NCERT Class 7 Ganita Prakash 2 Chapter 06 Constructions and Tilings PDF Download

Chapter 6: Constructions and Tilings

6.1 Geometric Constructions

Eyes

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 spatial 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 systematically 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 must have AX = BX. Since AX = AY and BX = BY, this means

AX = AY = BX = BY

[Figure: Eye construction with supporting line XY and centres A and B, See in your textbook]

How do we find such A and B?

From X and Y, draw arcs above and below XY, with the same radii. The two points at which the 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, and what is the angle formed between them?

We observe that AB passes through the midpoint of XY, and is also perpendicular to it. A division of a line, or any geometrical object, into two identical parts is called bisection. A line that bisects a given line and is perpendicular to it, is called the perpendicular bisector.

[Figure 6.1: Line segment XY with perpendicular bisector AB intersecting at point O, See in your textbook]

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?

This can be answered through congruence. Let us consider a line segment XY. Find points A and B such that AX = AY = BX = BY. Draw the lines AB, AX, AY, BX and 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 (that is, O is the midpoint of XY and AB is perpendicular to XY)?

If we show that \( \triangle AOX \cong \triangle AOY \), then OX = OY, and \( \angle AOX = \angle AOY \) because they are corresponding parts of congruent triangles. Since \( \angle AOX \) and \( \angle AOY \) together form a straight angle, we have \( \angle AOX + \angle AOY = 180° \). Thus, \( \angle AOX = \angle AOY = 90° \). This establishes that O is the midpoint of XY and AB is perpendicular to XY.

Teacher's Note

When you see the notation \( \triangle AOX \cong \triangle AOY \), the order of the letters matters. The first letter A corresponds to A, the second O to O, and the third X to Y. This means the sides and angles match up in that specific order: AO = AO, OX = OY, and AX = AY.

In \( \triangle AOX \) and \( \triangle AOY \), we already know that AX = AY, and AO is common to both triangles.

If we can show that \( \angle XAO = \angle YAO \) then, by the SAS congruence condition, we can conclude that the triangles are congruent.

To show this, we observe that \( \triangle ABX \cong \triangle ABY \). This is so because AX = AY, BX = BY, and AB is common to both the triangles.

Thus, we have \( \angle XAB = \angle YAB \), or \( \angle XAO = \angle YAO \) because they are corresponding parts of congruent triangles.

Hence, AB is the perpendicular bisector of XY.

Teacher's Note

The key insight is that if a point is equidistant from the endpoints of a line segment, then that point lies on the perpendicular bisector. In your constructions, always remember: to find the perpendicular bisector of any line segment, draw two arcs of equal radius from each endpoint, and the line through their intersection points will always be perpendicular and will bisect the segment.

We can have eyes of different shapes.

[Figure: Various eye shapes constructed using perpendicular bisectors, See in your textbook]

Key Points

• A bisector divides a line or geometric object into two identical parts.
• The perpendicular bisector of a line segment both bisects it and is perpendicular to it; it can be found by drawing equal arcs from both endpoints.
• If points A and B satisfy AX = AY = BX = BY for a line segment XY, then line AB is the perpendicular bisector of XY, which can be proven using triangle congruence (SAS condition).
• Congruent triangles have corresponding sides and angles equal, which allows us to deduce properties like the 90° angle in the construction.

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