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 $O$ and draw a line segment $OP_1$ of unit length. Draw a line segment $P_1P_2$ perpendicular to $OP_1$ of unit length (see Fig.). Now draw a line segment $P_2P_3$ perpendicular to $OP_2$. Then draw a line segment $P_3P_4$ perpendicular to $OP_3$. Continuing in this manner,you can get the line segment $P_{n-1}P_n$ by drawing a line segment of unit length perpendicular to $OP_{n-1}$. In this manner,you will have created the points $P_2, P_3, ..., P_n, ...$,and joined them to create a beautiful spiral depicting $\sqrt{2}, \sqrt{3}, \sqrt{4}, ...$.

(N/A) The construction of the square root spiral is based on the Pythagorean theorem,which states that in a right-angled triangle,the square of the hypotenuse is equal to the sum of the squares of the other two sides $(h^2 = a^2 + b^2)$.
$1$. Start with point $O$ and draw $OP_1 = 1$ unit.
$2$. Draw $P_1P_2$ perpendicular to $OP_1$ such that $P_1P_2 = 1$ unit. In $\triangle OP_1P_2$,by Pythagoras theorem,$OP_2 = \sqrt{OP_1^2 + P_1P_2^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$.
$3$. Draw $P_2P_3$ perpendicular to $OP_2$ such that $P_2P_3 = 1$ unit. In $\triangle OP_2P_3$,$OP_3 = \sqrt{OP_2^2 + P_2P_3^2} = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{2 + 1} = \sqrt{3}$.
$4$. Similarly,draw $P_3P_4$ perpendicular to $OP_3$ such that $P_3P_4 = 1$ unit. In $\triangle OP_3P_4$,$OP_4 = \sqrt{OP_3^2 + P_3P_4^2} = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{3 + 1} = \sqrt{4} = 2$.
$5$. By continuing this process,the length of the hypotenuse $OP_n$ will be $\sqrt{n}$.