ganit Prakash II textbook. Having access to accurate, worksheet-based NCERT solutions helps students in several important ways. First, it gives them a reliable reference to check whether their construction steps are correct — not just the final figure, but the logic behind it. Second, for justification questions involving congruence and angle properties, the solutions show exactly what level of explanation is expected at the Class 7 level. Third, tiling questions build mathematical reasoning skills that go far beyond geometry — they introduce students to ideas like parity, the colouring argument, and proof by contradiction in a fun, accessible way. Students who understand this chapter deeply will find it much easier to tackle Class 8 and Class 9 geometry topics. Regular practice with these NCERT solutions builds the precision and confidence that exams demand.
Help your child master geometric thinking and logical reasoning with structured, expert-guided learning support.