Question 641702
<pre>
It has four names: Root Spiral, Root Snail, Spiral of Theodorus, and Wheel of Theodorus.

{{{drawing(650,650,-4.5,4.5,-4.5,4.5,
triangle(0,0,0,1,1,1),
triangle(0,0,1.0,1.0,1.70710678,0.29289322),
triangle(0,0,1.70710678,0.29289322,1.87620876,-0.69270534),
triangle(0,0,1.87620876,-0.69270534,1.52985609,-1.63080972),
triangle(0,0,1.52985609,-1.63080972,0.80053581,-2.31498216),
triangle(0,0,0.80053581,-2.31498216,-0.1445517,-2.64179954),
triangle(0,0,-0.1445517,-2.64179954,-1.14305807,-2.58716413),
triangle(0,0,-1.14305807,-2.58716413,-2.05775872,-2.18303208),
triangle(0,0,-2.05775872,-2.18303208,-2.78543608,-1.4971125),
triangle(0,0,-2.78543608,-1.4971125,-3.25886462,-0.61628027),
triangle(0,0,-3.25886462,-0.61628027,-3.44468012,0.36630438),
triangle(0,0,-3.44468012,0.36630438,-3.33893715,1.36069788),
triangle(0,0,-3.33893715,1.36069788,-2.96154746,2.28675242),
triangle(0,0,-2.96154746,2.28675242,-2.35038717,3.07825928),
triangle(0,0,-2.35038717,3.07825928,-1.55558404,3.68512663),
triangle(0,0,-1.55558404,3.68512663,-0.63430238,4.07402264)

  )}}}

There are 16 triangles poosible before any overlapping occurs.

By using the Pythagorean theorem over and over, the first 
(shortest spoke, the vertical one) has length 2 cm and 
the last spoke (longest spoke) has length {{{2sqrt(17)}}}

The sequence of lengths of spokes radiating from center C going
clockwise is:
{{{2sqrt(1)}}}cm,{{{2sqrt(2)}}}cm, {{{2sqrt(3)}}}cm, {{{2sqrt(4)}}}cm, {{{2sqrt(5)}}}cm, {{{2sqrt(6)}}}cm, {{{2sqrt(7)}}}cm, {{{2sqrt(8)}}}cm, {{{2sqrt(9)}}}cm, {{{2sqrt(10)}}}cm, {{{2sqrt(11)}}}cm, {{{2sqrt(12)}}}cm, 
{{{2sqrt(13)}}}cm, {{{2sqrt(14)}}}cm, {{{2sqrt(15)}}}cm, {{{2sqrt(16)}}}cm,{{{2sqrt(17)}}}cm.

[This spiral is usually drawn with the first (vertical spoke being 1 unit
instead of 2 cm. In other words, we could let 1 unit be 2 cm. and then those
spoke lengths would be the square roots of the integers in units, not
centimeters. They would be {{{sqrt(1)}}}units,{{{sqrt(2)}}}units, etc., 
{{{sqrt(17)}}}units.
I'm sure the reason you were instructed to use 2 cm instead of 1 unit, was 
so that the area would be easier to calculate since there would be no
fractions].

The area of each triangle is found by {{{1/2}}}×(first leg)×(second leg)

The length of the first leg of the 1st triangle is 2
The length of the second leg of the 1st triangle is 2
The area of the 1st triangle is {{{1/2}}}×2×2 = 2 cm²

Area of first triangle = 2cm²

The length of the first leg of the 2nd triangle is {{{2sqrt(2)}}}
The length of the second leg of the 2nd triangle is 2
The area of the 2nd triangle is {{{1/2}}}×2{{{sqrt(2)}}}×2 = 2{{{sqrt(2)}}} cm²

Area of first two triangles = [2 + 2{{{sqrt(2)}}}] cm²

The length of the first leg of the 3rd triangle is {{{2sqrt(3)}}}
The length of the second leg of the 3rd triangle is 2
The area of the 3rd triangle is {{{1/2}}}×2{{{sqrt(3)}}}×2 = 2{{{sqrt(3)}}} cm²

Area of first three triangles = [2 + 2{{{sqrt(2)}}} + 2{{{sqrt(3)}}} ] cm²

So you can see that the area of the entire spiral is, after 
factoring out a 2:

A = {{{2(sqrt(1)+sqrt(2)+sqrt(3)+sqrt(4)+sqrt(5)+sqrt(6)+sqrt(7)+sqrt(8)+sqrt(9)+sqrt(10)+sqrt(11)+sqrt(12)+sqrt(13)+sqrt(14)+sqrt(15)+sqrt(16))}}}

[Notice that the last spoke, {{{2sqrt(17)}}} is not needed since it is
not used to find the area of the 16th triangle]

I get that total area to be 88.9383932 cm². Better check me on that. 

Edwin</pre>