CBSE Class 7 Mathematics Constructions and Tilings Assignment Set 01

Short Answer Questions

Question 1. Answer the following:
(i) What is meant by the bisection of a line segment?
(ii) What extra condition makes a bisecting line the perpendicular bisector of that segment?
Answer:
(i) Bisection of a line segment means dividing it into two identical (equal) parts; the dividing line passes through the midpoint.
(ii) If the bisecting line is also perpendicular (at 90°) to the segment, it is the perpendicular bisector.
In simple words: Bisecting cuts into equal halves; adding a right angle makes it the perpendicular bisector.

Teacher's Note:
1. Many lines pass through the midpoint, but only one is at 90°.
2. Keep the two ideas separate.

Question 2. Write the steps you would follow to construct the perpendicular bisector of a line segment XY of length 7 cm, using only an unmarked ruler and a compass.
Answer:
Step 1: Open the compass to a radius more than half of XY (say 5 cm). With centres X and then Y, draw two arcs above XY; mark their meeting point A.
Step 2: With the same radius, draw two arcs below XY from X and from Y; mark their meeting point B.
Step 3: Join A and B. AB is the required perpendicular bisector of XY.
In simple words: Draw equal arcs from both ends above and below, then join the two crossing points.

Teacher's Note:
1. The radius must be more than half of XY, or the arcs will not meet.
2. Do not change the radius within a pair.

Question 3. While constructing the ‘Eyes’ figure, the lower arc is drawn with centre A and the upper arc with centre B. Explain why the construction requires AX = AY = BX = BY for the eye to look symmetrical about the supporting line XY.
Answer:
For symmetry about XY, the upper and lower arcs must have the same radius.
The lower arc has centre A and passes through X and Y, so its radius is AX = AY.
The upper arc has centre B, so its radius is BX = BY.
Equal radii therefore need AX = AY = BX = BY.
In simple words: Both arcs must pass through X and Y with the same radius, so all four distances are equal.

Teacher's Note:
1. Each arc passes through both X and Y.
2. Equal radii give mirror-image arcs.

Question 4. A point C is marked such that CX = CY, where X and Y are the endpoints of a line segment.
(i) On which line must C lie?
(ii) State the property of the perpendicular bisector that you used.
Answer:
(i) C lies on the perpendicular bisector of XY.
(ii) Any point at the same distance from X and from Y lies on the perpendicular bisector of XY; since a segment has only one perpendicular bisector, C lies on that line.
In simple words: A point equally far from X and Y is always on the perpendicular bisector of XY.

Teacher's Note:
1. This property is the heart of the construction.
2. There is only one perpendicular bisector.

Question 5. Define tiling. Give two examples of tiling that you can see around you, and state the two conditions a covering must satisfy to be called a tiling.
Answer:
Tiling is covering a region using a set of shapes, without gaps or overlaps.
Examples: square tiles on a kitchen floor; bricks in a wall; hexagonal cells of a honeycomb.
Conditions: (a) no gaps (no part left uncovered) and (b) no overlaps (no shape on top of another).
In simple words: Tiling covers a region fully with no gaps and no overlaps, like floor tiles.

Teacher's Note:
1. Give examples you have actually seen.
2. State both conditions clearly.

Question 6. Write the steps to construct an angle of 60° at a point A on a ray \( \overrightarrow{AX} \) using a ruler and a compass, and explain in one sentence why the angle obtained is exactly 60°.
Answer:
Step 1: With centre A and any convenient radius, draw an arc cutting \( \overrightarrow{AX} \) at B.
Step 2: With the same radius and centre B, draw an arc cutting the first arc at C. Join AC. Then ∠CAX = 60°.
Reason: AB = AC = BC, so △ABC is equilateral and each of its angles is 60°.
In simple words: Equal arcs from A and B make an equilateral triangle, whose angle is 60°.

Teacher's Note:
1. Do not change the compass opening.
2. The triangle need not be drawn fully.

Question 7. Name the three regular polygons whose copies alone can tile the entire plane. Write the interior angle of each and show that it divides 360° exactly.
Answer:
The equilateral triangle, the square and the regular hexagon.
Equilateral triangle: angle 60°, and 360° ÷ 60° = 6.
Square: angle 90°, and 360° ÷ 90° = 4.
Regular hexagon: angle 120°, and 360° ÷ 120° = 3.
In each case the angles at a point add up exactly to 360°.
In simple words: Triangles, squares and hexagons tile because 60°, 90° and 120° divide 360°.

Teacher's Note:
1. Whole-number answers mean the shape tiles.
2. These are the only three regular ones.

Question 8. Describe how the perpendicular bisector construction can be used to draw a 90° angle at a point O lying on a given line. Why is only one pair of intersecting arcs enough here?
Answer:
Extend the line on both sides of O. With centre O and a convenient radius, cut the line at X and Y, so O is the midpoint of XY.
With a larger radius, the same for both, draw arcs from X and Y on one side, meeting at A. Join AO; then ∠AOX = 90°.
One pair of arcs is enough because O is already a point of the perpendicular bisector of XY, and two points fix a line.
In simple words: Make O the midpoint of XY, find one more equidistant point, and join it to O.

Teacher's Note:
1. O itself is one of the two points.
2. The second radius must be larger than OX.

Question 9. Write the steps to bisect a given angle ∠XOY with a ruler and a compass. Name the pair of triangles that become congruent and the congruence condition used.
Answer:
Step 1: With centre O and any radius, cut the arms at A and B, so OA = OB.
Step 2: With a large enough radius, the same for both, draw arcs from A and B meeting at C.
Step 3: Join OC; it bisects ∠XOY.
△OAC ≅ △OBC by SSS, since OA = OB, AC = BC and OC is common. So ∠AOC = ∠BOC.
In simple words: Equal cuts on the arms, equal arcs to C, and OC splits the angle in half.

Teacher's Note:
1. Name the triangles and the condition.
2. The arcs from A and B must have equal radii.

Question 10. Starting from the constructions of 60° and 90° and using repeated angle bisection:
(i) Explain how a 15° angle is obtained.
(ii) Explain why an angle of 65.5° cannot be obtained by these methods.
Answer:
(i) Construct 60° using an equilateral triangle, bisect it to get 30°, and bisect 30° to get 15°.
(ii) 60° = 4 × 15° and 90° = 6 × 15°. Bisecting or adding such angles only gives 15° multiplied by a whole number or by a half, a quarter, an eighth, and so on (for example 7.5°, 22.5°, 3.75°).
But 65.5° ÷ 15° = 4 \( \frac{11}{30} \), and \( \frac{11}{30} \) is not a half, a quarter, an eighth or any such fraction, so 65.5° cannot be obtained.
In simple words: Halving and adding 60° and 90° never produces the odd value 65.5°.

Teacher's Note:
1. The building block is 15°.
2. Repeated halving only gives denominators 2, 4, 8, 16, …

Question 11. A rectangular grid of unit squares is to be covered completely by 2 × 1 tiles, which may be placed horizontally or vertically.
(i) Is a 4 × 7 grid tileable? Give the strategy.
(ii) Is a 5 × 7 grid tileable? Give the reason.
Answer:
(i) Yes. The grid has 4 rows, an even number, so each of the 7 columns is filled with 2 vertical tiles: 14 tiles in all, with no square left bare.
(ii) No. The grid has 5 × 7 = 35 squares. Tiles always cover an even number of squares, and 35 is odd.
In simple words: 4 × 7 works column by column; 5 × 7 has an odd number of squares, so it fails.

Teacher's Note:
1. Look for an even side.
2. Odd total means no tiling.

Question 12. In the construction for copying an angle, an arc of the same radius is drawn from the vertex of the given angle and from the new vertex, and then the chord length is transferred with the compass. Explain why the copied angle must be equal to the original angle.
Answer:
Let the arc from vertex A cut the arms at B and C; the arc of the same radius from the new vertex X cuts the new arm at Z, and Y is marked with ZY = BC.
AB = XY and AC = XZ (same radius); BC = ZY (length transferred).
So △ABC ≅ △XYZ by SSS, and therefore ∠A = ∠X.
In simple words: Same radius and same chord give congruent triangles, so the angles are equal.

Teacher's Note:
1. Three pairs of equal sides: SSS.
2. Matching parts of congruent triangles are equal.

Question 13. A line l cuts a given line m at A, and B is a point on l. Explain how copying the angle between m and l at the point B produces a line through B parallel to m, and name the angle property that makes it work.
Answer:
The line l is a transversal of m and of the new line.
Copying the angle that m makes with l at A onto B, on the same side and in the same direction, makes the new line form an equal angle with l.
These are corresponding angles; since they are equal, the new line through B is parallel to m.
In simple words: Equal corresponding angles on a transversal give parallel lines.

Teacher's Note:
1. Copy on the same side of l.
2. Name the property: equal corresponding angles.

Question 14. Six angles with a common vertex O are placed one after another without overlapping. Their measures are 35°, 60°, 85°, 50° and 70°, and one gap is left.
(i) Find the gap angle.
(ii) Will an angle of 65° fit exactly into the gap? Give the reason.
Answer:
(i) Sum = 35° + 60° + 85° + 50° + 70° = 300°. Gap = 360° − 300° = 60°.
(ii) No. 65° is larger than the 60° gap, so it would overlap the angles beside it. Only exactly 60° fits.
In simple words: The gap is 360° minus the sum; 65° is too big for a 60° gap.

Teacher's Note:
1. The gap is the sixth angle.
2. Add carefully before subtracting.

Question 15. Show by calculation that copies of a regular pentagon alone cannot tile the plane, while copies of a regular hexagon can.
Answer:
Pentagon: each angle = \( \frac{(5-2)\times 180^\circ}{5} \) = 108°. 360° ÷ 108° = 3 \( \frac{1}{3} \), not a whole number, so a gap or overlap always remains: pentagons do not tile.
Hexagon: each angle = \( \frac{(6-2)\times 180^\circ}{6} \) = 120°. 360° ÷ 120° = 3, so three hexagons fit exactly at every point: hexagons tile.
In simple words: 108° does not divide 360° but 120° does.

Teacher's Note:
1. Use the interior-angle formula.
2. Whole number means it tiles.

Question 16. Six congruent equilateral triangles are placed around a point O to form the regular hexagon ABCDEF.
(i) Why do exactly six triangles fit around O without a gap or an overlap?
(ii) Why is each angle of the hexagon 120°?
Answer:
(i) Each triangle angle is 60° and 6 × 60° = 360°, the complete angle around O, so there is no gap and no overlap.
(ii) At each vertex of the hexagon, angles of two triangles meet, so each angle is 60° + 60° = 120°.
In simple words: Six 60° angles make 360° at O, and two 60° angles make each 120° corner.

Teacher's Note:
1. Look at the centre for (i) and a corner for (ii).
2. All six triangles are the same size.

Question 17. While constructing the perpendicular bisector of XY, a student uses one radius for the pair of arcs drawn above XY and a different (larger) radius for the pair drawn below XY. Will the line joining the two meeting points still be the perpendicular bisector? Justify your answer.
Answer:
Yes.
The two arcs above have the same radius, so their meeting point A has AX = AY.
The two arcs below have the same radius as each other, so BX = BY.
Both A and B are equidistant from X and Y, so both lie on the perpendicular bisector; two points fix a line, so AB is the perpendicular bisector.
In simple words: Each pair uses one radius, so both points still lie on the bisector.

Teacher's Note:
1. Only the shape of the figure changes.
2. The radii must match within each pair.

Question 18. The 4 × 4 grid shown below has two unit squares, at opposite corners, cut away. Explain, using the black-and-white colouring idea, why the remaining region cannot be tiled with 2 × 1 tiles, even though the number of unit squares left is even.

Answer:
The grid has 4 × 4 = 16 squares; after removing two, 14 remain. 14 is even, so counting alone does not forbid a tiling.
Colour the grid like a chessboard: 8 black and 8 white squares. Opposite corners always have the same colour, so both removed squares are of one colour.
The region left has 8 squares of one colour and only 6 of the other.
Every 2 × 1 tile covers one black and one white square, so a tiling needs equal numbers. Since 8 ≠ 6, the region cannot be tiled.
In simple words: Both removed squares have the same colour, leaving 8 of one colour and 6 of the other.

Teacher's Note:
1. Shade the grid to check opposite corners.
2. Even count is not enough on its own.

Question 19. At one vertex of a tiling, only equilateral triangles and squares meet, with no gap and no overlap.
(i) If three equilateral triangles meet there, how many squares must also meet there?
(ii) Show that four equilateral triangles and one square cannot meet at a vertex.
Answer:
(i) Three triangles give 3 × 60° = 180°. The squares must give 360° − 180° = 180°, so the number of squares = 180° ÷ 90° = 2.
(ii) 4 × 60° + 90° = 240° + 90° = 330°, which is less than 360°. A 30° gap is left, so such a vertex is impossible.
In simple words: Three triangles need two squares; four triangles and one square leave a 30° gap.

Teacher's Note:
1. Always compare the total with 360°.
2. Triangle 60°, square 90°.

Question 20. A line l is drawn and a point P is marked outside it. Describe how a perpendicular from P to l can be constructed with a ruler and a compass, and justify why the line obtained is perpendicular to l.
Answer:
Step 1: With centre P and a radius large enough, draw an arc cutting l at A and B.
Step 2: With equal radii, draw arcs from A and B on the side of l away from P, meeting at Q.
Step 3: Join PQ; it meets l at right angles.
Justification: PA = PB (same arc) and QA = QB (equal arcs), so P and Q are both equidistant from A and B. Hence PQ is the perpendicular bisector of AB, so PQ ⟂ l.
In simple words: Make AB on l with P equidistant from its ends; the bisector of AB passes through P.

Teacher's Note:
1. The arc from P must cut l twice.
2. P and Q are the two equidistant points.