The diagonals of the trapezoid ABCD intersect at P.
The area of the triangle ABP is 216 and the area of CDP is 150.
What is the area of the trapezoid ABCD?
This is nowhere near as neat as Bractals solution, but does establish the principle assumed in his first paragraph, that two of the triangles (i.e. those having a side in common with the lateral sides of the trapezoid) are equal:
1. Definition: A trapezoid is a quadrilateral with two unequal parallel sides.
2. Construction:
a) Connect the midpoints of the parallel sides of the trapezoid. Note that this line passes through P. This is a bisection of the trapezoid (essentially averaging the slopes of the nonparallel sides) so the two smaller trapezoids formed by the bisection are of equal area.
b) Construct 2 parallels to this line from the vertices of the shorter side passing through the longer side. Now we have a regular parallelogram and 2 triangular pieces, the 'end pieces'. The end pieces are of equal area, because they represent the difference between the areas of the parallelograms bisected by the midpoint bisector of the trapezoid, which areas are plainly congruent, and the two smaller trapezoids, which are of equal area.
(I) The sides of the end pieces along the long parallel side are plainly equal, because they represent two equal lengths (the bisection of the long side) from which two other equal lengths (the bisection of the short side, translated onto the long side) have been deducted.
c) The diagonals of the trapezoid intersect the lines we have just drawn. Construct the line through the points of intersection, noting that this is parallel to the long parallel side. Two small triangles lie between the long side, the points of intersection, and the diagonals of the trapezoid; the 'tiny pieces'.
(I) The sides of the tiny pieces between the line through the points of intersection and the long parallel side are equal, as those two lines are on parallels which are crossed by those two other parallel lines we have just drawn.
(II) The sides of the tiny pieces between the vertices of the trapezoid and the two parallels are the same, equal, lengths mentioned in 2(b)(I).
3. The equal triangles.
a) Consider the small parallelogram whose centre is at P. Bisecting the side of a triangle through the opposite angle cuts it into equal halves. Hence the four small triangles formed by the diagonals of the trapezoid and its midpoint bisector are of equal area. Clearly by construction these triangles are respectively similar to the tiny pieces formed by the parallels. Since the tiny pieces also have 2 pairs of sides of equal length (2(c)); so they are also of equal area to each other.
b) Now consider the areas of the two triangles formed by the diagonals of the trapezoid which do not have as a side one of the parallel sides of the trapezoid. They comprise:
(I) The end pieces; which are equal, see 2(b); plus
(II) The two larger triangles, not so far considered, of the small parallelogram whose centre is at P, which are obviously congruent; less
(III) The tiny pieces, which are of equal area, see 3(a).
(IV) So the areas of those two triangles are equal.
Result A: The areas of the triangles, formed by the diagonals of a trapezoid and its nonparallel sides, are equal.
4. The unequal triangles.
The two remaining triangles, formed by the diagonals of the trapezoid and its parallel sides, are clearly similar. Since the one is larger both in length and in height by the proportion of the lengths of the parallel sides, so their areas differ by the square of that proportion.
5. The area of the trapezoid.
a) The small parallelogram is divided, by the diagonals it shares with the trapezoid, into 4 equal triangular parts. (The bisector of the trapezoid divides two of those triangles into two further parts, also equal, as seen in 3(a).)
b) It is convenient to consider the triangle whose base is the short parallel side, since that triangle is also the smaller of the unequal triangles. Call its area 2a. Call the shorter parallel side of the trapezoid, b, for base. It follows that the height is (4a/b).
c) Call the longer parallel side of the trapezoid, c. Then the area of the larger unequal triangle is (c/b)^2*2a, and its height is (4ac)/b^2.
d) The area of a trapezoid is its height multiplied by the average of its two parallel sides. We have just established the height of the two unequal triangles, the sum of which is the same as the height of the trapezoid, to be (4a/b)+(4ac)/b^2. The area of the trapezoid is (2ac^2)/b^2+(4ac)/b+2a.
e) We can deduct the known areas 2a and (c/b)^2*2a from this: (2ac^2)/b^2+(4ac)/b+2a2a(c/b)^2*2a =(4ac)/b, which is the combined area of the two equal triangles, each of which is therefore (2ac)/b.
f) The square of this last is (4a^2c^2)/b^2. The product of the two unequal areas (c/b)^2*2a*2a is also (4a^2c^2)/b^2.
Result B: The square of the area of a triangle, formed by the diagonals of a trapezoid, and its nonparallel sides, is equal to the product of the areas of the two unequal triangles, formed by those diagonals and the parallel sides.
It follows immediately that the area of the trapezoid ABCD is 216+150+2*(216*150)^(1/2) =726.
Edited on February 10, 2011, 1:54 am

Posted by broll
on 20110210 01:43:03 