STRAIGHT-LINE AND COMPASS CONSTRUCTIONS.

Constructing arbitrary square roots

Suppose you're shown a line segment and told that it is (exactly) one inch long.

Bonus: Show how you can Construct a line segment sqrt(x) inches long, for any line of length x (not necessarily an integer).

Solution

1. Let the line of length x have end-points a and b.
2. Extend ab into the line segment abc such that bc has length 1.
3. Find the mid-point of ac (call it d).
4. Draw a circle, centre d, radius ad.
5. Drop a perpendicular of ab from b until it intersects the circle. Call the point of intersection e.
6. Note that angle cea is a right-angle. Therefore, triangles abe and aec are similar. Also, triangles cbe and cea are similar. Therefore, Triangles abe and ebc are similar.
7. The consequence of this is that the ratios of corresponding sides of abe and ebc are the same. In particular, ab / be = eb / bc. That is, x / be = be / 1. In other words, the distance be squared = x.
8. The length be is therefore the required value, Sqrt(x).