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Solutions

Overview

Step-by-step NCERT solutions for Constructions and Tilings (Chapter 14, NCERT Class 7 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Constructions and Tilings textbook chapter.

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What these solutions cover

All 34 questions in Constructions and Tilings are solved in the PDF. Here's what's inside, exercise by exercise:

Perpendicular Bisector

  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.]
  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.
  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.
  4. Recreate the 4-petalled flower design shown in the textbook (Fig on p.140) using only a ruler and compass. The design shows 4 petal shapes formed by arcs, with supporting construction lines visible.

Sulba-Sutras Rope Construction

  1. Justify why AB in Fig. 6.4 is the perpendicular bisector. In Fig. 6.4, a rope is looped at both ends and fixed to pegs at X and Y. The midpoint of the rope is pulled above XY to point A (with the two halves of the rope fully stretched), then below XY to point B (with the two halves fully stretched).

Angle Bisection

  1. Construct at least 4 different angles. Draw their bisectors.
  2. Construct the 8-petalled figure shown in Fig. 6.5 (p.143). The figure has 8 petals arranged around a centre, with 8 equally spaced supporting lines.
  3. In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side of the angle (as shown in Fig. on p.144, where C is on the opposite side of the angle from the usual position), will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.
  4. What are the other angles that can be constructed using angle bisection? Can you construct a 65.5° angle? [Math Talk]

Copying an Angle

  1. Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles using only a ruler and compass.
  2. Construct Fig. 6.6 (p.145 — a repeating fan-petal pattern where a single curved unit repeats in two orientations along a horizontal band).

Parallel Lines and Star Construction

  1. Construct 4 pairs of parallel lines in different orientations.
  2. Construct the 8-pointed star figure shown on p.148. The star has 8 sharp points (labelled A, B, C, D, E, F, G, H) and its construction uses a regular 8-sided outline with 4 pairs of parallel sides (labelled ST, UV, WX, YZ and the connecting diagonals).

Pointed Arch

  1. Use support lines in Fig. 6.11 (p.150) to construct a pointed arch. Make different arches by changing the radius of the arcs. Fig. 6.11 shows two line segments of equal length meeting at a point, with midpoints marked — these form the support lines for the pointed arch.

Regular Hexagon and Equilateral Triangles

  1. Consider Fig. 6.12 (p.152): six congruent equilateral triangles arranged around a common centre O, with vertices A, B, C, D, E, F on the outer hexagon. Will the 70° angle fit into the gap ∠AOI? We are given angles around O: 40° + 60° + 50° + 30° + 40° + 90° + ∠AOI = 360°. Find ∠AOI and determine whether a 70° shape fits.
  2. In Fig. 6.12 (six equilateral triangles arranged around centre O forming a regular hexagon with vertices A, B, C, D, E, F), can you explain why AOD, BOE, and COF are straight lines?
  3. Construct a regular hexagon with side length 4 cm using a ruler and a compass.
  4. Why is ∠CAX = 60° in the construction of a 60° angle? (Step 1: arc from A with radius AB gives point B on line AX. Step 2: arc from B with same radius meets the first arc at C.) Is there an equilateral triangle here?
  5. Construct a regular hexagon of side length 5 cm.

Related Constructions

  1. How will you construct 30° and 15° angles?
  2. In the 6-pointed star (Fig. p.154), six triangles form the star points: △AGH, △BHI, △CIJ, △DJK, △ELK, △FLG. Are these triangles equilateral? Find their angles to justify.
  3. Construct the following figures (p.154):
    • (a) An Inflexed Arc — two tall curved columns with inward-curving arcs at the top, like a gothic twin arch;
    • (b) A 6-petal flower (can also be constructed using only a compass);
    • (c) A regular hexagon inscribed in a circle;
    • (d) Five equal circles arranged in a ring (with each circle tangent to its neighbours);
    • (e) A hexagonal pattern made of overlapping…
  4. Optical Illusion (p.155): A figure made from incomplete arcs and angles creates the visual impression of a triangle, even though no triangle is drawn. How does this happen? Recreate it in your notebook.
  5. Construct the 6-pointed star figure shown on p.155 (a Star of David / hexagram). [Hint: Find the angles in this figure.]

Tiling a Grid — 2×1 Tiles

  1. Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles (rotation allowed)?
  2. Can a 4 × 7 grid be tiled using 2 × 1 tiles?
  3. Can a 5 × 7 grid be tiled using 2 × 1 tiles? Complete the justification.
  4. Is an m × n grid tileable with 2 × 1 tiles if both m and n are even? Give a general strategy.
  5. Is an m × n grid tileable with 2 × 1 tiles if one of m and n is even and the other is odd? Give a general strategy.
  6. Is an m × n grid tileable with 2 × 1 tiles if both m and n are odd? Give reasons.

Tiling with Coloring Argument — Fig. 6.13 and 6.14

  1. A 5 × 3 grid has a unit square removed (Fig. 6.13, p.159). The figure shows a staircase-shaped region formed by removing the top-right corner square from a 5 × 3 grid. The resulting region has an even number of cells (14). Is it tileable with 2 × 1 tiles? Use the black-and-white coloring argument.
  2. Use the black-and-white coloring idea to find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable with 2 × 1 tiles.

Figure it Out — Tiling Possible?

  1. Is the following tiling possible? (Fig. on p.160.) Region: an L-shaped region — a 2-cell-wide, 2-cell-tall block on top, sitting on a 4-cell-wide, 2-cell-tall block below (the block extends 2 columns further to the right), for a total of 12 unit squares. Tile: an L-shaped tromino covering 3 unit squares.
  2. Is the following tiling possible? (Fig. on p.160.) Region: a 7-column by 8-row block of unit squares that has been 'sheared' by one column — every row is 7 cells wide, but the whole block slants so the top-left corner and the bottom-right corner each stick out by one cell (a parallelogram-shaped region of 7 × 8 = 56 unit squares). Tile: a 2 × 1 domino.
Sample

Worked examples

3 questions from Constructions and Tilings, solved step by step.

Perpendicular Bisector

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.

Yes, for each pair of arcs that are meant to meet at a single point, both arcs in that pair MUST use the same radius.

Reason:

We need the intersection point P (above XY) to satisfy PX = PY, so that P lies on the perpendicular bisector.

If the arc from X uses radius r and the arc from Y uses radius s (r ≠ s), their intersection point P satisfies PX = r and PY = s.

Since r ≠ s, we get PX ≠ PY, so P does NOT lie on the perpendicular bisector.

Therefore, within a single pair of arcs (both drawn from X and Y to meet at one point), the radii must be equal.

Note: The two pairs can use different radii from each other (as established in Q1), but within each pair the radius must be the same.

Answer: Yes, within each pair of arcs, both arcs must use the same radius. If the radii differ, the meeting point is not equidistant from X and Y, so it does not lie on the perpendicular bisector.

Related Constructions

How will you construct 30° and 15° angles?

Constructing 30°:

Step 1: Construct a 60° angle at point A on ray AX (using the equilateral triangle method: arc from A, arc from B at same radius, they meet at C, giving ∠CAX = 60°).

Step 2: Bisect the 60° angle using the angle bisection method. The bisector makes a 30° angle with AX.

Result: ∠ = 60°/2 = 30°. ✓

Constructing 15°:

Step 1: Construct a 30° angle (as above).

Step 2: Bisect the 30° angle using the angle bisection method.

Result: ∠ = 30°/2 = 15°. ✓

General principle: Start from a constructible base angle and apply bisection repeatedly.

Answer: 30° = half of 60° (bisect the 60° angle constructed by the equilateral triangle method). 15° = half of 30° (bisect again).

Tiling a Grid — 2×1 Tiles

Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles (rotation allowed)?

Yes. A 4 × 6 grid has 4 × 6 = 24 unit squares.

Each 2 × 1 tile covers exactly 2 unit squares.

Number of tiles needed = 24 ÷ 2 = 12 (a whole number). ✓

One explicit tiling strategy: cover all 6 columns with vertical (2×1) tiles.

Each column has 4 cells. 4 is even, so each column can be covered by 2 vertical tiles stacked.

Total: 6 columns × 2 tiles = 12 tiles. ✓

Answer: Yes. The 4 × 6 grid has 24 cells; 24 ÷ 2 = 12 tiles. Cover each column with 2 vertical tiles (possible since 4 rows is even).

All 34 questions in Constructions and Tilings are solved step by step in the Solutions PDF above.

Keep solving

More solutions in Ganita Prakash

01Large Numbers Around Us02Arithmetic Expressions03A Peek Beyond the Point04Expressions using Letter Numbers05Parallel and Intersecting Lines06Number Play07A Tale of Three Intersecting Lines08Working with Fractions09Geometric Twins10Operations with Integers11Finding Common Ground12Another Peek Beyond the Point13Connecting the Dots15Finding the Unknown
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More NCERT Solutions for Class 7

Read the Constructions and Tilings textbook chapter / PDF, or browse all NCERT Class 7 Mathematics solutions.

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