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enter image description here

Given that $ABCD$ is an isosceles trapezoid and that $|EB|=24$, $|EC|=26$, and m(EBC)=$90^o$. Find $A(ABCD)= ?$

From the pythagorean theorem, I can find that $|BC|=|AD|=10$. Then, I can find the area of the triangle $EBC$. But after this point, I can't progress any further. How do I find the area of this trapezoid?

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    $\begingroup$ ABCD does not look very isosceles in the image ... $\endgroup$ – hmakholm left over Monica Dec 27 '18 at 19:03
  • $\begingroup$ Oh, Sorry, I didn't notice when I drew it $\endgroup$ – Eldar Rahimli Dec 27 '18 at 19:04
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Option 1. Consider the isosceles trapezoid with $BE=BD$ as diagonal:

$\hspace{3cm}$enter image description here

We find: $$BF=\frac{2S_{\Delta BCD}}{CD}=\frac{240}{26}=\frac{120}{13};\\ CF=\sqrt{BC^2-BF^2}=\sqrt{100-\frac{120^2}{13^2}}=\frac{50}{13};\\ S_{ABCD}=\frac{AB+CD}{2}\cdot BF=\frac{(26-2\cdot CF)+26}{2}\cdot \frac{120}{13}=\\ =\frac{34560}{169}\approx \color{red}{204.5}.\\ $$

Option 2. Consider the point $E$ as a midpoint:

$\hspace{3cm}$![enter image description here

We find: $$EH=\sqrt{BE^2+BH^2}=\sqrt{24^2+5^2}=\sqrt{601};\\ BI=\frac{2S_{\Delta BEH}}{EH}=\frac{2\cdot 60}{\sqrt{601}};\\ S_{ABCD}=EH\cdot BF=\sqrt{601}\cdot \frac{4\cdot 60}{\sqrt{601}}=\color{red}{240}.$$

Conclusion: The trapezoid is not unique.

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  • $\begingroup$ Interesting that the area in option 2 is exactly the same as when we take $A=E$. Perhaps that is what misled the problem setter? $\endgroup$ – hmakholm left over Monica Dec 27 '18 at 19:28
  • $\begingroup$ Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme. $\endgroup$ – farruhota Dec 27 '18 at 19:33
  • $\begingroup$ x @farruhota: But the extreme cases are $204.5$ and $240$. He would need to have considered one extreme and one in-the-middle case. (In your option 2, if we cut off triangle DEG and put it next to AE instead, we get a parallelogram that is obviously twice the area of BEC; the setter may have mistakenly thought that this generalizes). $\endgroup$ – hmakholm left over Monica Dec 27 '18 at 19:37
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    $\begingroup$ @farruhota This was a problem from a high school geometry textbook. It was given in the properties section that when $E$ is the middle point the area of the triangle is half of the trapezoid's. Indeed, I believe that is where problem setter made a mistake by not mentioning that $AE=ED$. Thanks for the answer, by the way. $\endgroup$ – Eldar Rahimli Dec 28 '18 at 15:30
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    $\begingroup$ @TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares. $\endgroup$ – farruhota Dec 29 '18 at 5:35
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The triangle $BCE$ is uniqely determined, but other points $D$ and $A$ are not so this trapezoid does not have fixed area, it depend on $A$ (and then is $D$ determined also).

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  • $\begingroup$ If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please? $\endgroup$ – Eldar Rahimli Dec 27 '18 at 18:40
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    $\begingroup$ Note that the trapezoid is isosceles. $\endgroup$ – Arthur Dec 27 '18 at 18:43
  • $\begingroup$ Try to play in some aplet, say Geogebra. You can move $AD$ through $E$ and you will see you don't get a single trapezoid. $\endgroup$ – Aqua Dec 27 '18 at 18:43
  • $\begingroup$ How does that affect? @Arthur $\endgroup$ – Aqua Dec 27 '18 at 18:44
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    $\begingroup$ You can tilt the sides. Change the $BCD$ angle. $\endgroup$ – Andrei Dec 27 '18 at 18:48
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As a concrete example of how the figure is not determined:

  • One option is that $A$ and $E$ coincide, and $ABCD$ is a rectangle of area $24\cdot 10=240$.

  • Another option is that $E$ and $D$ coincide, in which case the area of the trapezoid is $\frac{10\cdot 24}{26}(26-\frac{10\cdot 10}{26}) \approx 204$.

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  • $\begingroup$ thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240 $\endgroup$ – Eldar Rahimli Dec 27 '18 at 19:16
  • $\begingroup$ @EldarRahimli: "Coincide" means that $A$ and $E$ are the same point, so EC is simply the diagonal of the rectangle. It is true that $240$ is one possible answer, but based on the conditions you have disclosed it is not the only possible answer. $\endgroup$ – hmakholm left over Monica Dec 27 '18 at 19:21

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