In a trapezoid ABCD with Ab parallel to Cd the diagnals intersect at point E. Te area of triangle ABE is 32 and the area of triangle CDE is 50.Find the area of the trapezoid.
Thnks for all of the usefull help
I will not answer this (I've got no time) but I will point these out:
1) angleAEB is a vertical angle to CED
2) angle ABE = 180 - angleCDE
3) angle EAB = 180 - angle ECD
4) angle BAD = 180 - angle ADC
5) angle BCD = 180 - angle ABC
6) angle AED + angle BEC = 360 - ( angle AEB + angle CED )
I think what Quick was aiming for is this:
Triangle AEB is similar to triangle CED.
So we know the following information:
AB/CD = sqrt(32/50) and h1/h2 = sqrt(32/50) where h1 and h2 are the heights of triangles AEB and CED respectively.
Now, consider the area formula for the trapezoid:
A = (1/2)(AB + CD)h = (1/2)(AB + CD)(h1 + h2)
where h = h1 + h2 is the height of the trapezoid.
Now for some fun with factoring. We don't know ANY of these values, but let's factor a CD from (AB + CD):
A = (1/2)(CD)(1 + [AB/CD])(h1 + h2)
Now let's factor an h2 from (h1 + h2):
A = (1/2)(CD)(h2)(1 + [AB/CD])(1 + [h1/h2])
A = {(1/2)*CD*h2}{(1 + [AB/CD])(1 + [h1/h2])}
Now, look at these factors for a moment. We know [AB/CD] = [h1/h2] = sqrt(32/50), and what is (1/2)*CD*h2? Nothing less than the area of triangle CED, which we know!
Thus
A = 50*[1+sqrt(32/50)]^2
-Dan
The only thing I didn't really explain is why the two triangles are similar. If you need help with what I did on the factoring, just let me know.
My sketch of the trapeziod has A in the upper LH corner, and B C and D labelled in a clockwise fashion from this point. (My AB was shorter than my CD, which is probably what you drew as well.) We have two diagonals drawn: AC and BD, which meet at point E.
We know that AB and CD are parallel (defined as such in the problem). So we know that angles BAC and CDA are equal (opposite interior angles of a line cutting two parallel lines are equal). Similarly we know that angles ABC and CDB are equal (same reason.) We know that angles AEB and CED are equal (vertex angles are equal). Thus the three angles of triangle AEB are equal to the respective angles of triangle CED. Thus triangles AEB and CED are similar.
Oh, the other thing I didn't mention in case you aren't familiar with it. In two similar triangles 1 and 2 (actually any two similar shapes), we know that distances are proportional. (In the triangles in the problem: AB/CD = AE/CE = BE/DE = k where k is some constant.) But we are given the areas of the two triangles. Well, areas go by the square of this proportionality constant. (So in the problem, Area of AEB/Area of CED = k^2 where k is the same as the constant AB/CD = k.) In the problem you are solving, of course, we had the ratio of the areas, so when comparing linear disances I took the square root of the ratio.
If these concepts are not known to you, you might want a proof based on other ideas. (I admit I can't think of another way, but more than one other method likely exists.)
-Dan