# A table and its tablecloth

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Hey all,

I was browsing through our math book the other day, and I came upon an interesting question. Say you have a table with a diameter of 0.6m, and on that table a square meter-by-meter tablecloth. One corner of the tablecloth is hanging over by half a meter, and another by 0.3 meters. By how much are the other corners hanging over? I came up with 0.5 and 0.3 meters, but I'm not sure I'm right. Can anyone verify?

Gabe

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Is the "another corner" an adjacent or an opposing one? Does the table have to be completely covered by the tablecloth? Actually I'm not sure if even then there's just one correct answer.

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So far I've figured out that the only way the tablecloth is big enough is if it's opposite corners, but after that my sketch turned into an Escher-esque confusion of ever decreasing triangles.

Ah... got .4m each for the other two.... please hold for translation of method from gibberish.

Our tablecloth is square, ABCD clockwise from top left, our table, has it's center E on line AC. AE must equal 60cm, 30cm drop of corner A plus radius of table. AC is root 2. Or to be precise 1.4ish.

We need to find length ED and EB which are in symmetry about AC.

Drawing a line perpendicular to AD to E gives us one side of a triangle the hypotenuse of which is ED. Where it crosses AD is point F

EF forms a 45 degree triangle with AE, where the hypotenuse is 60cm, and therefore AE is square root of .6^2/2 = 0.424

The other side DF is 1-.424

So sqrt ( (1-.424)^2 + .424^2) = DE = .715m

We want drop from edge of table so taking diameter of table away .715 - .3 = .415 m

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I didn't finish reading, so I can't comment on the rest and I'm short on time, but I can already see what I perceive to be an error:

our table, has it's center E on line AC

I made the same mistake; we don't know that its center is on AC. In fact, I believe it isn't.

I'll post my solution to the problem once I have more time, sorry for not including it in my first post.

Cheers,

Gabe

EDIT: Oh shoot, I forgot to include a very important part of the problem; the table is a circle with a diameter of 0.6

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