Is this problem a trick question or something?

[chicaobscura88]

I have a summer calculus take-home test due the first day of school of 200+ problems. I got all of them but two-- this one and the one I posted in the "other math help" category. This one is geometry related. I have asked everyone I know to help me out with this problem and no one can get it. A few people have suggested that it may be unsolvable, or that my teacher left out some information so it is incomplete. If anyone can get an answer though, I would be forever in your debt. Here it goes:

A rectangle is inscribed in a circle of radius r. The horizontal dimension of the rectangle is x, the vertical dimension is y. Express the area of the rectangle as a function of the variable x only.

There is a diagram included as well, but the problem is pretty much self-explanatory. It is on a page of the test that is hand-written by my teacher, so it's possible that he messed it up, but if there is a way to solve it, I'm hoping that someone here can find it. HELP PLEASE! Thanks in advance for any information you can offer!

[Denis]

Draw a diagonal inside the rectangle;
this diagonal = 2r (or the diameter of the circle):
an inscribed rectangle in a circle has same center as circle,
so the diagonal goes through the circle's center.

So you have 2 right triangles, each with hypotenuse = 2r and legs = x and y.
So (2r)^2 = x^2 + y^2
4r^2 = x^2 + y^2
rearrange:
y = sqrt(4r^2 - x^2)

Since area = xy, then area = x[sqrt(4r^2 - x^2)]

That's the best you can do; professor probably had too much to drink!

[stapel]

I agree with Denis. I see no way of getting rid of the radius variable "r".

Eliz.

[chicaobscura88]

Yeah, that's about as far as I got. I tried a whole bunch of formulas I have for the areas of circles and triangles and whatnot and I can't seem to find any possible way to eliminate r from any of them. Oh well. If I only leave one blank, I can still hope to get a pretty good grade.

[Denis]

DO NOT leave it "blank"; there is an answer: CANNOT BE DONE.

[chicaobscura88]

Thanks again! Will do.