Class 7 Maths Ganita Prakash Part 2 Chapter 6 MCQs – Constructions and Tilings

Use these Constructions and Tilings Class 7 MCQ questions to revise ruler-and-compass geometry and the mathematics of tiling. Constructions and Tilings is Chapter 6 of Class 7 Maths Ganita Prakash Part 2 and it explains how to construct a perpendicular bisector, a 90° angle, an angle bisector, a copy of an angle and a parallel line. The questions also cover 60°, 30°, 45° and 120° angles, regular hexagons and covering a region with shapes without gaps or overlaps. Picture each construction before answering.

Constructions and Tilings Class 7 MCQ Questions with Answers

Class 7 Maths Part 2 Chapter 6 MCQs Set 1

Q1. A line is the perpendicular bisector of a segment XY. Which statement must be true?

[A]. It passes through X and is parallel to XY.
[B]. It passes through the midpoint of XY and meets XY at 90°.
[C]. It divides XY in the ratio 1:2.
[D]. It meets XY at 45° but need not pass through its midpoint.

Q2. The perpendicular bisector of a 10 cm segment XY meets it at O. What is OX?

[A]. 10 cm
[B]. 20 cm
[C]. 5 cm
[D]. 2.5 cm

Q3. A point P satisfies PX = PY for two distinct points X and Y. On which line must P lie?

[A]. The perpendicular bisector of XY
[B]. Any line parallel to XY
[C]. The line through X parallel to the perpendicular bisector of XY
[D]. A line through Y making 45° with XY

Q4. In constructing the perpendicular bisector of XY, arcs from X and Y meet at A above XY and at B below XY. Which condition is sufficient?

[A]. AX = BX only.
[B]. AX = 2AY and BX = 2BY.
[C]. AX = AY and BX = BY, even if AX ≠ BX.
[D]. AY = BY only.

Q5. Which pair of tools is sufficient to construct a perpendicular bisector without measuring the segment’s length?

[A]. A protractor and an eraser only
[B]. Two pencils without a ruler or compass
[C]. An unmarked ruler and a compass
[D]. An unmarked ruler and an eraser only

Class 7 Maths Part 2 Chapter 6 MCQs Set 2

Q6. To construct a perpendicular at O on a line, X and Y are marked on opposite sides of O with OX = OY. If AX = AY and A is off the line, which line is required?

[A]. AO
[B]. AX
[C]. XY
[D]. AY

Q7. A rope has its ends fixed at X and Y. Its marked midpoint is pulled to A until both halves are straight and taut. Which relation follows?

[A]. AX + AY = XY in every position
[B]. AX = 2AY
[C]. AX = AY
[D]. AX = XY in every position

Q8. Eight equally spaced rays start from a point and divide the full turn into equal angles. What is each angle?

[A]. 90°
[B]. 30°
[C]. 60°
[D]. 45°

Q9. A 90° angle is bisected. What is the measure of each resulting angle?

[A]. 60°
[B]. 30°
[C]. 180°
[D]. 45°

Q10. A 60° angle is bisected, and then one of the resulting angles is bisected again. What is the smallest angle obtained?

[A]. 15°
[B]. 20°
[C]. 30°
[D]. 45°

Class 7 Maths Part 2 Chapter 6 MCQs Set 3

Q11. For an angle AOB, OA = OB. A point C inside the angle satisfies AC = BC. Which congruence condition proves △OAC ≅ △OBC?

[A]. Equality of area alone
[B]. AAA alone
[C]. SSS
[D]. SSA as a general rule

Q12. While copying ∠BAC as ∠YXZ, the construction gives AB = XY, AC = XZ and BC = YZ. What follows?

[A]. ∠BAC and ∠YXZ must both be right angles.
[B]. ∠BAC = ∠YXZ
[C]. ∠BAC + ∠YXZ = 90° in every case
[D]. ∠BAC = 2∠YXZ

Q13. Two lines are cut by a transversal. One corresponding angle is 65°. What must the matching corresponding angle be to make the lines parallel?

[A]. 65°
[B]. 130°
[C]. 25°
[D]. 115°

Q14. A symmetrical trefoil arch uses support segments AB and CD with AB = CD and ∠BAD = ∠CDA. If AB = 4 cm and ∠BAD = 55°, what are CD and ∠CDA?

[A]. 4 cm and 55°
[B]. 8 cm and 55°
[C]. 2 cm and 110°
[D]. 4 cm and 125°

Q15. Which description defines a regular polygon?

[A]. All its sides are equal, with no condition on its angles.
[B]. All its sides are equal and all its interior angles are equal.
[C]. All its angles are equal, with no condition on its sides.
[D]. It has an even number of sides.

Class 7 Maths Part 2 Chapter 6 MCQs Set 4

Q16. How many congruent equilateral triangles, meeting at their 60° angles, fit exactly around one point to form a regular hexagon?

[A]. 6
[B]. 8
[C]. 4
[D]. 5

Q17. What is each interior angle of a regular hexagon formed from six congruent equilateral triangles?

[A]. 60°
[B]. 90°
[C]. 180°
[D]. 120°

Q18. Angles of 40°, 60°, 50°, 30°, 40° and 90° meet around a point without overlapping. What angle fills the remaining gap?

[A]. 60°
[B]. 50°
[C]. 70°
[D]. 40°

Q19. In a compass construction, AB = AC = BC. What is ∠CAB?

[A]. 45°
[B]. 60°
[C]. 120°
[D]. 90°

Q20. A 60° angle and another angle together form a straight angle. What is the other angle?

[A]. 30°
[B]. 60°
[C]. 120°
[D]. 240°

Class 7 Maths Part 2 Chapter 6 MCQs Set 5

Q21. All seven pieces of a tangram square are rearranged into a new shape without gaps or overlaps. How does the area change?

[A]. The area becomes half.
[B]. The area depends only on the new perimeter.
[C]. The area doubles.
[D]. The area remains the same.

Q22. Which arrangement is a tiling of a region?

[A]. Shapes cover only half of the region.
[B]. Shapes cover the entire region with no gaps or overlaps.
[C]. Shapes overlap but cover the boundary.
[D]. Shapes leave small gaps between every pair.

Q23. How many 2 × 1 domino tiles are needed to cover a 4 × 6 rectangular grid of unit squares?

[A]. 24
[B]. 10
[C]. 12
[D]. 6

Q24. Can a 4 × 7 rectangular grid be covered by 2 × 1 dominoes, with rotation allowed?

[A]. Yes, using 11 dominoes.
[B]. No, because both side lengths must be even.
[C]. No, because 7 is odd.
[D]. Yes, using 14 dominoes.

Q25. Why can a 5 × 7 rectangular grid not be tiled completely with 2 × 1 dominoes?

[A]. It has fewer than 40 unit squares.
[B]. A domino cannot be rotated.
[C]. It has 35 unit squares, an odd number.
[D]. Its side lengths are different.

Class 7 Maths Part 2 Chapter 6 MCQs Set 6

Q26. For positive whole numbers m and n, when can a complete m × n rectangular grid be tiled with 2 × 1 dominoes?

[A]. When both m and n are odd
[B]. Only when m = n
[C]. Only when both m and n are even
[D]. When at least one of m and n is even

Q27. A grid is coloured like a chessboard. What does each 2 × 1 domino cover when placed along the grid?

[A]. Two white squares only
[B]. One black square and one white square
[C]. Two black squares only
[D]. Either two black or two white squares, but never one of each

Q28. A region on a chessboard-coloured grid contains 8 white squares and 6 black squares. Can 2 × 1 dominoes cover it completely?

[A]. Yes, because its total area is even.
[B]. Yes, because 8 + 6 = 14.
[C]. No, because every tile must cover two white squares.
[D]. No, because the two colour counts are unequal.

Q29. In the usual tiling of the plane by equal regular hexagons, how many hexagons meet at each vertex?

[A]. 3
[B]. 5
[C]. 4
[D]. 6

Q30. In an edge-to-edge tiling by equal equilateral triangles, how many triangles meet at each vertex? A

[A]. 6
[B]. 4
[C]. 5
[D]. 4

Turn MCQ Practice into Confident Constructions

Most Constructions and Tilings MCQs reward knowing why a step works, not just the order of steps. Every point on the perpendicular bisector of a segment is equally distant from its two endpoints. A 60° angle comes from an equilateral triangle, bisecting it gives 30° and bisecting a 90° angle gives 45°. In tiling, the angles meeting at a point must add up to 360°; the angles of squares, equilateral triangles and regular hexagons fit this exactly, which is why each can tile a plane on its own. These constructions draw on congruence from Geometric Twins and on Parallel and Intersecting Lines from Part 1, so revise both chapters on the Class 7 Maths MCQ page. In combined numbering for both parts, this is Chapter 14.

Ganita Prakash Part 2 Chapter 6 is a hands-on chapter. Its activities and Figure it Out tasks ask you to construct designs such as eyes, trefoil and pointed arches, and regular hexagons, or to arrange tangram pieces into given shapes. The NCERT Solutions for Class 7 Ganita Prakash take you through the construction steps in order and explain why each arc is drawn where it is, so you understand the method instead of copying a figure. Keep a sharp pencil, ruler and compass ready, follow a solution once, then redo the construction on your own and check it with a protractor. You can also try the rope method for a perpendicular bisector, which the chapter takes from the Kātyāyana Śulbasūtra.