Class 7 Maths Chapter 6 Constructions and Tilings notes and book solutions from Ganita Prakash Part 2 with easy explanations, important concepts, examples and complete textbook question answers for revision and exam preparation.
Constructions and Tilings
Grade 7 Mathematics – Ganita Prakash Part II
Easy Notes, Concepts and Book Solutions
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Easy Concepts
Perpendicular Bisector
A line that cuts a line segment into two equal parts and makes a 90° angle with it.
A line that cuts a line segment into two equal parts and makes a 90° angle with it.
Bisection
Bisection means dividing something into two equal parts.
Bisection means dividing something into two equal parts.
Angle Bisector
A line that divides an angle into two equal angles.
A line that divides an angle into two equal angles.
60° Angle
A 60° angle can be made by constructing an equilateral triangle.
A 60° angle can be made by constructing an equilateral triangle.
Parallel Lines
Lines that never meet can be constructed by copying a corresponding angle.
Lines that never meet can be constructed by copying a corresponding angle.
Regular Hexagon
A regular hexagon has six equal sides and six equal angles.
A regular hexagon has six equal sides and six equal angles.
Tiling
Tiling means covering a region without gaps or overlaps.
Tiling means covering a region without gaps or overlaps.
Domino Tile
A 2 × 1 tile covers exactly two unit squares.
A 2 × 1 tile covers exactly two unit squares.
Important Construction Ideas
Perpendicular Bisector: Draw equal-radius arcs from X and Y. Their intersection points are equally distant from X and Y. Joining these points gives the perpendicular bisector.
Angle Bisection: Mark equal distances on the two arms of an angle. Draw equal arcs from those points. Join their intersection to the vertex.
60° Construction: Construct an equilateral triangle. Each angle of an equilateral triangle is 60°.
30° Construction: Bisect a 60° angle.
15° Construction: Bisect a 30° angle.
Regular Hexagon: Use the side length as the compass radius and step it around a circle six times.
Domino Tiling: Every 2 × 1 tile covers two unit squares. Therefore a region with an odd number of unit squares cannot be completely tiled using 2 × 1 tiles.
Book Questions and Easy Solutions
Q1
In Fig. 6.1, join A and B with a line. Where does AB intersect XY, and what is the angle formed between them?
Easy Answer:AB intersects XY at its midpoint.AB is perpendicular to XY.Therefore, the angle formed between AB and XY is 90°.
Q2
Which two triangles should be congruent for AB to be the perpendicular bisector of XY (that is, O is the midpoint of XY and AB is perpendicular to XY)?
Easy Answer:We should prove thatΔAOX ≅ ΔAOY.Then OX = OY and the two angles at O are equal.Since they form a straight angle, each angle is 90°.
Q3
Justify the following statement using the facts that we have established. Any point that has the same distance from X and Y lies on the perpendicular bisector of XY.
Easy Answer:Suppose a point P has equal distances from X and Y.So,PX = PY.A point that is equally distant from X and Y lies on the perpendicular bisector of XY.Therefore, P lies on the perpendicular bisector of XY.
Q4
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.
Easy Answer:No.The arcs above and below XY do not have to use the same radius.What is important is that each intersection point is equally distant from X and Y.Therefore, both points lie on the perpendicular bisector of XY.
Q5
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.
Easy Answer:No.Both pairs of arcs can be constructed on the same side.The two intersection points will still be equally distant from X and Y.Therefore, both points lie on the perpendicular bisector of XY.Joining the two points gives the required perpendicular bisector.
Q6
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.
Easy Answer:Yes.For the usual construction, the two arcs should have the same radius.Then their intersection point is equally distant from X and Y.Therefore, the point lies on the perpendicular bisector.
Q7
4. Recreate this design using only a ruler and compass—
Easy Answer:Use only the ruler and compass.First copy the given supporting lines.Then use the compass to copy the required lengths and arcs.Repeat the construction symmetrically until the complete design is obtained.
Q8
Justify why AB in Fig. 6.4 is the perpendicular bisector.
Easy Answer:A and B are equally distant from X and Y.Therefore, A and B both lie on the perpendicular bisector of XY.Hence, the line AB is the perpendicular bisector of XY.
Q9
Can you think of different methods to construct a 90° angle at a given point on a line using a rope?
Easy Answer:Yes.Mark two points on the line at equal distances from the given point.Use the rope to find a point equally distant from both marked points.Join this new point to the given point.The new line will be perpendicular to the original line and will make a 90° angle.
Q10
1. Construct at least 4 different angles. Draw their bisectors.
Easy Answer:Draw four different angles.For each angle:1. Draw an arc cutting both arms.
2. From the two points on the arms, draw equal arcs.
3. Mark their intersection.
4. Join this point to the vertex.The joining line is the angle bisector.
Q11
2. Construct the 8-petalled figure shown in Fig. 6.5.
Easy Answer:A full angle is 360°.The figure has 8 equal parts.Therefore,360° ÷ 8 = 45°.Construct a 90° angle and bisect it to obtain 45°.Repeat the 45° construction around the centre.Then draw the petals.
Q12
3. In Step 2 of angle bisection, if arcs of equal radius are drawn on the other side, as shown in the figure, will the line OC still be an angle bisector? Explore this through construction, and then justify your answer.
Easy Answer:Yes.The new point is equally distant from the two points on the arms.Therefore, the two triangles formed are congruent.Hence, OC still divides the angle into two equal angles.
Q13
4. What are the other angles that can be constructed using angle bisection? Can you construct 65.5° angle?
Easy Answer:Using angle bisection, we can construct angles such as:45°, 30°, 15°, 22.5°, 7.5° and many other halves of constructible angles.A 65.5° angle cannot be obtained simply by repeatedly bisecting the standard 90° and 60° constructions.
Q14
5. Come up with a method to construct the angle bisector using a rope.
Easy Answer:Use the rope like a compass.Mark equal distances on the two arms of the angle.From these two points, use equal rope lengths to locate another point.Join that point to the vertex.The joining line is the angle bisector.
Q15
6. Construct the following figure. How do we construct the petals so that they are of the maximum possible size within a given square?
Easy Answer:Use the square as the boundary.Choose the centres of the arcs according to the figure.Use the largest possible radius that keeps every petal inside the square.Repeat the same construction symmetrically.
Q16
Draw an angle. Create a copy of this angle using only a ruler and compass.
Easy Answer:1. Draw an arc from the original angle.
2. Draw an arc of the same radius from the new point.
3. Measure the distance between the two points made by the first arc.
4. Transfer this distance to the second arc.
5. Join the new point to the transferred point.The new angle is equal to the original angle.
Q17
1. Construct at least 4 different angles in different orientations without taking any measurement. Make a copy of all these angles.
Easy Answer:Draw four angles in different directions.For each angle, draw an arc through both arms.Draw the same arc from the new vertex.Transfer the distance between the first two arc points.Join the points.The copied angle will be equal to the original angle.
Q18
2. Construct the Fig. 6.6.
Easy Answer:Construct one basic unit first.Copy its arm lengths using the compass.Copy its angle using the angle-copying construction.Repeat the unit in the required orientations.
Q19
Construct a line parallel to the given line.
Easy Answer:Draw a transversal through the given line.Choose a point where the parallel line is required.Copy the corresponding angle at that point.Extend the new line.The two lines are parallel because their corresponding angles are equal.
Q20
1. Construct 4 pairs of parallel lines in different orientations.
Easy Answer:Draw a line and a transversal.Copy the corresponding angle at another point.Draw the second line.Repeat this in four different directions.
Q21
2. Construct the following figure.
Easy Answer:Use the given line segments as supporting lines.Copy the required lengths with the compass.Copy the required angles.Join the marked points in the same order as the textbook figure.
Q22
Construct this arch shape on a piece of paper.
Easy Answer:For a symmetric arch, make the two support sides equal.Construct equal angles at the two ends.Mark equal distances for the support points.Use the compass to draw matching arcs.
Q23
1. Use support lines in Fig. 6.11 to construct a pointed arch. Make different arches, by changing the radius of the arcs.
Easy Answer:Draw the two equal support line segments.Mark their midpoints.Use the marked points as centres.Draw arcs with the required radius.Changing the radius gives different pointed arches.
Q24
2. Make your own arch designs.
Easy Answer:Make equal support lines for a balanced design.Choose suitable arc centres.Use equal radii for matching sides.Try different radii to make different arch designs.
Q25
Construct a regular hexagon with a sidelength 4 cm using a ruler and a compass.
Easy Answer:1. Draw a circle with radius 4 cm.
2. Choose any point on the circle.
3. Keep the compass opening at 4 cm.
4. Step the compass around the circle six times.
5. Join the six points.You get a regular hexagon of side 4 cm.
Q26
Construct a regular hexagon of sidelength 5 cm.
Easy Answer:1. Draw a circle of radius 5 cm.
2. Select one point on the circle.
3. Keep the compass opening at 5 cm.
4. Mark six points around the circle.
5. Join consecutive points.The result is a regular hexagon with side 5 cm.
Q27
1. Construct 30° and 15° angles.
Easy Answer:First construct 60°.Bisect 60° to get 30°.Then bisect 30° to get 15°.
Q28
2. Do you see a hexagon here? Are the six triangles forming the 6 points of the star — ∆AGH, ∆BHI, ∆CIJ, ∆DJK, ∆ELK, ∆FLG — equilateral? Why? [Hint: Find the angles.]
Easy Answer:Yes, a hexagon can be seen in the middle.The six triangles forming the points have angles of 60° and equal corresponding sides in the construction.Therefore, the triangles are equilateral triangles.
Q29
1. Construct the following figures: (a) An Inflexed Arc (b) The figure shown in the book.
Easy Answer:Follow the support lines shown in the textbook.Use the compass to copy the required lengths.Draw the arcs using the indicated centres.Join the required points to complete each figure.
Q30
2. Optical Illusion: Do you notice anything interesting about the following figure? How does this happen? Recreate this in your notebook.
Easy Answer:The repeated lines and patterns make some parts of the picture appear bent or shifted.This happens because our eyes compare the lines with the surrounding pattern.Copy the same lines, angles and spacing in your notebook to recreate the illusion.
Q31
3. Construct this figure. [Hint: Find the angles in this figure.]
Easy Answer:First identify the angles in the given figure.Construct the required basic angles using the ruler and compass.Copy or bisect the angles wherever necessary.Finally join all the marked points.
Q32
4. Draw a line l and mark a point P anywhere outside the line. Construct a perpendicular to the given line l through P. [Hint: Find a line segment on l whose perpendicular bisector passes through P.]
Easy Answer:1. With P as centre, draw an arc cutting line l at X and Y.
2. Since PX = PY, P is equally distant from X and Y.
3. Therefore P lies on the perpendicular bisector of XY.
4. Construct the perpendicular bisector of XY.
5. The perpendicular bisector passes through P and is perpendicular to l.
Q33
How can the tangram pieces be rearranged to form each of the following figures?
Easy Answer:Use all seven tangram pieces.Move and rotate the pieces until their outer edges form the required figure.There should be no gaps and no overlapping pieces.
Q34
Can a 4 × 6 grid be tiled using multiple copies of 2 × 1 tiles?
Easy Answer:Yes.A 4 × 6 grid has:4 × 6 = 24 unit squares.Each tile covers 2 squares.24 ÷ 2 = 12 tiles.Therefore, it can be tiled.
Q35
Can a 4 × 7 grid be tiled using 2 × 1 tiles?
Easy Answer:Yes.The grid has:4 × 7 = 28 unit squares.28 is even, so the squares can be paired.Therefore, the grid can be tiled using 2 × 1 tiles.
Q36
What about a 5 × 7 grid?
Easy Answer:No.The grid has:5 × 7 = 35 unit squares.Each 2 × 1 tile covers 2 squares.35 is odd, so the squares cannot be completely paired.Therefore, it cannot be tiled.
Q37
Complete the justification. Is an m × n grid tileable with 2 × 1 tiles, if both m and n are even? If yes, come up with a general strategy to tile it.
Easy Answer:Yes.If both m and n are even, the number of rows and columns can both be paired.One simple method is to cover each column with vertical 2 × 1 tiles.Since the number of rows is even, this can be done without gaps.
Q38
Is an m × n grid tileable with 2 × 1 tiles, if one of m and n is even and the other is odd? If yes, come up with a general strategy to tile it.
Easy Answer:Yes.Use the even dimension for pairing.For example, if m is even, place the 2 × 1 tiles vertically.If n is even, place them horizontally.
Q39
Is an m × n grid tileable with 2 × 1 tiles, if both m and n are odd? Give reasons.
Easy Answer:No.Odd × odd gives an odd number of unit squares.Each 2 × 1 tile covers exactly 2 squares.An odd number of squares cannot be completely divided into pairs.Therefore, the grid cannot be tiled.
Q40
Here is a 5 × 3 grid, with a unit square removed. Now, it has an even number of unit squares. Is it tileable with 2 × 1 tiles?
Easy Answer:Not necessarily.Having an even number of squares is necessary, but it is not always enough.We can use a black-and-white colouring to check whether the region can be tiled.Each 2 × 1 tile must cover one black and one white square.
Q41
Is the following region tileable with 2 × 1 tiles?
Easy Answer:Colour the region like a chessboard.Every 2 × 1 tile covers one black square and one white square.If the region has unequal numbers of black and white squares, it cannot be tiled.
Q42
What about this one? Fig. 6.13. Were you able to tile this? How can we be sure that this is not tileable?
Easy Answer:Use a black-and-white colouring.The book’s coloured region has 8 white squares and 6 black squares.Every 2 × 1 tile covers one square of each colour.Because the numbers are unequal, the region cannot be tiled.
Q43
Can you find another unit square that, when removed from a 5 × 3 grid, makes it non-tileable?
Easy Answer:Yes.Remove a square in such a position that the remaining black and white squares are unequal when the grid is coloured like a chessboard.Then the remaining region cannot be tiled with 2 × 1 tiles.
Q44
Are the following tilings possible?
1. Region to be tiled / Tile
2. Region to be tiled / Tile
Easy Answer:Check each region carefully.A tiling is possible only when:1. There are no gaps.
2. There are no overlaps.
3. Every part of the region is covered.
4. The number and arrangement of squares allow the tiles to fit.For these textbook figures, use the black-and-white colouring method wherever required.
Q45
Can you think of a shape whose copies can tile the entire plane?
Easy Answer:Yes.Squares can tile the entire plane.Equilateral triangles can also tile the plane.Regular hexagons can also tile the plane.
Extra Concept Questions from the Chapter
Q46
Will the line joining the two points at which the arcs meet, above and below XY, always be the perpendicular bisector of XY, i.e., when XY is of any length, and the arcs are drawn using a radius of any length?
Easy Answer:Yes, provided the arcs from X and Y in each pair intersect.The intersection points are equally distant from X and Y.Therefore, both points lie on the perpendicular bisector of XY.The line joining them is the perpendicular bisector.
Q47
How do we construct a 60° angle?
Easy Answer:Construct an equilateral triangle.All three sides of an equilateral triangle are equal.All three angles are 60°.Therefore, the required angle is 60°.
Chapter Summary
| Concept | Remember |
|---|---|
| Bisection | Dividing something into two equal parts. |
| Perpendicular Bisector | It cuts a line segment into two equal parts at 90°. |
| Equal Distances | A point equally distant from the endpoints of a segment lies on its perpendicular bisector. |
| Angle Bisector | It divides an angle into two equal angles. |
| 60° | Construct an equilateral triangle to obtain a 60° angle. |
| Regular Hexagon | It has six equal sides and six equal angles. |
| Tiling | Covering a region without gaps or overlaps. |
| 2 × 1 Tile | Each tile covers exactly two unit squares. |
⁷
Prepare for your exams with this easy and student-friendly worksheet based on Class 7 Maths Chapter 6 from Ganita Prakash Part 2. Practice all the questions carefully to revise the chapter, strengthen your concepts, and improve your mathematical understanding.
CLASS 7 • MATHEMATICS PART 2
Class 7 Maths Part 2 Chapter 6 Worksheet
Practice worksheet for Class 7 Mathematics Part 2 Chapter 6 with important questions, extra practice and answer key.
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