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The figure below is a trapezoid, what is the length of the red line?

Thank you very much in advance!

enter image description here

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The use of subjective judgements like "hard", "easy", "tricky" in the title rarely contributes valuable information to the title. Here, as in most cases when such words are used, the perceived difficulty will vary widely among different users. Further, "geometry/trigonometry problem" is very general and conveys no information beyond that already contained in the tags, which are visible wherever the title appears. A more informative title might be, for example, "Finding a diagonal in a trapezoid given the other diagonal and three sides". – joriki Nov 9 '12 at 20:26
I sincerely accept your suggestion – John Lee Nov 9 '12 at 20:37
So what possibly useful facts have you inferred so far? It's a good strategy to make a catalog of everything you can find out so you can see if any of those facts help. – Harald Hanche-Olsen Nov 9 '12 at 20:45
up vote 2 down vote accepted

There are 2 circles, one of radius $49$ , with center $(0,0)$the other with radius $28$ with center $(49,0)$, solving the system: $$x^2+y^2=49^2$$ $$(x-49)^2+y^2=28^2$$

Gives you the intersection point $(41,12\sqrt{5})$. Then you solve the system $$y=12\sqrt{5}$$ $$(x-49)^2+y^2=28^2$$ and cuts the circle with radius $28$ at $x=57$, but we have also $y=12\sqrt{5}$ so the lenght is $\sqrt{57^2+(12\sqrt{5})^2}=63$

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Nice work, but perhaps you can clarify for the OP what you are taking to be the origin? (or perhaps, where you are centering the circles?) @dotdot: +1 – amWhy Nov 9 '12 at 21:06
very clever, thanks so much! – John Lee Nov 9 '12 at 21:16

Well, I think I see a method of proceeding. Let $x$ be one of the 2 equal angles of the isoceles triangle to the left. It's possible to use the law of cosines to calculate $\cos x$.

It turns out the left triangle is congruent to the right triangle. At least if you assume the drawing is accurate enough that the top and bottom are the 2 parallel lines of the trapezoid and not left and right. The 2 equal angles of the left and right isoceles triangles are equal due to alternate interior angles (had to look up the name, been a while). So the third angle of the triangle on the right is $180^o-2x$ and the lower right angle of the trapezoid is $x+180^o-2x=180^o-x.$

Now you can use the law of cosines along with $\cos(180^o-x)=-\cos x$ to solve for the length of the red line.

Not sure it's as elegant as dot dot's answer, but let's make sure it gets the same result. We get that

$$\cos x=\frac{49^2+28^2-49^2}{2\times49\times28}=\frac27$$

$$c^2=49^2+28^2-2\times49\times28\times-\frac27=$$ $$49\times7^2+16\times7^2+16\times7^2=81\times7^2$$ $$c=9\times7=63$$

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It's also nice. +1 – dot dot Nov 9 '12 at 22:00

Since the two triangles in figure are similar, the length of the upper base of the trapezoid is $28\cdot 28/49 = 16$, so the heigth is $\sqrt{28^2-8^2}=12\sqrt{5}$ and the length of the red line is $\sqrt{(12\sqrt{5})^2+(49+8)^2}=63.$

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Where exactly does the 8 come from? – Mike Nov 10 '12 at 3:23
$8$ is half of the length of the upper base. – Jack D'Aurizio Nov 11 '12 at 11:05

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