Construct the square root spiral up to $\sqrt{6}$.

(N/A) To construct a square root spiral up to $\sqrt{6}$, follow these steps:
$1$. Draw a line segment $OA$ of length $1 \text{ unit}$.
$2$. At point $A$, draw a perpendicular line $AB$ of length $1 \text{ unit}$. Join $OB$. By the Pythagorean theorem, $OB = \sqrt{1^2 + 1^2} = \sqrt{2}$.
$3$. At point $B$, draw a perpendicular line $BC$ of length $1 \text{ unit}$ to $OB$. Join $OC$. Then $OC = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}$.
$4$. At point $C$, draw a perpendicular line $CD$ of length $1 \text{ unit}$ to $OC$. Join $OD$. Then $OD = \sqrt{(\sqrt{3})^2 + 1^2} = \sqrt{4} = 2$.
$5$. At point $D$, draw a perpendicular line $DE$ of length $1 \text{ unit}$ to $OD$. Join $OE$. Then $OE = \sqrt{(\sqrt{4})^2 + 1^2} = \sqrt{5}$.
$6$. At point $E$, draw a perpendicular line $EF$ of length $1 \text{ unit}$ to $OE$. Join $OF$. Then $OF = \sqrt{(\sqrt{5})^2 + 1^2} = \sqrt{6}$.