Nixon as a young boy, long before he became President, was a Quaker and evidently his parents told him to never take the easy way of doing any task. How well he applied that advice is best left for historians. This is a maths investigation.
The easy way to find the area of a parallelogram is simply multiply the base times the height, just as you would do for a rectangle. A justification - though perhaps not a 'proof'- is offered pictorially below. Please note that by height, I most definitely do not mean the slant height, but the height that is perpendicular to the base.
In a slightly more difficult method we make use of the substitution:
height = slant height × sin Θ.
The overall formula for the area of the partallelogram is now the still very reasonable:
Area = base × (slant height × sin theta). Again the justification – for those who know a little trigonometry- is a picture.
And then, there is the following complication!
The parallelogeram is formed by the vector addition u + v. The coordinates of u are (a, c) and the coordinates of v are (b, d). The entire shape is a rectangle. To find the area of the parallelogram we will cut away two triangles and two trapeziums. Note that the diagonal of the parallelogram separates the diagram into two congruent halves.
parallelogram = rectangle – 2 congruent triangles – 2 congruent trapeziums
= (a + b) × (c + d) – 2 × ½ (bd) – 2 × ½ [(a + b) + b ] c
(ac + ad + bc + bd) – bd – (a + 2b)c
= ac + ad + bc + bd – bd – ac – 2bc
= ad – bc (Q.E.D. Is sometimes put here. Originally Q.E.D. was short for quod erat demonstrandum, which means "That which was to be demonstrated," although I prefer the somewhat cheeky, “Quite Easily Done.”)
The matrix M is made from u and v written as column vectors
The determinant of the matrix is, det M = product of main diagonal – product of off main
det M = ad – bc, which is exactly the area of the parallelogram, thus justifying one description of the determinant of a matrix as being or representing an area scaling factor.
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