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ABCD is a trapezoid in which AD: BC=3:5. If the area of
triangle AMD is 315 cm2, find the area of the trapezoid, in cm2.
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I will give another solution, to get you totally different view on this problem.
Triangles AMD and BMC are similar (since their three angles are congruent, in pairs).
The similarity coefficient is k = 5/3, from the greater triangle to the smaller.
Lat a = AM, b = BM, c = CM and D = DM.
Then b = ; c = , due to similarity.
For any triangle with the side p and q and the concluded angle ,
the area of this triangle is .
So, for the area of triangle AMD we have = = 315 cm^2.
Here M is the angle between sides a = AM and d = DM .
Next, = = ,
where N is the supplementary angle to M.
Since sin(N) = sin(M), we can re-group the formula above and to get
= . = *area(AMD) = = = 25*35 = 875.
This calculation re-tells us a well known fact that the areas
of similar triangles do relate as the square of the similarity coefficient.
Now let's consider the most interesting things - the areas of triangles adjacent to lateral sides.
= = = = = 5*105 = 525,
= = = = = 5*105 = 525.
Thus the total area of the trapezoid ABCD is 315 + 875 + 525 + 525 = 2240 cm^2.
At this point, the problem is solved completely.
The lesson to learn from this my solution:
if you are given a trapezoid, divided by its diagonals in four triangles, then
(a) the triangles, adjacent to the bases, always are similar;
(b) the triangles adjacent to lateral sides always have equal areas.
If, in addition, you are given the area "a" of a triangle adjacent to a base and the base-to-base ratio "k", then
(c) the areas of all triangles of the division can be easily found using the similarity coefficient:
- the area of the opposite triangle is ;
- the areas of the triangles adjacent to lateral sides are k*a;
- the whole area of the trapezoid, in terms of " a " and " k " is = .