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raghu
Senior Member

USA
37 Posts

Posted - 04/30/2008 :  08:34:25  Show Profile  Reply with Quote
a circular table is pushed into a corner in a rectangular room so that it touches both walls. A point on the edge of the table between the two points of contact is 2 inches from one wall and 9 inches from the othre. What is the radius of the table
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 04/30/2008 :  09:17:24  Show Profile  Reply with Quote
quote:
Originally posted by raghu

a circular table is pushed into a corner in a rectangular room so that it touches both walls. A point on the edge of the table between the two points of contact is 2 inches from one wall and 9 inches from the othre. What is the radius of the table



Please show your work - so that we know where to begin to help you.

1) Draw a sketch of the problem.

2) denote the walls as the x & y axes - write the equation of the circle (table) with radius 'r'.

3) P(2,9) is a point on the circle.

4) solve for 'r' - using (2) and (3)

5) Why did the problem statement include "A point on the edge of the table between the two points of contact ..."?

Edited by - Subhotosh Khan on 04/30/2008 09:24:51
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raghu
Senior Member

USA
37 Posts

Posted - 04/30/2008 :  10:44:02  Show Profile  Reply with Quote
This is a problem that appeared in Georgia tech math competition. The answer suggested is 17 inches. I got sqrt(85) as the answer,
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 04/30/2008 :  13:26:30  Show Profile  Reply with Quote
17" is the correct answer.
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raghu
Senior Member

USA
37 Posts

Posted - 04/30/2008 :  15:00:21  Show Profile  Reply with Quote
But, How? sqrt(85) <> 17
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 04/30/2008 :  20:50:56  Show Profile  Reply with Quote
quote:
Originally posted by raghu

But, How? sqrt(85) <> 17<<<< because (85) is not the correct answer

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raghu
Senior Member

USA
37 Posts

Posted - 05/01/2008 :  07:40:49  Show Profile  Reply with Quote
Got it. Thank you. it results in a quadratic (a-5)(a-12)= and it has to be a = 17 because a = 5 is not feasible. Thanks for your help.
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Subhotosh Khan
Advanced Member

USA
9117 Posts

Posted - 05/01/2008 :  08:49:10  Show Profile  Reply with Quote
quote:
Originally posted by raghu

Got it. Thank you. it results in a quadratic (a-5)(a-17)= and it has to be a = 17

because a = 5 is not feasibleCorrect ... that is where the part "A point on the edge of the table between the two points of contact ..." comes into play


. Thanks for your help.

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