trapezium divided to 4 triangle what is the area of trapezium
Answers
Step-by-step explanation:
How can I find the area of a trapezium when the areas of opposite triangles which are made by the diagonals are known?
Since Doug Dillon is graciously letting me use his diagram I might as well post my solution (w/ less geometry but possibly more algebra than his).
Suppose a trapezium MNPQ with bases a & b, of height h, has 2 known areas A and B. Triangles A and B are similar (parallel lines ⇒ pairs of equal angles). Their sides are therefore proportional, in the ratio a/b, so the areas A and B are in a ratio of proportionality a2/b2 (if you’re wondering why: keep in mind that the heights of similar triangles have the same ratio of proportionality as the sides, and to get areas you multiply bases x heights.)
Triangles MNP and QPN have the same area (same base a, same height h) so there are 2 equal areas in these, squeezed between the diagonals, at left & right which we will call C. We then have C+A= ah/2. From triangle NMQ we also have C + B = bh/2. Subtracting these equations yields:
B−A=bh−ah2⟹h=2(B−A)b−a
The total area of a trapezium is T=(a+b)h/2, but:
AB=a2b2⟹ba=B−−√A−−√
Putting it all together:
T=(a+b)(B−A)b−a=(1+ba)(B−A)ba−1=
=(1+B√A√)(B−A)B−−√A−−√−1
Multiplying throughout by A−−√ :
T=(B−−√+A−−√)(B−A)B−−√−A−−√=(B−−√+A−−√)2