Same Areas of Trapezoid within Triangle, why?

chrisoutwrigh

Good day,
while reading up on an elementary math study book, i have encountered that a proof is build upon the following (see attachment for the figure).
Are the Areas CDE and BED really the same? I tried to calculate this from abstraction, not sure where I could have made a mistake..
g1 is the falling diagonal, g2 the rising, referenced to the start (0,0), the intercepts c1 is yielded when g1(x=b); c2 when g2(x=0).
The formula for the area is constructed via (two triangles at the points: [C-(b,c1)-D) and (b,c1)-D-E]) the other vice versa):
CDE: (h-c1)*b*0.5 + 0.5*h*r*m - 0.5*c1*r*m
BDE: (h-c2)*b*0.5 + 0.5*h*r - 0.5*c2*r
I made the whole figure and the formula, so it could all be wrong!
Although I calculated the intercepts with the drawn perpendicular and they came out correct with quite a big figure and m number, I think the rug is in the surface area formula..
Also upon rough calculation I think formula for BED is correct! so CDE must be revisited!!
Thank you for your response!

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mfb

They have the same base (DE) and the same height (h), so they have the same area 1/2 b h.

I don't see the reason for that complicated calculation.

How do you get (b+r,0) as coordinate for C? In general, the difference in x-direction between D and C is not r. This would require that BA and CA have the same length.

chrisoutwrigh

Hi!
You are right! I' might have used a different letter for it. Actually it should be root[(e*m)^2-h^2] .. i will calculate it accordingly!

chrisoutwrigh

I have now replaced the wrong x-coordinate value for the point C but my formula still was phrased correctly, I just made a transfer typo..some other ideas why this seems wrong?
Does it suffice when the base and height is the same? the angle isn't so it shouldn't be so obvious? Or do I overcomplicate this simple exercise?

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mfb

I still don't see the reason for that complicated calculation.

How did you get the area CDE with g₁(x)?

chrisoutwrigh

Sadly I had no time to properly end this seedless quest.
To answer your question:
The intercepts of the linear equations (diagonals) are needed and then one can split up the triangles (upwards from the diagonal / beneath the diagonal) in two triangles with right angles.
The reason for this complicated formula is following: I only want to use the segments stated in the picture, so as few information as possible.
@ mfb: why is it obvious that they the same base/height? One side they share right! but the angles are different and the other side needed has different length..for me it's not at all obvious..

mfb

I don't see what you mean.
Easy: Both triangles have the area 1/2 b h. Nothing else was required.

They share the same base, and their height is the same (h) as well. Use the standard formula for the area of a triangle, and you are done. This general formula can be derived from the formula of the area of a right triangle in two simple steps, if you like.