Question (2)

ASAP Please help.

Construct 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 [tex]\rm{P_{1} P_{2}}[/tex] perpendicular to [tex]\rm{OP_{1}}[/tex] of unit length. Now draw a line segment [tex]\rm{P_{2} P_{3}}[/tex] perpendicular to [tex]\rm{OP_{2}}[/tex] . Then draw a line segment [tex]\rm{P_{3} P_{4}}[/tex] perpendicular to [tex]\rm{OP_{3}}[/tex]. Continuing in this matter, you get line segment of unit length perpendicular to [tex]\rm{OP_{n-1}}[/tex]. In this manner, you will have created the points [tex]\rm{P_{2}, P_{3},...... P_{n}....}[/tex], and joined them to create a beautiful spiral depicting [tex]\rm{\sqrt{2}, \sqrt{3}, \sqrt{4},...}[/tex]

The "square root spiral," donned the "Spiral of Theodorus" was created by Theodorus to visualize a set of 17 isosceles triangles where [tex]n[/tex] is equal to a value between one and seventeen.

The central angle is attached to a central point, and the side opposite of the central angle is always equal to 1.
The hypotenuse of the triangle is equivalent to [tex]\sqrt{n+1}[/tex]. The hypotenuse becomes a leg for the next triangle.

I have attached an image of the Square Root spiral below.