Important Questions Class 7 Maths Part 2 Chapter 6: Constructions and Tilings

Important Questions Class 7 Maths Part 2 Chapter 6 focus on geometric construction, the exact drawing of figures using defined tools and steps. Students use a ruler, compass, angle bisection, perpendicular bisectors, hexagons, tangrams, and tiling rules to solve shape-based problems.

Geometry becomes exact when students draw shapes through reasoning instead of estimation. Important Questions Class 7 Maths Part 2 Chapter 6 help students practise perpendicular bisectors, 90° angles, angle bisectors, copied angles, parallel lines, regular hexagons, tangrams, and tilings. The CBSE 2026 chapter uses ruler-compass methods, congruent triangles, repeated designs, arch patterns, domino tiling, and plane tiling to connect construction with visual proof.

Key Takeaways

• Perpendicular Bisector: It divides a line segment into two equal parts at 90°.
• Angle Bisection: It divides a given angle into two equal angles.
• 60° Construction: An equilateral triangle creates a 60° angle.
• Tiling: A tiling covers a region without gaps or overlaps.

Important Questions Class 7 Maths Part 2 Chapter 6 Structure 2026

Important Questions Class 7 Maths Part 2 Chapter 6 with Answers

Constructions in this chapter depend on exact steps, equal distances, and congruent triangles.
Students must explain why a construction works, instead of only drawing the final figure.
These class 7 maths part 2 chapter 6 important questions follow the NCERT 2026 style for construction and tiling.

1. What does Important Questions Class 7 Maths Part 2 Chapter 6 test in Constructions and Tilings?

Important Questions Class 7 Maths Part 2 Chapter 6 test ruler-compass constructions, angle construction, repeated designs, and tiling logic. The chapter also checks proof through congruent triangles.

1. Construction Skill: Draw perpendicular bisectors, angle bisectors, and copied angles.
2. Design Skill: Use repeated angles and arcs in patterns.
3. Tiling Skill: Check whether shapes cover a region without gaps.
4. Final Result: The chapter tests exact drawing and geometric reasoning.

2. Why is a supporting line used in the eye construction?

A supporting line helps create symmetry in the eye construction. It acts as the line across which the upper and lower arcs match.

1. Given Data: The eye has endpoints X and Y.
2. Construction Idea: Centres A and B stay equally distant from X and Y.
3. Reason: AX = AY = BX = BY.
4. Final Result: The supporting line helps locate symmetrical arc centres.

3. What is bisection in Class 7 geometry?

Bisection means dividing a line, angle, or geometric object into two identical parts. A midpoint bisects a line segment.

1. Term: Bisection
2. Line Example: O bisects XY when OX = OY.
3. Angle Example: OC bisects ∠XOY when ∠XOC = ∠COY.
4. Final Result: Bisection creates two equal parts.

Class 7 Maths Part 2 Chapter 6 Constructions and Tilings

The NCERT chapter begins with exact geometric constructions and then moves to tiling.
Students first construct lines, angles, and repeated designs before testing tileable regions.
This class 7 maths part 2 chapter 6 constructions and tilings section links construction logic with visual reasoning.

4. What is a perpendicular bisector in Class 7 Maths?

A perpendicular bisector is a line that divides a segment into two equal parts at 90°. It passes through the midpoint of the segment.

1. Given Data: XY is a line segment.
2. Condition 1: OX = OY.
3. Condition 2: AB ⟂ XY.
4. Final Result: AB is the perpendicular bisector of XY.

5. Why does the line through two arc-intersection points bisect XY?

The line through two arc-intersection points bisects XY because both points are equally distant from X and Y. Such points lie on the perpendicular bisector.

1. Given Data: AX = AY and BX = BY.
2. Proof Idea: Join A, B, X, and Y.
3. Congruence Used: Equal sides form congruent triangles.
4. Final Result: AB passes through the midpoint of XY.

6. Why is AB perpendicular to XY in the perpendicular bisector construction?

AB is perpendicular to XY because the angles at the intersection are equal and form a straight angle. Each angle becomes 90°.

1. Given Data: AB meets XY at O.
2. Proof Step: ∆AOX ≅ ∆AOY.
3. Angle Fact: ∠AOX + ∠AOY = 180°.
4. Calculation:
   ∠AOX = ∠AOY
   ∠AOX = 180° ÷ 2 = 90°
5. Final Result: AB ⟂ XY.

Perpendicular Bisector Class 7 Questions

Perpendicular bisector questions test equal distances and midpoint logic.
The compass creates points that remain equally distant from both endpoints of a segment.
These perpendicular bisector class 7 questions focus on construction steps and justification.

7. How do you construct the perpendicular bisector of a line segment XY?

The perpendicular bisector of XY is constructed using equal arcs from X and Y. The line through the arc intersections is the required bisector.

1. Draw line segment XY.
2. From X, draw arcs above and below XY with the same radius.
3. From Y, draw arcs above and below XY with the same radius.
4. Mark the arc intersections as A and B.
5. Join A and B.
6. Final Result: AB is the perpendicular bisector of XY.

8. Is the same radius needed for arcs above and below XY?

The same radius is needed for each pair of arcs from X and Y. The radius above and below may differ if each pair uses equal radii.

1. Above XY: Arcs from X and Y must use the same radius.
2. Below XY: Arcs from X and Y must use the same radius.
3. Reason: Each intersection point must stay equally distant from X and Y.
4. Final Result: Equal radii are needed within each pair.

9. Can both arc-intersection points lie on the same side of XY?

Yes, both arc-intersection points can lie on the same side of XY. Two points on the perpendicular bisector determine the same line.

1. Given Data: Two arc intersections lie above XY.
2. Condition: Each point is equally distant from X and Y.
3. Rule: Any two such points define the perpendicular bisector.
4. Final Result: The construction still works on one side of XY.

Construction of 90 Degree Angle Class 7

A 90° angle can be constructed by adapting the perpendicular bisector method.
The given point becomes the midpoint of a segment on the original line.
These construction of 90 degree angle class 7 questions use compass marks instead of a protractor.

10. How do you construct a 90° angle at a point O on a line?

A 90° angle at O is constructed by making O the midpoint of a segment. Its perpendicular bisector passes through O.

1. Draw a line and mark point O on it.
2. With O as centre, mark X and Y on the line such that OX = OY.
3. Draw arcs from X and Y with equal radius above the line.
4. Mark their intersection as A.
5. Join AO.
6. Final Result: ∠AOX = 90°.

11. Why does the perpendicular through O make a right angle?

The perpendicular through O makes a right angle because it bisects XY at O. A perpendicular bisector meets the segment at 90°.

1. Given Data: O is the midpoint of XY.
2. Construction: AO is the perpendicular bisector of XY.
3. Angle Fact: A perpendicular line forms right angles.
4. Final Result: ∠AOX = ∠AOY = 90°.

12. Why is a compass more accurate than measuring the midpoint with a scale?

A compass is more accurate because it creates equal distances directly. Measurement marks on a scale can cause small errors.

1. Scale Method: It depends on reading and marking length.
2. Compass Method: It uses equal radii from endpoints.
3. Construction Fact: Equal arcs locate exact bisector points.
4. Final Result: Compass construction gives a more reliable midpoint.

Angle Bisector Class 7 Questions

Angle bisection divides one angle into two equal angles.
The method uses equal arm lengths and equal arcs to create congruent triangles.
These angle bisector class 7 questions explain why the constructed line gives equal angles.

13. How do you bisect an angle using a ruler and compass?

An angle is bisected by creating two equal triangles inside it. The joining line from the vertex divides the angle equally.

1. Draw ∠XOY.
2. Mark A on OX and B on OY such that OA = OB.
3. From A and B, draw equal arcs that meet at C.
4. Join OC.
5. Final Result: OC bisects ∠XOY.

14. Why does OC bisect ∠XOY?

OC bisects ∠XOY because ∆OAC and ∆OBC are congruent. Their corresponding angles at O are equal.

1. Given Data: OA = OB and AC = BC.
2. Common Side: OC is common.
3. Congruence Used: ∆OAC ≅ ∆OBC by SSS.
4. Final Result: ∠AOC = ∠BOC.

15. How do you construct a 45° angle using angle bisection?

A 45° angle is constructed by bisecting a 90° angle. Half of 90° equals 45°.

1. Construct a 90° angle at O.
2. Use the angle bisection method on the 90° angle.
3. Draw the bisector OC.
4. Calculation:
   90° ÷ 2 = 45°
5. Final Result: ∠AOC = 45°.

Ruler and Compass Construction Class 7

Ruler and compass construction uses only straight lines and arcs.
The chapter uses these tools to draw perpendiculars, angle bisectors, copied angles, parallel lines, and designs.
These ruler and compass construction class 7 questions focus on exact construction without protractor-based guessing.

16. How do you copy an angle using a ruler and compass?

An angle is copied by constructing a congruent triangle at a new point. Equal sides make the copied angle equal to the original angle.

1. Draw an arc from the original vertex A.
2. Let the arc meet the arms at B and C.
3. Draw the same-radius arc from the new vertex X.
4. Transfer length BC onto the new arc.
5. Join the new vertex to the transferred point.
6. Final Result: The new angle equals the original angle.

17. Why does the copied angle equal the original angle?

The copied angle equals the original angle because the constructed triangles are congruent. SSS congruence fixes the angle measure.

1. Original Triangle: ∆ABC is formed by the first arc.
2. Copied Triangle: ∆XYZ is formed using equal transferred lengths.
3. Congruence Used: ∆ABC ≅ ∆XYZ by SSS.
4. Final Result: ∠A = ∠X.

18. How do you construct a line parallel to a given line using angle copying?

A parallel line is constructed by copying a corresponding angle. Equal corresponding angles make the lines parallel.

1. Draw a transversal line l cutting the given line m at A.
2. Choose point B on l.
3. Copy the angle made by m and l at A to point B.
4. Draw the new line n through B.
5. Final Result: m ∥ n.

Construction of 60 Degree Angle Class 7

A 60° angle comes from an equilateral triangle.
The compass marks equal distances from the same radius, creating three equal sides.
These construction of 60 degree angle class 7 questions connect angle construction with triangle properties.

19. How do you construct a 60° angle at point A on a line?

A 60° angle is constructed by making an equilateral triangle on the line. Each angle of an equilateral triangle is 60°.

1. Draw line segment AX.
2. With A as centre, draw an arc that cuts AX at B.
3. With B as centre and the same radius, cut the first arc at C.
4. Join AC.
5. Final Result: ∠CAX = 60°.

20. Why is ∠CAX equal to 60°?

∠CAX equals 60° because triangle ABC is equilateral. All angles in an equilateral triangle are 60°.

1. Given Data: AB = AC = BC by equal compass radii.
2. Triangle Type: ∆ABC is equilateral.
3. Angle Rule: Each angle of an equilateral triangle = 60°.
4. Final Result: ∠CAX = 60°.

21. How do you construct 30° and 15° angles?

A 30° angle is made by bisecting 60°. A 15° angle is made by bisecting 30°.

1. Construct a 60° angle.
2. Bisect it to get 30°.
3. Bisect the 30° angle to get 15°.
4. Calculation:
   60° ÷ 2 = 30°
   30° ÷ 2 = 15°
5. Final Result: 30° and 15° angles come from repeated bisection.

Regular Hexagon Construction Class 7

A regular hexagon has six equal sides and six equal angles.
NCERT connects a regular hexagon with six equilateral triangles around one point.
These regular hexagon construction class 7 questions use the 60° angle and equal compass lengths.

22. Why can six equilateral triangles form a regular hexagon?

Six equilateral triangles can form a regular hexagon because six 60° angles fill 360° around a point. Their outer sides become equal.

1. Given Data: Each equilateral triangle has a 60° angle at the centre.
2. Calculation:
   6 × 60° = 360°
3. Shape Fact: Equal outer sides form a regular hexagon.
4. Final Result: Six equilateral triangles form a regular hexagon.

23. How do you construct a regular hexagon of side 5 cm?

A regular hexagon of side 5 cm is constructed using six equal steps around a circle. Each side equals the radius used.

1. Draw a circle with centre O and radius 5 cm.
2. Mark point A on the circle.
3. Using the same radius, mark six points around the circle.
4. Join consecutive points.
5. Final Result: The figure is a regular hexagon of side 5 cm.

24. What is each interior angle of a regular hexagon in this chapter?

Each interior angle of a regular hexagon is 120°. It comes from two 60° angles of adjacent equilateral triangles.

1. Given Data: Six equilateral triangles meet at the centre.
2. Each Triangle Angle: 60°.
3. Interior Angle: 60° + 60° = 120°.
4. Final Result: Each interior angle is 120°.

Class 7 Maths Constructions Questions

Constructions in Chapter 6 also include repeated designs, arches, and stars.
Students must identify support lines before drawing arcs or repeated units.
These class 7 maths constructions questions focus on the hidden geometry behind designs.

25. How do repeated units help in constructing a design?

Repeated units help build a design by copying the same angle and arm length. Each copy keeps the pattern uniform.

1. Given Data: A design has one repeating unit.
2. Construction Need: Each unit needs equal arm lengths.
3. Angle Need: Each unit needs equal angles.
4. Final Result: Angle copying creates exact repeated units.

26. How do you construct an 8-petalled design using angle bisection?

An 8-petalled design uses eight equal angles around a point. Each angle measures 45°.

1. Given Data: Full angle around a point = 360°.
2. Calculation:
   360° ÷ 8 = 45°
3. Construction Step: Construct 90° angles and bisect them.
4. Final Result: Each adjacent support line is 45° apart.

27. How is a trefoil arch constructed using support lines?

A trefoil arch is constructed using equal side lengths and equal angles. Support lines guide the arcs.

1. Mark two base points A and D.
2. Construct equal angles at A and D.
3. Mark B and C such that AB = CD.
4. Draw arcs using suitable centres and radii.
5. Final Result: Equal support lines create a symmetric trefoil arch.

Class 7 Maths Tiling Questions

Tiling means covering a region with shapes without gaps or overlaps.
Chapter 6 starts with tangrams and then studies grids, domino tiles, and plane tilings.
These class 7 maths tiling questions test area, parity, black-white colouring, and shape arrangement.

28. What is tiling in Class 7 Maths?

Tiling is covering a region using shapes without gaps or overlaps. The shapes must fit the given region exactly.

1. Term: Tiling
2. Condition 1: No gaps.
3. Condition 2: No overlaps.
4. Example: Unit squares tile a rectangular grid.
5. Final Result: Tiling covers a region exactly.

29. Can a 4 × 6 grid be tiled using 2 × 1 tiles?

Yes, a 4 × 6 grid can be tiled using 2 × 1 tiles. The grid has an even number of unit squares.

1. Given Data: Grid size = 4 × 6.
2. Area: 4 × 6 = 24 unit squares.
3. Tile Area: 2 × 1 = 2 unit squares.
4. Calculation:
   24 ÷ 2 = 12 tiles
5. Final Result: The grid can be tiled using 12 domino tiles.

30. Can a 5 × 7 grid be tiled using 2 × 1 tiles?

No, a 5 × 7 grid cannot be tiled using 2 × 1 tiles. It has an odd number of unit squares.

1. Given Data: Grid size = 5 × 7.
2. Area: 5 × 7 = 35 unit squares.
3. Tile Area: 2 unit squares.
4. Reason: 35 cannot be divided exactly by 2.
5. Final Result: The 5 × 7 grid cannot be tiled with dominoes.

Tiling with Dominoes Class 7

A 2 × 1 tile covers exactly two adjacent unit squares.
Some regions with even area still cannot be tiled, because the shape arrangement also matters.
These tiling with dominoes class 7 questions use area checks and black-white colouring arguments.

31. Is an m × n grid tileable with 2 × 1 tiles if both m and n are even?

Yes, an m × n grid is tileable when both m and n are even. Each column or row can be covered by dominoes.

1. Given Data: m and n are even.
2. Tile Size: Each tile covers 2 unit squares.
3. Strategy: Cover each column using vertical 2 × 1 tiles.
4. Final Result: The grid is tileable with dominoes.

32. Is an m × n grid tileable if one of m and n is even?

Yes, the grid is tileable if one dimension is even. The even dimension can be split into pairs.

1. Given Data: One of m and n is even.
2. Tile Size: 2 × 1.
3. Strategy: Pair the units along the even direction.
4. Final Result: The grid is tileable when at least one dimension is even.

33. Why is a region with 8 white squares and 6 black squares not domino-tileable?

The region is not domino-tileable because each domino covers one black and one white square. Unequal colour counts make tiling impossible.

1. Given Data: 8 white squares and 6 black squares.
2. Domino Rule: One domino covers one black and one white square.
3. Problem: White and black counts are unequal.
4. Final Result: The region cannot be tiled with dominoes.

Tangram Class 7 Maths Questions

Tangrams use seven pieces made by dividing a square.
The same pieces can be rearranged to make figures like arrows and other shapes.
These tangram class 7 maths questions focus on rearrangement without gaps or overlaps.

34. What is a tangram in Class 7 Maths?

A tangram is a puzzle made of seven pieces from a square. The pieces can form different figures by rearrangement.

1. Given Shape: A square.
2. Pieces: Seven tangram pieces.
3. Rule: Use all pieces without gaps or overlaps.
4. Final Result: A tangram forms new shapes from seven fixed pieces.

35. Why does a tangram activity connect to tiling?

A tangram activity connects to tiling because the pieces cover a region exactly. They must not leave gaps or overlap.

1. Given Data: Seven tangram pieces.
2. Tiling Rule: Cover the target region completely.
3. Condition: Every piece must fit inside the boundary.
4. Final Result: Tangram rearrangement is a tiling activity.

NCERT Class 7 Maths Part 2 Chapter 6 Questions for Mixed Practice

Mixed practice in this chapter combines construction steps and tiling tests.
Students must choose whether the question needs equal arcs, congruent triangles, angle copying, or area reasoning.
These NCERT class 7 maths part 2 chapter 6 questions match the reasoning style used in the 2026 chapter.

36. How do you construct a perpendicular to a line through a point outside it?

A perpendicular through an outside point can be built using a perpendicular bisector. The outside point must lie on that bisector.

1. Draw line l and mark point P outside it.
2. Find two points X and Y on l such that PX = PY.
3. Construct the perpendicular bisector of XY.
4. The bisector passes through P.
5. Final Result: The line through P is perpendicular to l.

37. Why can squares tile the entire plane?

Squares can tile the entire plane because their angles fit exactly around a point. Four right angles make 360°.

1. Given Data: Each square angle = 90°.
2. Calculation:
   4 × 90° = 360°
3. Tiling Rule: Shapes must meet without gaps.
4. Final Result: Squares tile the plane exactly.

38. Why can equilateral triangles tile the plane?

Equilateral triangles can tile the plane because six 60° angles meet around a point. Their angles total 360°.

1. Given Data: Each equilateral triangle angle = 60°.
2. Calculation:
   6 × 60° = 360°
3. Tiling Rule: Full angle around a point = 360°.
4. Final Result: Equilateral triangles tile the plane exactly.

39. Why can regular hexagons tile the plane?

Regular hexagons can tile the plane because three 120° angles meet around a point. Their angles total 360°.

1. Given Data: Each regular hexagon angle = 120°.
2. Calculation:
   3 × 120° = 360°
3. Tiling Rule: No gap remains around the meeting point.
4. Final Result: Regular hexagons tile the plane exactly.

40. Why do honeycombs show tiling in nature?

Honeycombs show tiling because hexagonal cells cover space without gaps. The cells store eggs, larvae, pupae, and food.

1. Observed Shape: Hexagonal cells.
2. Tiling Feature: No space is wasted between cells.
3. Use: Bees and wasps use cells for storage and safety.
4. Final Result: Honeycombs use hexagonal tiling.

Class 7 Maths Part 2 Chapter 6 Questions and Answers for One-Mark Practice

One-mark questions often test exact definitions from the chapter.
Students should know bisection, perpendicular bisector, angle bisector, copying angles, and tiling.
These class 7 maths part 2 chapter 6 questions and answers cover the core NCERT terms.

41. What tools are used in ruler and compass construction?

An unmarked ruler and a compass are used in ruler and compass construction. The ruler draws lines, and the compass draws arcs.

1. Tool 1: Unmarked ruler.
2. Tool 2: Compass.
3. Use: Lines and arcs create exact geometric figures.
4. Final Result: The chapter uses an unmarked ruler and compass.

42. What is an angle bisector?

An angle bisector is a line that divides an angle into two equal angles. It starts from the vertex of the angle.

1. Term: Angle bisector.
2. Condition: ∠AOC = ∠BOC.
3. Example: A 90° angle becomes two 45° angles.
4. Final Result: An angle bisector creates two equal angles.

43. What is the meaning of copying an angle?

Copying an angle means constructing another angle equal to the given angle. The method uses equal arcs and congruent triangles.

1. Original Angle: ∠A.
2. Copied Angle: ∠X.
3. Condition: ∠A = ∠X.
4. Final Result: The copied angle has the same measure.

44. What is a regular hexagon?

A regular hexagon is a six-sided polygon with equal sides and equal angles. Each interior angle is 120°.

1. Sides: 6 equal sides.
2. Angles: 6 equal angles.
3. Interior Angle: 120°.
4. Final Result: A regular hexagon has six equal sides and six equal angles.

CBSE Class 7 Maths Important Links