StudyOSStudyOS CBSE Study Material, Class 5–12

Constructions and Tilings

NCERT Class 7 · Mathematics Based on NCERT Class 7 Mathematics textbook · Free CBSE study kit

Everything you need to ace this chapter, in one place.
Notes · Mind Map · Quiz · Assertion-Reason · Case Studies · Question Bank · Past Papers

Get Full Study Kit →
Studying with friends? Send them this chapter.

Chapter Notes

Overview
This chapter teaches students how to construct geometric figures like perpendicular bisectors, angles, and parallel lines using only an unmarked ruler and a compass. It also introduces the concept of tiling, which is covering a region without gaps or overlaps using shapes like squares, triangles, and hexagons. The big idea is that precise geometric constructions rely on congruence of triangles, and tiling problems can be solved using creative thinking like colouring grids. A student must walk away understanding that geometry is not just about drawing but about logical reasoning and proof.
Key Concepts (13)
Bisection
A division of a line, or any geometrical object, into two identical parts is called bisection.
e.g. When you cut a piece of roti exactly in half, you have bisected it.
Perpendicular bisector
A line that bisects a given line and is perpendicular to it is called the perpendicular bisector.
e.g. The line that cuts a pencil exactly in the middle and stands straight up at 90° is its perpendicular bisector.
Angle bisector
A ray that divides an angle into two equal parts is called the angle bisector.
e.g. If you fold a piece of paper with a corner angle so that the two sides meet, the fold line is the angle bisector.
Congruence
Two triangles are congruent if their corresponding sides and angles are equal.
e.g. Two identical chapatis are congruent because they have the same shape and size.
SSS congruence condition
If three sides of one triangle are equal to three sides of another triangle, the triangles are congruent.
e.g. If two triangles have all three sides of the same length, they are congruent by SSS.
SAS congruence condition
If two sides and the included angle of one triangle are equal to two sides and the included angle of another triangle, the triangles are congruent.
e.g. If two sides of a triangle and the angle between them match another triangle, they are congruent by SAS.
Regular polygon
A regular polygon has equal sides and equal angles.
e.g. A square is a regular polygon with 4 equal sides and 4 right angles.
Equilateral triangle
A triangle with all three sides equal and all three angles equal to 60°.
e.g. A samosa cut into a triangle shape with all sides equal is an equilateral triangle.
Tiling
Covering a region using a set of shapes, without gaps or overlaps, is called tiling.
e.g. The floor of your kitchen covered with square tiles without any gaps is a tiling.
Transversal
A line that intersects two or more lines at distinct points.
e.g. A road crossing two railway tracks is a transversal.
Corresponding angles
Angles that are in the same position relative to the transversal and the two lines.
e.g. When a line cuts two parallel lines, the angles on the same side of the transversal and in the same position are corresponding angles.
Śulba-Sūtras
Geometric texts of Vedic period dealing with the construction of fire altars for rituals.
e.g. The Śulba-Sūtras describe how to construct a perpendicular bisector using a rope.
Tangram
A puzzle that uses 7 pieces obtained by dividing a square, used to form different shapes.
e.g. You can rearrange the 7 tangram pieces to make an arrow or a cat shape.
Formulas & Equations (4)
∠AOX + ∠AOY = 180°
Reads as: Angle AOX plus angle AOY equals 180 degrees
Variables: ∠AOX and ∠AOY are angles formed by line AO with line XY at point O
Use when: When proving that a line is perpendicular to another line using straight angle property
⚠ Students forget that the sum of angles on a straight line is 180°, not 360°
Sum of angles around a point = 360°
Reads as: The sum of all angles around a point is 360 degrees
Variables: Each angle is measured in degrees
Use when: Checking if multiple angles can fit around a point without gaps, e.g., six 60° angles
⚠ Students think the sum is 180° because they confuse it with a straight line
Each angle of an equilateral triangle = 60°
Reads as: Each angle of an equilateral triangle equals 60 degrees
Variables: No variables — it is a fixed property
Use when: Constructing a 60° angle or proving that six equilateral triangles fit around a point
⚠ Students think equilateral triangles have 90° angles like a square
Interior angle of a regular hexagon = 120°
Reads as: Interior angle of a regular hexagon equals 120 degrees
Variables: No variables — derived from 60° + 60° = 120°
Use when: Constructing a regular hexagon or understanding its properties
⚠ Students think the interior angle of a hexagon is 60° because of the equilateral triangles inside
How It Works
Construction of perpendicular bisector of a line segment
  1. Step 1: Taking some fixed radius, from X and then Y, construct two sufficiently long arcs above XY. Name the point where the arcs meet as A.
  2. Step 2: Using the same radius, from X and then Y, construct two sufficiently long arcs below XY. Name the point where the arcs meet as B.
  3. Step 3: AB is the required perpendicular bisector.
💡 The radius for arcs above and below must be the same, and the arcs must be long enough to intersect.
Construction of a 90° angle at a given point on a line
  1. Step 1: Draw a line and mark a point O on it.
  2. Step 2: Extend the line on either side of O.
  3. Step 3: Using a compass, mark two points X and Y at equal distance from O, so that O is the midpoint of XY.
  4. Step 4: Construct the perpendicular bisector of XY (only one pair of arcs needed since O is already a point on the bisector).
  5. Step 5: The line through O and the intersection of arcs is perpendicular to the given line, forming a 90° angle.
💡 You only need one pair of arcs because O is already one point on the perpendicular bisector.
Angle bisection
  1. Step 1: Mark points A and B on the arms of the angle such that OA = OB.
  2. Step 2: Choosing any sufficiently long radius, cut arcs from A and B, keeping the radius same. Mark the point of intersection as C.
  3. Step 3: OC bisects ∠AOB.
💡 The arcs from A and B must have the same radius to ensure BC = AC, making triangles congruent by SSS.
Copying an angle
  1. Step 1: Draw an arc from the vertex A of the given angle, giving points B and C on the arms.
  2. Step 2: Draw an arc of the same radius from the new vertex X.
  3. Step 3: Measure BC using a compass and transfer this length on the arc from Z to get YZ = BC.
  4. Step 4: By SSS congruence, ∆ABC ≅ ∆XYZ, so ∠A = ∠X.
💡 The radius for the first arc must be the same for both angles, and the distance BC must be transferred accurately.
Construction of a line parallel to a given line
  1. Step 1: Draw a transversal line l that intersects the given line m at point A.
  2. Step 2: Choose a point B on l through which the parallel line will pass.
  3. Step 3: Construct arcs of equal radius from A and B.
  4. Step 4: Transfer the length from the arc at A to the arc at B to copy the corresponding angle.
  5. Step 5: Draw the line through B and the transferred point — this line is parallel to m.
💡 The key is to copy the corresponding angle using the compass; the transversal must intersect both lines.
Construction of a 60° angle
  1. Step 1: Draw a line segment AX with A as the vertex.
  2. Step 2: Construct an arc with centre A and any radius, meeting AX at B.
  3. Step 3: With the same radius, cut another arc from B that meets the first arc. Let C be the point where the arcs meet.
  4. Step 4: ∠CAX = 60° because triangle ABC is equilateral.
💡 The radius for both arcs must be the same to form an equilateral triangle.
Construction of a regular hexagon
  1. Step 1: Construct a 60° angle at a point.
  2. Step 2: Use the 60° angle to construct a 120° angle (since 180° - 60° = 120°).
  3. Step 3: Draw six sides of equal length, each turning by 120° at the vertices.
  4. Alternatively: Draw a circle and mark six points on it at 60° intervals using the compass radius equal to the circle's radius.
💡 A regular hexagon can be divided into six equilateral triangles, so the side length equals the radius of the circumscribed circle.
Tiling a grid with 2×1 tiles
  1. Step 1: Check if the total number of unit squares in the grid is even (since each tile covers 2 squares).
  2. Step 2: If both dimensions are even, tile each column with vertical tiles.
  3. Step 3: If one dimension is even and the other odd, tile the even dimension first.
  4. Step 4: If both dimensions are odd, tiling is impossible because the total number of squares is odd.
  5. Step 5: For irregular regions, use black-and-white colouring to check if the number of black and white squares is equal.
💡 A 2×1 tile always covers one black and one white square in a chessboard-coloured grid, so the region must have equal numbers of black and white squares.
Exam Traps — Don't Make These Mistakes
⚠ Using different radii for arcs in perpendicular bisector construction
✗ Students write that the arcs above and below XY can have different radii.
→ If the radii are different, the intersection points A and B will not be equidistant from X and Y, so AB will not be the perpendicular bisector.
✓ The arcs above and below XY must be drawn with the same radius to ensure that the intersection points are equidistant from X and Y.
⚠ Confusing angle bisector with perpendicular bisector
✗ Students write that the angle bisector of an angle is the same as the perpendicular bisector of the opposite side.
→ An angle bisector divides an angle into two equal parts, while a perpendicular bisector divides a line segment into two equal parts at 90°. They are different concepts.
✓ The angle bisector is a ray that divides an angle into two equal angles, while the perpendicular bisector is a line that divides a line segment into two equal parts and is perpendicular to it.
⚠ Forgetting that arcs must intersect in angle bisection
✗ Students draw arcs from A and B that do not intersect and then draw a line from O to the midpoint of AB.
→ The arcs must intersect to define point C; otherwise, there is no unique point to determine the bisector. The midpoint of AB does not guarantee equal angles.
✓ Draw arcs from A and B with the same radius so that they intersect at a point C. Then OC is the angle bisector.
⚠ Assuming any odd×odd grid can be tiled if one square is removed
✗ Students think that removing any one square from a 5×7 grid makes it tileable with 2×1 tiles.
→ Only if the removed square is of the same colour as the majority colour in a chessboard pattern will the numbers of black and white squares become equal. Removing a square of the minority colour makes tiling impossible.
✓ For a 5×7 grid, there are 18 black and 17 white squares (or vice versa). Removing a black square leaves 17 black and 17 white, making tiling possible. Removing a white square leaves 18 black and 16 white, making tiling impossible.
⚠ Thinking a 60° angle can be constructed without an equilateral triangle
✗ Students try to construct a 60° angle by bisecting a 120° angle or using a protractor.
→ The chapter specifies using a ruler and compass only. The correct method is to construct an equilateral triangle, which gives a 60° angle naturally.
✓ Construct an equilateral triangle by drawing two arcs of equal radius from the endpoints of a base. The angle at the base is 60°.
Compare & Contrast
What they divide
Perpendicular bisector
Angle bisector
Perpendicular bisector divides a line segment into two equal parts at 90°, while angle bisector divides an angle into two equal angles.
Possibility of tiling
Tiling with 2×1 tiles on an even×even grid
Tiling with 2×1 tiles on an odd×odd grid
An even×even grid can always be tiled because the total number of squares is even. An odd×odd grid cannot be tiled because the total number of squares is odd, and each tile covers 2 squares.
Exam Focus
Marks pattern: 2 MCQs (1 mark each), 1 SA (3 marks) on construction steps, 1 LA (5 marks) on tiling or combined construction
Most asked: Construction of perpendicular bisector · Construction of angle bisector · Construction of 60° and 90° angles · Tiling problems with 2×1 tiles and colouring arguments
Don't skip: Construct a perpendicular bisector and justify using congruence; Tiling a grid with a missing square using black-and-white colouring
One-Line Revision (read aloud before the exam)
  1. A perpendicular bisector divides a line segment into two equal parts at 90°.
  2. Any point equidistant from the endpoints of a line segment lies on its perpendicular bisector.
  3. To construct a perpendicular bisector, draw arcs of equal radius above and below the line segment from both endpoints.
  4. A 90° angle at a point on a line can be constructed by making that point the midpoint of a segment and drawing its perpendicular bisector.
  5. An angle bisector divides an angle into two equal angles.
  6. To bisect an angle, mark equal distances on the arms, then draw intersecting arcs of equal radius from those points.
  7. A 60° angle is constructed by drawing an equilateral triangle using arcs of equal radius.
  8. A regular hexagon has interior angles of 120° and can be divided into six equilateral triangles.
  9. Tiling means covering a region without gaps or overlaps.
  10. A 2×1 tile covers exactly two unit squares.
  11. A grid with an odd number of squares cannot be tiled with 2×1 tiles.
  12. Colouring a grid like a chessboard helps determine if a region can be tiled with 2×1 tiles — the numbers of black and white squares must be equal.
  13. The Śulba-Sūtras are ancient Indian texts describing geometric constructions using ropes.
  14. Tangrams use 7 pieces from a square to form different shapes.

MCQs — 10 Questions with Answers

Q1. What is the main purpose of constructing a perpendicular bisector using arcs from X and Y?

  • A. To find two points equidistant from both X and Y ✓
  • B. To measure the length of XY
  • C. To draw circles at X and Y
  • D. To mark random points on the line

Answer: A — The perpendicular bisector is found by locating points A and B that satisfy AX = AY = BX = BY, making them equidistant from both endpoints.

Q2. Which congruence condition is used to prove that the line AB is the perpendicular bisector of XY?

  • A. AAA (Angle-Angle-Angle)
  • B. SAS (Side-Angle-Side) ✓
  • C. ASA (Angle-Side-Angle)
  • D. RHS (Right angle-Hypotenuse-Side)

Answer: B — SAS congruence is used because we show that AX = AY, angle XAO = angle YAO, and AO is common to both triangles AOX and AOY.

Q3. What angle is formed between the perpendicular bisector and the line segment it bisects?

  • A. 45°
  • B. 60°
  • C. 90° ✓
  • D. 180°

Answer: C — By definition, a perpendicular bisector forms a 90° angle with the line segment it bisects.

Q4. In the ancient Śulba-Sūtras, what tool was used instead of a compass to draw arcs and lines?

  • A. A straightedge
  • B. A rope ✓
  • C. A protractor
  • D. A wooden rod

Answer: B — The Śulba-Sūtras used ropes to draw circles or arcs and to stretch straight lines during construction of fire altars.

Q5. If you want to construct a 45° angle, which of these steps would you follow?

  • A. Directly measure 45° with a protractor
  • B. First construct a 90° angle, then bisect it ✓
  • C. Draw two lines at random and measure the angle
  • D. Use a rope to mark 45° on the ground

Answer: B — A 45° angle is constructed by first creating a 90° angle and then using angle bisection to divide it into two 45° angles.

Q6. In an 8-petalled flower design, what is the angle between each adjacent pair of petals?

  • A. 30°
  • B. 45° ✓
  • C. 60°
  • D. 90°

Answer: B — Since 360° is divided equally among 8 petals, each angle is 360° ÷ 8 = 45°.

Q7. To bisect an angle XOY, you mark points A and B on the rays OX and OY such that OA = OB. What should be true about point C where the arcs from A and B intersect?

  • A. C should be on ray OX
  • B. C should be on the angle bisector of angle XOY ✓
  • C. C should be equidistant from the ruler
  • D. C should be on ray OY

Answer: B — When AC = BC and OA = OB, by SSS congruence, triangles OAC and OBC are congruent, so OC bisects angle XOY.

Q8. Anika wants to find the midpoint of a wooden stick XY using construction method instead of measuring. Which method is more accurate?

  • A. Measuring with a marked scale and dividing by 2
  • B. Constructing the perpendicular bisector using compass and ruler ✓
  • C. Folding the stick in half
  • D. Marking the stick visually by eye

Answer: B — The perpendicular bisector construction method is more accurate than using a marked scale because it doesn't depend on measurement precision.

Q9. In the rope method from Śulba-Sūtras, after fastening the rope at pegs X and Y, what do you do to mark the perpendicular bisector?

  • A. Pull the midpoint of the rope upward and downward at different positions to mark points A and B ✓
  • B. Cut the rope at its midpoint
  • C. Measure the rope length and divide by 2
  • D. Stretch the entire rope tightly along XY

Answer: A — Pulling the rope's midpoint above and below XY with the rope fully stretched creates points A and B that form the perpendicular bisector.

Q10. When constructing the perpendicular bisector of XY by drawing arcs, is it necessary that both pairs of arcs (above and below XY) use the same radius?

  • A. Yes, both pairs must use identical radii
  • B. No, the radii can differ between pairs, but must be equal within each pair ✓
  • C. Yes, because the arcs must meet at the same point
  • D. No, all four arcs can use different radii

Answer: B — Only points equidistant from X and Y matter; so arcs from X and Y in each pair must use the same radius, but different pairs can differ.

Flashcards

What is a perpendicular bisector?

A line that passes through the midpoint of a line segment and is perpendicular (90°) to it.

Which congruence condition proves perpendicular bisector construction?

SAS (Side-Angle-Side) congruence, showing that triangles AOX and AOY are congruent.

How many points do you need to find if they are equidistant from X and Y?

You need two points (A and B) that are equidistant from both X and Y to construct the perpendicular bisector.

What is the angle formed by a perpendicular bisector with the original line segment?

The angle is 90° (a right angle), by definition of perpendicular.

What does 'bisection' mean in geometry?

Bisection means dividing a geometric object into two identical or equal parts.

Name the ancient Indian texts that contain construction methods.

The Śulba-Sūtras, which are Vedic geometric texts dealing with construction of fire altars.

How do you construct a 45° angle using ruler and compass?

First construct a 90° angle, then bisect it by finding two points equidistant from the vertex and drawing equal arcs.

What is the relationship between 360° and an 8-petalled figure?

360° is divided into 8 equal parts, creating angles of 45° between adjacent petals.

What must be true about OA and OB to bisect angle XOY?

OA and OB must be equal in length, so point A and B are at equal distances from vertex O.

Can you construct a perpendicular bisector using arcs on only one side of the line segment?

Yes, if you mark two different points using arcs on the same side, any two such points determine the perpendicular bisector.

Important Board Questions

Define the term 'perpendicular bisector' in your own words. [1 mark]

State that it bisects (divides into two equal parts) a line segment and meets it at a 90° angle.

Explain with a reason why point C lies on the perpendicular bisector of segment XY if CX = CY. [2 marks]

Use the fact that any point equidistant from X and Y must lie on the perpendicular bisector; refer to congruent triangles or the geometric property directly.

Describe the steps to construct a perpendicular bisector of a line segment XY using only a ruler (unmarked) and a compass. What geometric principle ensures this method works? [3 marks]

Steps: (1) Draw equal arcs from X and Y above XY meeting at A; (2) Draw equal arcs from X and Y below XY meeting at B; (3) Join AB. Principle: Points A and B are equidistant from X and Y, so they lie on the perpendicular bisector; use congruence (SAS or SSS) to prove AB ⊥ XY at the midpoint.

A designer wants to create an 8-petalled flower design where all petals are equally spaced around a central point O. Explain how you would construct the rays that support these 8 petals. Show all steps and state the angle between adjacent rays. (Draw and label a diagram showing at least 3 of the 8 rays.) [5 marks]

Total angle around O is 360°; divide by 8 to get 45° between adjacent rays. Steps: (1) Draw any ray; (2) Construct a 90° angle using perpendicular bisector method; (3) Bisect the 90° angle to get 45°; (4) Repeat bisection or use compass to mark 45° intervals around O. Diagram should show O at centre with rays at 45° intervals and labels for angles; congruence ensures all angles are equal.

Assertion-Reason Questions for Constructions and Tilings

Q1. Assertion (A):

The perpendicular bisector of a line segment is unique.

Reason (R):

A line segment can have only one midpoint, and through a given point, only one perpendicular can be drawn to a given line.

Answer: A — Both A and R are true, and R is the correct explanation of A — Both statements are true, and the reason correctly explains why the perpendicular bisector is unique: the midpoint is unique (bisection is unambiguous) and the perpendicular through that point is unique (by geometric axioms). Together they guarantee a single perpendicular bisector.

Q2. Assertion (A):

Any point that is equidistant from two endpoints of a line segment must lie on the perpendicular bisector of that segment.

Reason (R):

All points at equal distance from a segment's endpoints form a circle centered at the segment's midpoint.

Answer: C — A is true but R is false — The assertion is true and is the fundamental property proven through congruent triangles. However, the reason is false: points equidistant from two endpoints form a perpendicular bisector (a line), not a circle. A circle is formed by points at a fixed distance from a single point (center).

Q3. Assertion (A):

When constructing a 90° angle at a point O on a line, we must first create a segment with O as its midpoint.

Reason (R):

Because the perpendicular bisector of any segment passes through the midpoint at a right angle.

Answer: B — Both A and R are true, but R is NOT the correct explanation of A — Both statements are true. The assertion describes the correct procedure, and the reason accurately describes the geometric property. However, the reason is not the direct explanation for why the procedure works—rather, it supports the method. The reason explains what happens, but the assertion works because we're specifically constructing O as the midpoint of XY, ensuring the perpendicular bisector passes through O.

Unlock more Assertion-Reason questions in Premium — unlock →

Case-Based Questions (CBQ) for Constructions and Tilings

Case Study:

A sports event organizer is setting up a relay race track with two parallel walls on either side. She wants to place a timing sensor that is equidistant from both walls. The walls form a 90° angle at the corner of the venue. To find all possible positions where the timing sensor can be placed equidistant from both walls, the organizer consults a geometry expert. The expert suggests constructing a specific geometric line using ruler and compass that will give all valid sensor positions.

Q1. What geometric construction should the organizer use to find all points equidistant from the two walls?

  • A. The perpendicular bisector of the segment joining the two walls
  • B. The angle bisector of the 90° angle formed by the two walls ✓
  • C. A circle centered at the corner of the venue
  • D. A line parallel to both walls at equal distance

Answer: B

Q2. If one wall runs north-south and the other runs east-west, in which direction does the line of equidistant points travel?

  • A. Along one of the walls
  • B. Perpendicular to both walls
  • C. At a 45° angle to both walls, dividing the right angle into two 45° angles ✓
  • D. In a curved path away from both walls

Answer: C

Q3. Why are all points on this geometric line equidistant from the two walls?

  • A. Because they are all the same distance from the corner
  • B. Because this line is the locus of points with equal perpendicular distances to both walls ✓
  • C. Because the line is parallel to the walls
  • D. Because the walls determine the endpoints of the line

Answer: B

Q4. If the organizer wants to place multiple sensors at different positions along this line, will they all be equidistant from the walls?

  • A. No, only the first sensor will be equidistant
  • B. No, the distance changes as you move along the line
  • C. Only if they are placed at equal intervals
  • D. Yes, every point on the angle bisector maintains equal distance from both walls ✓

Answer: D

Unlock more case studies in Premium — unlock →

Next chapterFinding the Unknown →
Found this useful? Share it with your class group.

You've seen the free preview. Open the full interactive study kit — mind map, instant-graded quiz, progress tracking — free, no sign-in required.

Open Full Study Kit — Free →

Premium unlocks the full Question Bank, Daily Practice Papers, Chapter Quiz and Full Mock Papers for every chapter.