If a square is 1/q units on a side, where q is an integer, then q2 such squares combine to build the unit square, hence the area is 1/q2.
Three postulates define area in the plane.
1. The area of a square, one unit on a side, is one.
2. The area of a shape is unchanged as the shape is moved about, rotated, or reflected.
3. The area of a shape is the sum of the areas of its components. Thus the area of a 2×1 rectangle is 2, because it is two unit squares pasted together.
These "axioms" can actually be proved if area is defined as an integral. But we must remember that the concept of integration was developed in a manner that would satisfy the above criteria. So it is no surprise that integral = area.
|
The area of an m by n rectangle is m×n,
because the rectangle can be broken up into an m by n grid of unit squares.
If a square is 1/q units on a side, where q is an integer, then q2 such squares combine to build the unit square, hence the area is 1/q2. |
|
If the height and width of a rectangle are rational numbers, let q be the least common denominator. Separate the rectangle into a grid of tiny squares, 1/q units on a side, and the area of the rectangle is the product of the two rational numbers. This extends to real numbers by continuity. Therefore the area of a rectangle is base times height.
| The area of a right triangle with base b and height h is bh/2. This is clear because two copies of this triangle form a rectangle with base b and height h. |
|
|
Given any triangle,
spin it about so that the apex is the largest of the three angles.
By convention, the apex is placed at the top, like the point of a roof.
Drop a perpendicular from the apex to the base,
cutting the triangle into two right triangles.
This descending perpendicular is called the altitude.
The area of each right triangle is given by the formula above.
Add them together and the area of this triangle
is bh/2, where b is the base and h is the height from base to apex.
|
| |
| If the apex of the triangle is not directly over the base, slanted left for instance, extend the base out to the left, and drop a perpendicular from the apex to this line. This altitude creates two right triangles, using the two lines of the original triangle. Subtract the smaller right triangle from the larger to get the desired area. This leads to the same formula, bh/2. |
|
| If a parallelogram slants to the right, cut off the right triangle and paste it onto the left side, giving a rectangle with the same base and height. Thus the area of a parallelogram is bh, where b is the base and h is the height from floor to ceiling. |
|
| Two copies of a trapezoid combine to form a parallelogram of the same height. If the base of the trapezoid is b, and the top is t, and the height from floor to ceiling is h, the area of the composite parallelogram is (b+t)h. Hence the area of the trapezoid is (b+t)h/2. |
|
| The area of a kite is half the product of the diagonals. I'll leave this as an exercise for you. This generalizes to any quadrilateral whose diagonals are perpendicular. |
|