#2

Re: #2

If I have this right, I will be using the √ symbol three times in my answer.

Make that "twice"

My goal, before I reach the age of 90, is to learn to check my work BEFORE giving my answer! But I'm almost 2/3 of the way there, and haven't been able to learn that yet! ;-)

Re: Make that "twice"

What do you mean by "twice"?

Are you using the square root symbol six times?

Re: Make that "twice"

What do you mean by "twice"?

Are you using the square root symbol six times, or does your result have an overall factor of two?

Re: #2

The result is simple, but the proof is fairly difficult.

Sum of three simple terms, one of them containing a square root.

Re: Make that "twice"

I meant twice instead of three times. But I probably agree with your answer, mine could be restated as sum of three terms, one of which involves a square root.

Re: Make that "twice"

The latter, which multiplied out gives your answer (I suspect).

Re: #2

I'd say hard to see, but easy to prove once you do see it.

Re: #2

The area of the trapezoid is (square root of A + square root of B)^2.

Re: #2

That's the form of the answer I had (after correcting my work!), but I think I like Niko's version, multiplying it out better:

A + B + 2*√(AB)

This also tells us that the side triangles are each √(AB), and interesting result.

Solution/proof

Triangle A is similar to triangle B (opposite angles identical, and alternating interior angles of line crossing two parallel lines, if I remember my terminology correctly). I.e., take triangle B, rotate it 180 degrees around the point where they meet, then supersize it, and you have triangle A.

Since the ratio of the areas is A:B, the ratio of linear dimensions (specifically the altitude and base) is √A : √B.

Now consider the triangles composed of A+C and B+D Those triangles have the same bases as A and B, but their altitude is the sum of the altitude of A and the altitude of B. So the areas are A * (√A + √B) / √A and B * (√A + √B) / √B. Straight algebra gets you to (√A + √B) ^2.

And while not needed for the above, C and D have the same area, which surprised me. Consider the triangles A+C and A+D. Each has the same base and altitude. QED

Re: Solution/proof *LINK*

My solution, although not identical, is along the same lines, i.e. working with ratios of areas instead of the areas themselves.

I you want to see how the areas A, B, C, and D are calculated using the linear dimensions of the trapezoid, follow the link.


http://mathworld.wolfram.com/Trapezoid.html