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Class 7 Maths: Constructions and Tilings, 8 practice questions with answers

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These are 8 practice questions on Chapter 6, Constructions and Tilings, of the new NCERT Class 7 Maths textbook, Ganita Prakash, Part 2. Each has a worked answer and the textbook page it comes from. They cover the places equally far from two wells, building new angles from 60° and 90°, bisecting a garden corner, a hexagon from its longest diagonal, cutting a cake and fitting angles round a point, which regular polygons tile the plane, boards for 3 × 1 tiles, and a colouring argument for a board with two inside squares removed. Every construction comes with lengths you can check with a ruler, to the nearest millimetre. Try each one on paper before you open the answer.

These questions are our own, drafted with AI help. Every number was re-computed by a test script and none of the chapter's own examples is reused, but no practising teacher has reviewed them yet. If you find a mistake, please tell us.

What the chapter says you need

Facts from the NCERT textbook (Ganita Prakash, Grade 7, Part 2, Chapter 6, printed pages 136 to 163):

Bisectors and angles

Question 1. On a map, two wells are 9 cm apart. A tap is to be placed so that it is equally far from both wells. (a) Which line holds all such places, and how do you construct it with ruler and compass? (b) Which of those places is nearest to the straight line joining the wells, and how far is it from each well? (c) A tap on that line is 6 cm from each well. Measure (or work out) how far it is from the line joining the wells.

Answer: (a) the perpendicular bisector of the line joining the wells; (b) the midpoint, 4.5 cm from each well; (c) about 4.0 cm. (a) Every point that is the same distance from two points lies on the perpendicular bisector of the line between them. Join the wells, then with any radius more than half of 9 cm draw arcs from each well, above and below the line; the line through the two crossing points is the perpendicular bisector. (b) It crosses the joining line at its midpoint, at a right angle, so that place is 4.5 cm from each well. (c) The tap, the midpoint and one well make a right-angled triangle with sides 6 cm (to the well) and 4.5 cm (half the gap); measuring the third side on the drawing gives about 4.0 cm. (The perpendicular bisector, pages 136 to 140.)

Question 2. Using only a ruler and compass, starting from 60° and 90° and bisecting or adding angles, which of these can you construct: 75°, 105°, 135°, 165°? Describe how to construct 105°.

Answer: all four; 105° = 90° + 15°. Bisecting angles gives 30° (from 60°), 15° (from 30°) and 45° (from 90°), and angles can be placed side by side: 75° = 60° + 15°, 105° = 90° + 15°, 135° = 90° + 45° and 165° = 90° + 60° + 15°. For 105°: on a line, construct a 90° angle at O. Then, starting from the upright arm, construct a 60° angle away from the line, bisect it to get 30° and bisect again to get 15°. The angle from the line to the new arm is 90° + 15° = 105°. Check with a protractor. (A 90° angle at a point, page 141; bisecting angles, pages 143 and 144; 60°, 30° and 15° angles, pages 153 and 154.)

Question 3. A garden has a straight fence on each side of a corner, and the angle between the two fences is 76°. (a) A straight path is to start at the corner and stay equally far from both fences. How do you construct it, and what angle does it make with each fence? (b) The gardener wants to split the 76° corner into four equal parts with three paths. What is the angle of each part, and how many times must you bisect?

Answer: (a) by bisecting the 76° angle; 38° with each fence; (b) 19° each, by bisecting twice. (a) A path that stays equally far from both fences runs along the bisector of the corner angle. With the corner as centre, draw an arc that cuts both fences; from those two points draw equal arcs that meet inside the angle; the line from the corner through that meeting point is the path, and it makes 76° ÷ 2 = 38° with each fence. (b) Bisecting each 38° half again gives four equal parts of 76° ÷ 4 = 19°, so you bisect twice: first the whole angle, then each half. (Angle bisection, pages 143 and 144.)

Hexagons and tilings

Question 4. A regular hexagon is to be drawn so that the line joining two opposite corners is exactly 12 cm. (a) What length must each side be, and why? (b) Construct it. What is its perimeter?

Answer: (a) 6 cm; (b) 36 cm. (a) A regular hexagon is made of six equilateral triangles meeting at its centre, and the line joining two opposite corners passes through the centre and is made of two of their sides. So 12 cm is two sides, and each side is 12 ÷ 2 = 6 cm. (b) With the compass set to 6 cm, mark six points round a centre and join them; the perimeter is 6 × 6 = 36 cm. (Regular hexagons and equilateral triangles, pages 151 to 153.)

Question 5. (a) A round cake is cut into five slices whose angles at the centre are 35°, 75°, 48°, 62° and 80°. The rest of the cake is to be cut into two equal slices. What is the angle of each? (b) Which three of these angles fit exactly round a point: 100°, 130°, 120°, 140°, 110°? Find every answer.

Answer: (a) 30° each; (b) 100°, 120° and 140°, or 110°, 120° and 130°. (a) The angles round the centre add up to 360°, and 35 + 75 + 48 + 62 + 80 = 300, so the rest of the cake is 360° − 300° = 60°, and two equal slices are 30° each. (b) Three angles fit exactly round a point when they add up to 360°. Trying the sets of three: 100 + 120 + 140 = 360 and 110 + 120 + 130 = 360 work; every other set of three adds up to something else (for example 100 + 130 + 140 = 370). (Angles round a point, page 152.)

Question 6. Copies of one regular polygon are to meet at a point with no gaps or overlaps. Which of these work: equilateral triangle, square, regular pentagon, regular hexagon, regular octagon? Can two regular octagons and one square meet at a point?

Answer: triangles, squares and hexagons work; pentagons and octagons alone do not; two octagons and a square do fit. The angle of a regular polygon is 180° minus 360° divided by the number of sides: 60°, 90°, 108°, 120° and 135°. (This rule, and the pentagon's and octagon's angles, are from outside this chapter.) The angles at a point must add up to exactly 360°. 360 ÷ 60 = 6, 360 ÷ 90 = 4 and 360 ÷ 120 = 3 are whole numbers, so triangles, squares and hexagons work. Three pentagons make 324°, leaving a 36° gap, and a fourth would overlap. Two octagons make 270°, leaving 90°, too small for a third. But that 90° gap is exactly a square's angle: 135° + 135° + 90° = 360°, so two octagons and a square fit. (Tiling the plane, pages 160 to 162; angles round a point, page 152.)

Grids and dominoes

Question 7. Which of these boards can be covered exactly with straight 3 × 1 tiles (which may be turned): (a) 6 × 6, (b) 5 × 5, (c) 3 × 7? Explain.

Answer: (a) yes; (b) no; (c) yes. Each tile covers 3 squares, so the number of squares must be a multiple of 3. (a) 6 × 6 = 36 squares, a multiple of 3, and each row of 6 squares takes 2 flat tiles: yes. (b) 5 × 5 = 25 squares, not a multiple of 3, so it cannot be covered. (c) 3 × 7 = 21 squares, and each of the 7 columns of 3 squares takes one upright tile: yes. (The same counting idea as tiling with 2 × 1 tiles, pages 157 and 158.)

Question 8. Colour a 4 × 4 board like a chessboard and number the rows and columns 1 to 4. (a) Two inside squares are removed: row 2, column 2 and row 3, column 3. Can the rest be tiled with dominoes? (b) Which other inside square could be removed instead of row 3, column 3, so that the rest can be tiled? Show a tiling.

Answer: (a) no; (b) row 2, column 3 (or row 3, column 2). Every domino covers one dark and one light square, so a tileable region needs equal numbers of each. (a) Squares (2, 2) and (3, 3) are the same colour (row + column is even for both), so removing them leaves 6 squares of that colour and 8 of the other: no tiling is possible. (b) Square (2, 3) is the other colour. Removing (2, 2) and (2, 3) leaves 7 of each, and a tiling exists: two flat dominoes in row 1, two in row 4, one flat domino on (3, 2) and (3, 3), and upright dominoes on (2, 1) with (3, 1) and on (2, 4) with (3, 4). Removing (3, 2) instead also works. (Colouring grids to decide tiling, pages 159 and 160.)

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Sources

All opened and read on 1 November 2026. Page numbers are the printed page numbers of the textbook; printed page = PDF page + 135 for this chapter.

The rules are from the textbook pages above. The questions, worked answers and checks are our own, and every number was re-computed by script. This guide is general information; follow your teacher and the textbook.