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Same Areas of Trapezoid within Triangle, why?

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chrisoutwrigh
#1
Dec25-12, 07:12 AM
P: 5
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
#2
Dec25-12, 09:34 AM
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P: 11,589
Are the Areas CDE and BED really the same?
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
#3
Dec25-12, 10:43 AM
P: 5
Quote Quote by mfb View Post
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.
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
#4
Dec25-12, 11:01 AM
P: 5
Same Areas of Trapezoid within Triangle, why?

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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maths.jpg  
mfb
#5
Dec25-12, 12:22 PM
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P: 11,589
I still don't see the reason for that complicated calculation.

How did you get the area CDE with g1(x)?
chrisoutwrigh
#6
Mar18-13, 12:59 PM
P: 5
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
#7
Mar21-13, 09:49 AM
Mentor
P: 11,589
Quote Quote by chrisoutwrigh View Post
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.
I don't see what you mean.
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.
Easy: Both triangles have the area 1/2 b h. Nothing else was required.

@ 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..
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.


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