SOLUTION: ABCD IS A PARALLELOGRAM.TRIANGLE DEC IS DRAWN SUCH THAT BE=1/3 AE.SUM OF THE AREAS OF TRIANGLES ADE AND BEC IS?

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Question 823877: ABCD IS A PARALLELOGRAM.TRIANGLE DEC IS DRAWN SUCH THAT BE=1/3 AE.SUM OF THE AREAS OF TRIANGLES ADE AND BEC IS?
Answer by KMST(5345) About Me  (Show Source):
You can put this solution on YOUR website!
Since the problem does not state any side measurements (just a ratio), the most that we could do is calculate that sum of areas relative to the area of the parallelogram.
Without a picture, I am not quite sure I can interpret the situation intended, but I will try.

We could calculate that sum of areas as %281%2F2%29%2AareaofABCD for the situation below.

In that case, BE=%281%2F3%29AE ,
areaofBCE=%281%2F2%29b%2A%28h%2F4%29
areaofADE=%281%2F2%29b%2A%283h%2F4%29
areaofABCD=b%2Ah
area of ADE + area of BCE =%281%2F2%29b%2A%283h%2F4%29%2B%281%2F2%29b%2A%28h%2F4%29
area of ADE + area of BCE =%281%2F2%29b%2A%283h%2F4%2Bh%2F4%29
area of ADE + area of BCE =%281%2F2%29b%2Ah%29
It really does not matter what fraction of AE is BE,
because the heights of ADE and BCE would add up to h anyway.

It really does not matter where E is located, as long as it is between the lines AD and BC,
because we could still calculate the areas of ADE and BCE
based on b (the length of AD and BC) as the base for both,
and the distances from E to AB and to BC as the heights.
And those heights would add up to h anyway.
The ratio of AE is BE, does not matter either,
because the heights of ADE and BCE would add up to h anyway.