I found that you could represent square roots of whole numbers as line lengths, given the basic unit length.
The first is done by constructing a 45 45 90 triangle of 1 by 1 by sqrt 2.
Then, you construct a rt triangle of sqrt 2 by 1 by sqrt 3, and so on. If these are built, one upon the other, using each new result as the hypotenuse of the next, an interesting spiral pattern arises.
What about the fourth root of a number?
If one constructs a right triangle of sqrt a by sqrt b, the hypotenuse is sqrt (a + b) . So, one can make sums of square roots.
If one could construct the square root of a sum of 2 numbers, as a hypotenuse, then one could get sides equal to one of the two other numbers, if one is known.
Take, for example, finding the fourth root of 5:
First, construct sqrt 5 as the hypotenuse of rt triangle 2 by 1. (You get the sqrt(sqrt4 ^2 plus 1^2) but you could alternatively construct sqrt2 by sqrt3.)
Then, add 1 to the sqrt 5 length. the square of that is 5 + 2sqrt5 + 1, or 6 + 2sqrt5.
Construct a rt triangle with that as the hypotenuse, and one side equal to sqrt6.
The other side is thus sqrt(2sqrt5), or sqrt2 times the fourth root of 5..
To divide by sqrt 2, simply construct the 45 45 90 triangle with your found length as the hypotenuse.
The short side will be your fourth root of 5.
You can divide a line A by any square root x by constructing a rt triangle of the form A, sqrt x, 1, using A as the hypotenuse.
First, construct a line perpendicular to your original, meeting it at one end. Then, from the other end, mark off an arc of the length of your divisor plus one, crossing your perpendicular. The perp is the result.
Or, one could make the original line A, into the diameter of a circle. Mark off the length sqrt(x - 1) on the circle.
The result is a rt triangle of 1,sqrt( x -1), A, where the side A is sqrtx times as long as the side 1.
The first is done by constructing a 45 45 90 triangle of 1 by 1 by sqrt 2.
Then, you construct a rt triangle of sqrt 2 by 1 by sqrt 3, and so on. If these are built, one upon the other, using each new result as the hypotenuse of the next, an interesting spiral pattern arises.
What about the fourth root of a number?
If one constructs a right triangle of sqrt a by sqrt b, the hypotenuse is sqrt (a + b) . So, one can make sums of square roots.
If one could construct the square root of a sum of 2 numbers, as a hypotenuse, then one could get sides equal to one of the two other numbers, if one is known.
Take, for example, finding the fourth root of 5:
First, construct sqrt 5 as the hypotenuse of rt triangle 2 by 1. (You get the sqrt(sqrt4 ^2 plus 1^2) but you could alternatively construct sqrt2 by sqrt3.)
Then, add 1 to the sqrt 5 length. the square of that is 5 + 2sqrt5 + 1, or 6 + 2sqrt5.
Construct a rt triangle with that as the hypotenuse, and one side equal to sqrt6.
The other side is thus sqrt(2sqrt5), or sqrt2 times the fourth root of 5..
To divide by sqrt 2, simply construct the 45 45 90 triangle with your found length as the hypotenuse.
The short side will be your fourth root of 5.
You can divide a line A by any square root x by constructing a rt triangle of the form A, sqrt x, 1, using A as the hypotenuse.
First, construct a line perpendicular to your original, meeting it at one end. Then, from the other end, mark off an arc of the length of your divisor plus one, crossing your perpendicular. The perp is the result.
Or, one could make the original line A, into the diameter of a circle. Mark off the length sqrt(x - 1) on the circle.
The result is a rt triangle of 1,sqrt( x -1), A, where the side A is sqrtx times as long as the side 1.
