Question

Draw square root spiral , also draw any spiral found in nature (Compare them and write few lines about beauty of Mathematics in nature). This needs to be added in the portfolio.​

Answer 1

Materials Required

Adhesive

Geometry box

Marker

A piece of plywood

Prerequisite Knowledge

Concept of number line.

Concept of irrational numbers.

Pythagoras theorem.

Theory

A number line is a imaginary line whose each point represents a real number.

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.

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.

Therefore, AC² = AB² +BC²

where, AC = hypotenuse, AB = perpendicular and BC = base

Procedure

Take a piece of plywood having the dimensions 30 cm x 30 cm.

Draw a line segment PQ of length 1 unit by taking 2 cm as 1 unit, (see Fig. 1.2)



Construct a line QX perpendicular to the line segment PQ, by using compasses or a set square, (see Fig. 1.3)

From Q, draw an arc of 1 unit, which cut QX at C(say). (see Fig. 1.4)



Join PC.

Taking PC as base, draw a perpendicular CY to PC, by using compasses or a set square.

From C, draw an arc of 1 unit, which cut CY at D (say).

Join PD. (see Fig. 1.5)



Taking PD as base, draw a perpendicular DZ to PD, by using compasses or a set square.

From D, draw an arc of 1 unit, which cut DZ at E (say).

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

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

Similarly, we will have

PE= √4

=> PF=√5

=> PG = √6 and so on.