The ancient Greeks, being masters of geometric manipulation, often tried their hand at "squaring" various shapes. This involved using only the most fundamental rules of geometry to construct a square whose area equals the area of the original shape.
Can you follow in their footsteps and square a simple triangle?
The solution must hold for all types of triangles.
Call the triangle ABC.
Construct a perpendicular bisector to BC, marking the midpoint of BC as M. Then erect a perpendicular from A to the perpendicular bisector, calling the point of intersection D.
Place the point of the compass on BC at its midpoint and measure off distance equal to MD on line BC (extended if necessary), in the same direction from M as C, marking the new point E. Now segment AE is equal to the height of the triangle plus half the base. Bisect this segment, and construct a semicircle with center at this midpoint and going through A and E. Where it intersects MD (extended if necessary) mark point N. Segment MN has length that is the geometric mean between the height of the original triangle and half the base; its square will have the same area as the original triangle, so just construct perpendiculars at the endpoints of MN and mark off lengths on these equal to that of MN, and all the corners of the square are now marked.
Posted by Charlie
on 2004-09-22 15:01:07