Question

To show the detail construction of a square root spiral representing √8 by making use of Pythagoras’ theorem.

Answer 1

Answer:

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.

Observations

On actual measurement, we get

PC = …….. ,

PD = …….. ,

PE = …….. ,

PF = …….. ,

PG = …….. ,

√2 = PC = …. (approx.)

√3 = PD = …. (approx.)

√4 = PE = …. (approx.)

√5 = PF = …. (approx.)