# A line segment with a length of 24 makes a 90-degree angle with one of the legs of an isosceles trapezoid. What is the area of this Trapezoid?

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?

• ABCD does not look very isosceles in the image ... – hmakholm left over Monica Dec 27 '18 at 19:03
• Oh, Sorry, I didn't notice when I drew it – Eldar Rahimli Dec 27 '18 at 19:04

Option 1. Consider the isosceles trapezoid with $$BE=BD$$ as diagonal:

$$\hspace{3cm}$$

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:

$$\hspace{3cm}$$

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.

• 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? – hmakholm left over Monica Dec 27 '18 at 19:28
• Perhaps, the setter considered the two extreme cases and concluded. Though, the option 1 was the extreme. – farruhota Dec 27 '18 at 19:33
• 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). – hmakholm left over Monica Dec 27 '18 at 19:37
• @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. – Eldar Rahimli Dec 28 '18 at 15:30
• @TheGreatDuck, here it states "at least one pair of parallel sides" and shows special types of trapezoids that are parallelograms, rectangles, rhombi and squares. – farruhota Dec 29 '18 at 5:35

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).

• If $|BCE|$ is fixed, then doesn't this fix A and D, too? Can you elaborate on your answer, please? – Eldar Rahimli Dec 27 '18 at 18:40
• Note that the trapezoid is isosceles. – Arthur Dec 27 '18 at 18:43
• 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. – Aqua Dec 27 '18 at 18:43
• How does that affect? @Arthur – Aqua Dec 27 '18 at 18:44
• You can tilt the sides. Change the $BCD$ angle. – Andrei Dec 27 '18 at 18:48

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 $$\frac{10\cdot 24}{26}(26-\frac{10\cdot 10}{26}) \approx 204$$.

• thanks for the answer. can you explain what do you mean by coincide,please? The answer is indeed 240 – Eldar Rahimli Dec 27 '18 at 19:16
• @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. – hmakholm left over Monica Dec 27 '18 at 19:21