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Same Areas of Trapezoid within Triangle, why? 
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#1
Dec2512, 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)DE]) the other vice versa): CDE: (hc1)*b*0.5 + 0.5*h*r*m  0.5*c1*r*m BDE: (hc2)*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! 


#2
Dec2512, 09:34 AM

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P: 11,589

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 xdirection between D and C is not r. This would require that BA and CA have the same length. 


#3
Dec2512, 10:43 AM

P: 5

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


#4
Dec2512, 11:01 AM

P: 5

Same Areas of Trapezoid within Triangle, why?
I have now replaced the wrong xcoordinate 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? 


#5
Dec2512, 12:22 PM

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I still don't see the reason for that complicated calculation.
How did you get the area CDE with g_{1}(x)? 


#6
Mar1813, 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.. 


#7
Mar2113, 09:49 AM

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