Question

Classroom Activity (Constructing the 'square root spiral'): Take a large sheet of paper and construct the 'square root spiral' in the following fashion. Start with a point $\mathrm{O}$ and draw a line segment ${\mathrm{OP}}_1$ of unit length. Draw a line segment ${\mathrm{P}}_1{\mathrm{P}}_2$ perpendicular to ${\mathrm{OP}}_1$ of unit length. Now draw a line segment ${\mathrm{P}}_2{\mathrm{P}}_3$ perpendicular to ${\mathrm{OP}}_2$. Then draw a line segment ${\mathrm{P}}_3{\mathrm{P}}_4$ perpendicular to ${\mathrm{OP}}_3$. Continuing in this manner, you can get the line segment ${\mathrm{P}}_{n-1}{\mathrm{P}}_n$ by square root spiral drawing a line segment of unit length perpendicular to ${\mathrm{OP}}_{n-1}$. In this manner, you will have created the points ${\mathrm{P}}_2,{\mathrm{P}}_3,.....,{\mathrm{P}}_{\mathrm{n}},.....$, and joined them to create a beautiful spiral depicting $\sqrt{2},\sqrt{3},\sqrt{4},.....$

Answer 1

Construct the “square root spiral”.

Step 1. On a large sheet of paper, mark a point $\mathrm{O}$ and draw a line segment ${\mathrm{OP}}_1$ of unit length.

Step 2. Draw a line segment ${\mathrm{P}}_1{\mathrm{P}}_2$ perpendicular to ${\mathrm{OP}}_1$ of unit length and join ${\mathrm{OP}}_2$.

Here, ${\mathrm{OP}}_2$ is of length $\sqrt{2}$ units.

As from Pythagoras theorem:${\left(O{P}_2\right)}^2={\left(P_1{P}_2\right)}^2+{\left(O{P}_1\right)}^2$

Step 3. Draw a line segment ${\mathrm{P}}_2{\mathrm{P}}_3$ perpendicular to ${\mathrm{OP}}_2$ of unit length and join ${\mathrm{OP}}_3$.

Here, ${\mathrm{OP}}_3$ is of length $\sqrt{3}$ units.

As from Pythagoras theorem: ${\left(O{P}_3\right)}^2={\left(P_2{P}_3\right)}^2+{\left(O{P}_2\right)}^2$

Step 4. Repeat the steps to get the points ${\mathrm{P}}_4,{\mathrm{P}}_5,{\mathrm{P}}_6,.....$ and join them with $\mathrm{O}$ to obtain $\sqrt{4},\sqrt{5},\sqrt{6},.....$

Hence, the 'square root spiral' is constructed as shown in the above figure.