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6.1 Geometric Constructions — The “Eye”

🎓 Class 7 Mathematics CBSE Theory Ch 6 — Geometric Constructions ⏱ ~19 min
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This MCQ module is based on: 6.1 Geometric Constructions — The “Eye”

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6.1 Geometric Constructions — The "Eye"

Do you recall the 'Eyes' construction we did in Grade 6? Of course, eyes can be drawn freehand — but we wanted to construct them so that the lower arc and upper arc of each eye look symmetrical. We relied on our special estimation to determine the two centres, A and B (see the figure), from which we drew the lower arc and upper arc respectively.

The arcs define a line XY that 'supports' the drawing, though it is not part of the final figure. We can start with this supporting line and symmetrically find the centres A and B. For the eye to be symmetrical, or for the supporting line to be the line of symmetry, the upper and lower arcs should have the same radius. In other words, we need AX = AY and BX = BY, which means AX = AY = BX = BY.

Fig. 6.1 — Symmetric eye construction X Y A B perp. bisector of XY AX AY BX BY
Fig. 6.1 — A and B lie on the perpendicular bisector of XY. The two arcs have equal radii, giving a symmetric eye shape with AX = AY = BX = BY.

How do we find A and B?

From X and Y, draw arcs above and below XY of the same radius. The two points at which these arcs meet, above and below XY, give us A and B respectively. Use this to construct an eye.

In Fig. 6.1, join A and B with a line. Where does AB intersect XY? At the midpoint, O. We observe that AB passes through the midpoint of XY, and also AB is perpendicular to XY. A division of a line, or any geometrical object, into two identical parts is called a bisection?. A line that bisects a given line and is perpendicular to it, is called the perpendicular bisector?.

Will the line joining the two points A and B always be the perpendicular bisector of XY?

Yes. Whatever radius we choose (larger than half of XY), the arcs drawn from X and Y will always meet on the perpendicular bisector of XY. This can be reasoned through congruence. Consider a line segment XY. Point A is such that AX = AY = BX = BY. Let O be the point of intersection between AB and XY.

Which two triangles should be congruent for AB to be the perpendicular bisector of XY? If we show that △AOX ≅ △AOY, then OX = OY and ∠AOX = ∠AOY. Since OX + OY = XY and ∠AOX + ∠AOY = 180°, this gives OX = OY = XY/2 and ∠AOX = ∠AOY = 90°. Thus AB is the perpendicular bisector of XY.

In △AOX and △AOY we already know that AX = AY, and AO is common to both triangles. By the SAS congruence condition, we can conclude that the triangles are congruent. Similarly △BOX ≅ △BOY. Thus, we have ∠XAB = ∠YAB and ∠XAO = ∠YAO because they are corresponding parts of congruent triangles. Hence, AB is the perpendicular bisector of XY. We can have eyes of different shapes too.

Eye Variations
How do we get these different shapes? Try! One way is to choose two other points C and D such that CX = CY = DX = DY. An eye of a different shape can be drawn using these points. Will C and D lie on the perpendicular bisector AB? Yes — the points C and D are at the same distance from both X and Y just like A and B, so they also lie on the perpendicular bisector of XY. Points C and D must lie on the line AB.

6.2 Construction of the Perpendicular Bisector

Given a line segment XY, how do we draw its perpendicular bisector using only an unmarked ruler and a compass? We have seen that joining any two points above XY and below XY yields a perpendicular bisector, provided that they are at an equal distance from X and Y. This gives the perpendicular bisector. Here is a method to construct the perpendicular bisector.

  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. 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. AB is the required perpendicular bisector.
X Y A B O
Construction of the perpendicular bisector of XY using only ruler and compass.
🔵 Figure it Out
  1. When constructing the perpendicular bisector, is it necessary to have the same radius for the arcs above and below XY? Explore through construction, and then justify your answer. (Hint: Any point that is at the same distance from X and Y lies on the perpendicular bisector.)
  2. We can draw the whole line if any two of its points are known. Is it necessary to construct the pairs of arcs on the same side of XY? Instead, can we construct both pairs of arcs on the same side of XY? Explore through construction, and then justify your answer.
  3. While constructing one pair of intersecting arcs, is it necessary that we use the same radii for both of them? Explore through construction, and then justify your answer.
  4. Recreate this design using only a ruler and compass (the flower pattern made from arcs).
Flower design using only a ruler and compass — arcs all the same radius.

6.3 Construction of a 90° Angle at a Given Point

Can we extend the method of constructing the perpendicular bisector to construct a 90° angle at any point on a line? Draw a line and mark a point O on it. Construct a 90° angle at point O.

Find a segment of this line for which O is the midpoint. Extend the line on either side of O. Using a compass, mark two points X and Y at equal distance from O on it. The perpendicular bisector of XY will pass through O and is perpendicular to the line. In this case, we do not need to draw two pairs of arcs — just one pair is enough (above, say) to draw the perpendicular bisector of XY. We have already drawn the perpendicular bisector.

Fig. 6.2 — 90° at point O X Y O A
Fig. 6.2 — Perpendicular at a given point O on a line.

Extending the line on either side (Fig. 6.3)

Find a segment of this line for which O is the midpoint. Extend the line on either side of O. In this case, we need to draw two pairs of arcs. Again, we already know how to do it — using the perpendicular bisector. Figures 6.2 and 6.3 describe the steps to construct a 90° angle at a given point on a line.

Fig. 6.3 — 90° at interior point O X Y O
Fig. 6.3 — Perpendicular constructed by arcs above and below the line.

6.4 Construction Methods in Sulba-Sutras

History Note
Ancient mathematicians from different civilisations — including India — knew exact procedures to construct perpendiculars and perpendicular bisectors. In India, the earliest known texts containing these methods are the Sulba-Sutras. These are geometric texts of Vedic period dealing with the construction of fire altars for rituals. The Sulba-Sutras are part of one of the six Vedāngas (a term that literally means 'limbs of the Vedas'). The Sulba-Sutras contain the methods that we developed earlier to construct a perpendicular and the perpendicular bisector.

All the construction methods in the Sulba-Sutras make use of a different kind of compass from what you would have used — a rope. A rope can be used to draw circles or arcs. It can also be stretched to form a straight line. In addition, the Sulba-Sutras also contain other methods to construct perpendicular lines. Here is an interesting construction of the perpendicular bisector using a rope (Kātyāyana-Śulbasūtra 1.2).

Let XY be the given line segment, drawn on the ground, for which we need to construct a perpendicular bisector. Fix a small pole or peg vertically into the ground at each point X and Y.

  1. Take a sufficiently long rope. Make two loops at its ends. Without taking into account the parts of the rope that has gone into the loops, fold the rope into half and mark its midpoint.
  2. Fasten the two loops at the ends of the rope to the poles at X and Y.
  3. Pull the midpoint of the rope above XY, as shown in the figure, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as A.
  4. Now similarly pull the midpoint of the rope below XY, as shown in the figure, such that the two parts of the rope on either side are fully stretched. Mark this position of the midpoint as B.
  5. AB is the required perpendicular bisector.
Fig. 6.4 — Rope-and-peg method X Y A B
Fig. 6.4 — A rope fastened between pegs at X and Y is pulled tight above and below to locate A and B; AB is the perpendicular bisector.
🔵 Figure it Out
  1. Justify why AB in Fig. 6.4 is the perpendicular bisector.
  2. Can you think of different ways to construct a 90° angle at a given point on a line using a rope?
Activity — Construct the Perpendicular Bisector Yourself
L3 Apply
Materials: Ruler (unmarked), pair of compasses, sharp pencil, plain paper.
Predict first: If you open your compass to a radius less than half of XY, can the two arcs still meet? Guess before trying.
  1. Draw a line segment XY of length 6 cm.
  2. Open the compass to a radius of 4 cm (more than half of XY).
  3. Place the compass tip at X and draw an arc above and below XY.
  4. Without changing the radius, place the tip at Y and draw arcs again above and below XY.
  5. Label the upper intersection A and the lower one B. Use the ruler to join A to B.
  6. Measure XO and OY where O is on XY — are they equal? Measure ∠AOX with a protractor — is it 90°?
Expected: XO = OY = 3 cm, ∠AOX = 90°. If the compass radius was less than 3 cm (half of XY), the arcs would not meet — because each arc would stop short of the midpoint.

Competency-Based Questions

A craftsperson is laying out a decorative 'eye' motif on a stone slab. The length XY is 12 cm. She needs points A (above) and B (below) to draw the two arcs that complete the eye, and must justify her layout using congruent triangles.
Q1. Which SAS equality in △AOX and △AOY allows us to conclude ∠AOX = ∠AOY = 90°?
L3 Apply
AX = AY (same radius arcs), AO = AO (common side), so by SAS △AOX ≅ △AOY. Thus ∠AOX = ∠AOY. Since they lie on a straight line, ∠AOX + ∠AOY = 180°, so each = 90°.
Q2. If the compass radius chosen is less than half of XY, what happens to the two arcs, and why can the perpendicular bisector not be drawn?
L4 Analyse
Each arc stops short of the midpoint of XY — the two arcs never intersect. Without an intersection point A (or B), we have no second point to define the perpendicular line through the midpoint of XY.
Q3. The craftsperson uses a rope tied to pegs at X and Y (the Sulba-Sutra method). Prove that the point A where the taut midpoint is pulled above XY satisfies AX = AY.
L5 Evaluate
Each half of the rope — from X to the midpoint A, and from Y to the midpoint A — has the same length (the rope was folded in half). Therefore AX = AY. The same argument gives BX = BY for the mirror position. Hence A and B lie on the perpendicular bisector.
Q4. Design an original "double-eye" logo: two eyes side by side of different sizes, each constructed correctly with ruler and compass. Describe your steps.
L6 Create
Example: Draw segment X₁Y₁ = 4 cm; construct perpendicular bisector with compass radius 3 cm → upper point A₁, lower point B₁; draw arcs through X₁ Y₁ using A₁ and B₁ as centres. Repeat with a larger segment X₂Y₂ = 6 cm (radius 4 cm). Two eyes of different sizes lie on a common horizontal line.

Assertion–Reason Questions

Assertion (A): If AX = AY and BX = BY, then A and B both lie on the perpendicular bisector of XY.
Reason (R): Every point equidistant from X and Y lies on the perpendicular bisector of XY.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — R is exactly the defining property of the perpendicular bisector and directly justifies A.
Assertion (A): To construct a perpendicular bisector we must use the same radius at X and at Y.
Reason (R): Equal radii ensure that the intersection point A is equidistant from X and Y.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (a) — Using equal radii from X and Y is exactly what makes AX = AY (and BX = BY), which is the reason the line AB is perpendicular-bisector.
Assertion (A): The Sulba-Sutras used a compass identical to the modern geometry-box compass.
Reason (R): The Sulba-Sutras describe geometric constructions for fire altars.
(a) Both true, R explains A.
(b) Both true, R doesn't explain A.
(c) A true, R false.
(d) A false, R true.
Answer: (d) — A is false: Sulba-Sutra constructions used a rope, not a steel compass. R is true: yes, they describe altar geometry.

Frequently Asked Questions — Chapter 6

What is Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool in NCERT Class 7 Mathematics?

Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool is a key concept covered in NCERT Class 7 Mathematics, Chapter 6: Chapter 6. This lesson builds the student's foundation in the chapter by explaining the core ideas with worked examples, definitions, and step-by-step methods aligned to the CBSE curriculum.

How do I solve problems on Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool step by step?

To solve problems on Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool, follow the NCERT method: identify the given quantities, choose the relevant formula or theorem, substitute values carefully, and simplify. Class 7 exercises gradually increase in difficulty — start with solved NCERT examples before attempting exercise questions, and always verify your answer by substitution or diagram.

What are the most important formulas for Chapter 6: Chapter 6?

The essential formulas of Chapter 6 (Chapter 6) are listed in the chapter summary and highlighted throughout the lesson in formula boxes. Memorise them and practise at least 2–3 problems per formula. CBSE board exams frequently test direct application as well as combined use of multiple formulas from this chapter.

Is Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool important for the Class 7 board exam?

Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool is part of the NCERT Class 7 Mathematics syllabus and appears in CBSE board exams. Questions typically include short-answer, long-answer, and competency-based items. Review the NCERT examples, exercise questions, and previous-year board problems on this topic to prepare confidently.

What mistakes should students avoid in Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool?

Common mistakes in Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool include skipping steps, misapplying formulas, sign errors, and losing track of units. Write each step clearly, double-check algebraic manipulations, and re-read the question after solving to verify that your answer matches what was asked.

Where can I find more NCERT practice questions on Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool?

End-of-chapter NCERT exercises for Part 1 — Geometric Constructions: Eye, Perpendicular Bisector & 90° | Class 7 Maths Ch 6 | MyAiSchool cover all difficulty levels tested in CBSE exams. After completing them, try the examples again without looking at the solutions, attempt the NCERT Exemplar questions for Chapter 6, and solve at least one previous-year board paper to consolidate your understanding.

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