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6.10 Related Constructions — 30° and 15° Angles

🎓 Class 7 Mathematics CBSE Theory Ch 6 — Geometric Constructions ⏱ ~21 min
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This MCQ module is based on: 6.10 Related Constructions — 30° and 15° Angles

This mathematics assessment will be based on: 6.10 Related Constructions — 30° and 15° Angles
Targeting Class 7 level in General Mathematics, with Basic difficulty.

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6.10 Related Constructions — 30° and 15° Angles

How will we construct 30° and 15° angles with only ruler and compass? Start from a 60° angle (from an equilateral triangle). Bisect it to obtain 30°. Bisect 30° to obtain 15°. So every angle of the form 60°/2n can be constructed.

6-Pointed Star

Construct a 6-pointed star. Note that it has a rotational symmetry. Hint: Do you see a hexagon here? Are the six triangles forming the 6 points of the star — △AGH, △BHI, △CIJ, △DJK, △ELK, △FLG — equilateral? Why? (Hint: Find the angles.)

6-pointed star (Star of David)
6-pointed star made from two overlapping equilateral triangles. Six small triangles on the points are also equilateral (each angle 60°).
🔵 Figure it Out
  1. Construct the following figures: (a) An Inflexed Arc — a single arc from a circle. (b) A fun 6-petal flower (can be constructed using only a compass). Do you see how? (c) A regular 7-gon (impossible with ruler-compass — why?). (d) A 6-pointed star made of arcs. (e) A regular hexagon of sidelength 3 cm.
  2. Optical Illusion — do you notice anything interesting about the figure? How does this happen? Recreate this in your notebook.
  3. Construct the figure given in Fig. 6.5 (8-petal flower).
  4. Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the line l through P. (Hint: Find a line segment on l whose perpendicular bisector passes through P.)

6.11 Tiling

The Tangram is a puzzle that originated from China. They make use of 7 pieces obtained by dividing a square as shown. For the problems ahead, we need these 7 tangram pieces. These are provided at the end of the book. Or, by looking at the figure, you could make cardboard cutouts of the pieces.

ABCDEFG
7-piece Tangram: 2 large right triangles (A, B), 1 medium (F), 2 small (E, G), a square (D) and a parallelogram (C).

We can form interesting pieces by rearranging the tangram pieces. Here is an arrow (shown in NCERT). How can the tangram pieces be rearranged to form each of the following figures — the letters C, M, a cat, a boat, a duck, a fish, a person and a face?

6.12 Tiling Grids with 2×1 Tiles

Covering a region using a set of shapes, without gaps or overlaps, is called tiling?. Consider a rectangular grid made of unit squares. We call this a 4 × 6 grid, since it has 4 rows and 6 columns.

A 4 × 6 grid of unit squares (24 cells).

Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles? We are allowed to rotate a 2 × 1 tile and use it. Yes — many tilings are possible. Here is one way (horizontal tiles filling the 4 × 6 grid row by row, 12 tiles). Obviously this is not the only tiling possible.

Which grids can be tiled with 2 × 1 tiles?

Can a 4 × 7 grid be tiled using 2 × 1 tiles? What about a 5 × 7 grid? To see that there is no way to tile a 5 × 7 grid using 2 × 1 tiles, observe that this grid has 35 unit squares. Each tile covers exactly 2 unit squares. An odd number of unit squares cannot be tiled by 2-square tiles — so the answer is No.

General rule for 2×1 tileability of an m×n grid
An m × n grid is tileable by 2 × 1 tiles if and only if at least one of m or n is even. When both are odd, the total number of unit squares is odd, and no 2-tile covering is possible.
🔵 Figure it Out
  1. Is an m × n grid tileable with 2 × 1 tiles if both m and n are even? Come up with a general strategy and justify.
  2. Is an m × n grid tileable with 2 × 1 tiles if one of m and n is even and the other is odd? Come up with a general strategy.
  3. Is an m × n grid tileable with 2 × 1 tiles if both m and n are odd? Give reasons.
  4. Here is a 5 × 3 grid with a unit square removed (top right). Now, it has an even number of unit squares. Is it tileable with 2 × 1 tiles?
  5. Is the region in Fig. 6.13 (L-shape of 15 unit squares) tileable with 2 × 1 tiles?

A Colouring Trick — Black and White Grids

For any tiling problem of this kind, we can create a similar problem with the unit squares coloured black and white like a chessboard, and check if the 2 × 1 tile must cover one black square and one white square. In Fig. 6.14, there are 8 white squares and 6 black squares — not equal. Since the tile must cover one of each, no matter how you place it, the region cannot be tiled.

Fig. 6.14 — 5×3 grid with colouring 8 white, 6 black → not tileable
Two-colour argument: each 2 × 1 tile covers exactly 1 black + 1 white square. Unequal counts ⇒ no tiling.

6.13 Tiling the Entire Plane

So far we have seen how to tile a given region. What about tiling the entire plane? Can you think of a shape whose copies can tile the entire plane? Clearly, squares can. Are there other regular polygons that can tile the plane? What about equilateral triangles?

Tiling with equilateral triangles shows the possibility of tiling with another regular polygon. A plane can be tiled using regular hexagons as well. A plane can also be tiled using more than one shape, and using non-regular polygons. The great Dutch artist M. C. Escher (1898–1972) — whose works revealed mathematical themes such as tiling — have come up with creative ways of tiling a plane with animal shapes!

Squares Equilateral triangles Regular hexagons
Three regular polygons that tile the plane: squares, equilateral triangles and regular hexagons.

Mathematicians are still exploring various ways of tiling the plane! Tiling (b) was found as recently as 2023. Have you seen tilings in daily life? They are often used in buildings and in designs. Tilings are found in nature too — the front face of bee hives and some wasp nests are tiled using hexagonal cells. These cells are used by the insects to keep their eggs, larvae and pupae safe, as well as to store food. Because the region is tiled, no space is wasted. Scientists still wonder how bees and wasps are able to make hexagonal cells. Next time you see any tiling, pay closer attention to it! Tiling is still one of the most exciting and active areas of research in geometry.

Activity — Tile a 4 × 6 Rectangle
L4 Analyse
Materials: Graph paper, 12 paper strips each 2 × 1 units, pencil, scissors.
Predict first: How many different tilings of a 4 × 6 rectangle by 2 × 1 tiles do you think exist? (More than 10? More than 50?)
  1. Draw a 4 × 6 grid on paper (24 unit squares).
  2. Cover the grid with 12 strips (2 × 1). All horizontal first.
  3. Next, try alternating rows: two horizontal, two vertical.
  4. Try a mixed pattern with 6 horizontal on the left-half and 6 vertical on the right-half.
  5. Count the different tilings you can find.
There are actually 281 different tilings of a 4 × 6 rectangle by 2 × 1 dominoes! Mathematicians study these counts as "domino-tiling numbers" — they grow rapidly with grid size.

6.14 Figure It Out — Exercises

Q1. Construct a 6-pointed star using compass-and-ruler only. Are the 6 small triangles at the points equilateral? Why?
Draw a circle, step the radius 6 times around to mark 6 vertices. Connect every second vertex to form one equilateral triangle; connect the others similarly to form a second equilateral triangle. Overlapping the two produces a 6-pointed star. Each small point triangle has angles of 60° (same as the big equilateral triangles), so yes, they are equilateral.
Q2. Construct the 6-petal flower using only a compass (no ruler needed!). Explain how this is possible.
Draw a circle C. Without changing the radius, place the compass tip on C and draw a second circle through the centre of C. Keep stepping the compass around, each new centre on the previous circle. After 6 steps, the petals formed by the overlapping arcs create a 6-petal flower. No ruler is needed!
Q3. Construct a regular hexagon of sidelength 3 cm.
Draw a circle of radius 3 cm. Mark any point P₁ on it. With the compass still at radius 3 cm, place the tip at P₁ and cut the circle at P₂. Step around from P₂ → P₃ → … → P₆, each time with the compass at 3 cm. Join P₁P₂, P₂P₃, …, P₆P₁. This is a regular hexagon of side 3 cm.
Q4. Draw a line l and mark a point P outside it. Construct a perpendicular to l through P.
With centre P and a suitable radius, draw an arc cutting l at two points A and B. The perpendicular bisector of AB (from P) passes through P — so construct the perpendicular bisector of AB. This line is perpendicular to l and passes through P.
Q5. Is a 4 × 7 grid tileable using 2 × 1 tiles? Justify.
Yes. 4 × 7 = 28 unit squares (even). Place all tiles horizontally in each row (each row = 7 is odd, so horizontal won't work in a single row). Better: place 7 vertical 2 × 1 tiles each across a pair of consecutive rows — use rows 1&2 with 7 vertical tiles, rows 3&4 with 7 vertical tiles. Total 14 tiles. Tileable.
Q6. A 5 × 3 grid with one unit square removed — tileable by 2 × 1 tiles?
14 unit squares — even, but colour it like a chessboard: the counts of black and white depend on which square was removed. If a white square was removed, the region has 7 white + 7 black → may be tileable; but check carefully — for a 5×3 board with corner removed, test concretely. In general: an even count is necessary but not always sufficient. Try to actually construct a tiling — if you cannot, a colouring argument can prove impossibility.
Q7. Are the following tilings possible? (a) An L-shape of 8 unit squares tiled by 2 × 1 tiles? (b) A larger L-shape of 7 unit squares tiled?
(a) 8 is even — tileable; actually place the 4 tiles along the arms of the L. (b) 7 is odd — impossible, since each tile covers 2 squares.
Q8. Show how to rearrange all 7 tangram pieces into the letter "C" and the letter "M".
Cut out the 7 tangram pieces and use trial-and-error. For "C": the 2 large triangles form the top and bottom horizontal arms, the square sits in the middle-left, and the smaller pieces fill the gap. Verify: total area should equal the original square because all 7 pieces are used.
Q9. Is it possible to tile the plane using regular pentagons? Explain using interior angles.
Interior angle of a regular pentagon = 108°. At every vertex of a tiling the angles around a point must sum to 360°. But 360° ÷ 108° ≈ 3.33, not a whole number. Three pentagons give 324° (leaves a gap), four give 432° (overlap). So regular pentagons cannot tile the plane.
Q10. List all regular polygons that tile the plane. Justify using interior-angle sums.
Only equilateral triangles (60° × 6 = 360°), squares (90° × 4 = 360°) and regular hexagons (120° × 3 = 360°). These are the only regular polygons whose interior angles divide evenly into 360°.

Summary — Constructions and Tilings

★ Key Ideas from Chapter 6 ★

  • A division of a line, line segment, or any geometrical quantity, into two identical parts is called bisection.
  • Any point that is at equal distance from the two endpoints of a given line segment lies on its perpendicular bisector. This property can be used to construct the perpendicular bisector of a line segment using a ruler and compass.
  • The method of constructing the perpendicular bisector can be modified to draw a 90° angle at any point on a line using a ruler and compass.
  • An angle can be bisected and copied using the congruence properties of triangles (SSS).
  • A 60° angle can be constructed using a ruler and compass by constructing an equilateral triangle. Combined with bisection, we can construct 30°, 15°, 7.5°, 90°, 45°, 22.5°, 120°, etc.
  • Covering a region using a set of shapes, without gaps or overlaps, is called tiling.
  • A 2 × 1 tiling of an m × n grid exists if and only if the total number of unit squares is even — that is, at least one of m or n is even.
  • Among regular polygons, only equilateral triangles, squares, and regular hexagons can tile the plane by themselves.

Competency-Based Questions

A town planner is designing a public square. The central floor is a 6 × 8 grid of unit concrete squares. She wants to tile the floor with 2 × 1 rectangular stones. She also wants a circular hexagonal fountain of side 5 m at the centre, inscribed in a circle of radius 5 m.
Q1. How many 2 × 1 stones will fully cover the 6 × 8 grid? Justify using the tileability rule.
L3 Apply
Grid has 6 × 8 = 48 unit squares. Both 6 and 8 are even, so tileable. Number of 2 × 1 stones = 48 ÷ 2 = 24.
Q2. If one corner stone is removed (creating an irregular region of 47 unit squares), can the region be tiled by 2 × 1 tiles? Justify.
L4 Analyse
No — 47 is odd, and each 2 × 1 tile covers exactly 2 squares, so any tiling must cover an even number of squares.
Q3. For the hexagonal fountain (side 5 m), describe the compass-and-ruler construction and compute the interior angle.
L5 Evaluate
Draw a circle of radius 5 m. Step off the radius 6 times around the circle to mark vertices. Connect adjacent vertices. Interior angle at each vertex = (n−2)·180°/n = 4·180°/6 = 120°.
Q4. Design a mixed tiling for the 6 × 8 floor using both 2 × 1 rectangles and 1 × 1 squares. What is the smallest number of 1 × 1 squares you can use if 3 corners are marked as "must be 1 × 1"?
L6 Create
Place 1 × 1 tiles on the 3 required corners — that uses 3 squares. Remaining area = 48 − 3 = 45 cells (odd!). To make the remainder tileable by 2 × 1, use one more 1 × 1 square somewhere, leaving 44 cells = 22 rectangular 2 × 1 tiles. Minimum = 4 unit squares (3 corners + 1 extra).

Assertion–Reason Questions

Assertion (A): A 5 × 5 square grid cannot be tiled by 2 × 1 rectangles.
Reason (R): 25 is an odd number, and each 2 × 1 tile covers exactly 2 unit squares.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — An odd total of unit squares can never be covered by 2-square tiles.
Assertion (A): Regular pentagons can tile the plane.
Reason (R): The interior angle of a regular pentagon is 108°, and 108° is not a divisor of 360°.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (d) — A is false: regular pentagons do NOT tile the plane. R is true and gives the reason why.
Assertion (A): A 6-pointed star can be drawn by overlapping two equilateral triangles.
Reason (R): Every equilateral triangle has 60° angles and three-fold rotational symmetry.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (b) — Both are true. R describes a property of equilateral triangles, but it is the 180° rotation between the two overlapping triangles (not 3-fold symmetry) that creates the 6 points. So R doesn't directly explain A.

Frequently Asked Questions — Chapter 6

What is Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool in NCERT Class 7 Mathematics?

Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 7 Mathematics, Chapter 6: Chapter 6. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool step by step?

To solve problems on Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 6: Chapter 6?

The essential formulas of Chapter 6 (Chapter 6) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool important for the Class 7 board exam?

Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool?

Common mistakes in Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool?

End-of-chapter NCERT exercises for Part 3 — Tiling, Tangrams, Exercises & Summary | Class 7 Maths Ch 6 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.

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