Construct the square root spiral of …

OBJECTIVE

To construct a square root spiral.

Materials Required

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

Prerequisite Knowledge

1. Concept of number line.
2. Concept of irrational numbers.
3. Pythagoras theorem.

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)
   
   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)
   
3. Construct a line QX perpendicular to the line segment PQ, by using compasses or a set square, (see Fig. 1.3)
   
4. From Q, draw an arc of 1 unit, which cut QX at C(say). (see Fig. 1.4)
   
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)
   
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
A square root spiral has been constructed.