Activity & Practical on Construct a Square Root Spiral | Class 9th Mathematics-EduGrown

With the help of explained activity, existence of irrational numbers can be illustrated.

OBJECTIVE

To construct a square root spiral.

MATERIALS REQUIRED

  1. Adhesive
  2. Geometry box
  3. Marker
  4. A piece of plywood

THEORY

  1. A number line is a imaginary line whose each point represents a real number.
  2. The numbers which cannot be expressed in the form p/q where q ≠ 0 and both p and q are integers, are called irrational numbers, e.g. √3, π, etc.
  3. According to Pythagoras theorem, in a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of other two sides containing right angle. ΔABC is a right angled triangle having right angle at B. (see Fig. 1.1)NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 1
  4. Therefore, AC² = AB² +BC²
    where, AC = hypotenuse, AB = perpendicular and BC = base

PROCEDURE

    1. Take a piece of plywood having the dimensions 30 cm x 30 cm.
    2. Draw a line segment PQ of length 1 unit by taking 2 cm as 1 unit, (see Fig. 1.2)
      NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 2
    3. Construct a line QX perpendicular to the line segment PQ, by using compasses or a set square, (see Fig. 1.3)NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 3
    4. From Q, draw an arc of 1 unit, which cut QX at C(say). (see Fig. 1.4)
      NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 4
    5. Join PC.
    6. Taking PC as base, draw a perpendicular CY to PC, by using compasses or a set square.
    7. From C, draw an arc of 1 unit, which cut CY at D (say).
    8. Join PD. (see Fig. 1.5)
      NCERT Class 9 Maths Lab Manual - Construct a Square Root Spiral 5
    9. Taking PD as base, draw a perpendicular DZ to PD, by using compasses or a set square.
    10. From D, draw an arc of 1 unit, which cut DZ at E (say).
    11. Join PE. (see Fig. 1.5)

    Keep repeating the above process for sufficient number of times. Then, the figure so obtained is called a ‘square root spiral’.

Demonstration

  1. In the Fig. 1.5, ΔPQC is a right angled triangle.
    So, from Pythagoras theorem,
    we have PC² = PQ² + QC²
    [∴ (Hypotenuse)² = (Perpendicular)² + (Base)²]
    = 1² +1² =2
    => PC = √2
    Again, ΔPCD is also a right angled triangle.
    So, from Pythagoras theorem,
    PD² =PC² +CD²
    = (√2)² +(1)² =2+1 = 3
    => PD = √3
  2. Similarly, we will have
    PE= √4
    => PF=√5
    => PG = √6 and so on.

Observations

On actual measurement, we get
PC = …….. ,
PD = …….. ,
PE = …….. ,
PF = …….. ,
PG = …….. ,
√2 = PC = …. (approx.)
√3 = PD = …. (approx.)
√4 = PE = …. (approx.)
√5 = PF = …. (approx.)

Result

  1. A square root spiral has been constructed.

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