T O P I C R E V I E W 
raghu 
Posted  04/30/2008 : 08:34:25 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 
7 L A T E S T R E P L I E S (Newest First) 
Subhotosh Khan 
Posted  05/01/2008 : 08:49:10 quote: Originally posted by raghu
Got it. Thank you. it results in a quadratic (a5)(a17)= 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.

raghu 
Posted  05/01/2008 : 07:40:49 Got it. Thank you. it results in a quadratic (a5)(a12)= and it has to be a = 17 because a = 5 is not feasible. Thanks for your help. 
Subhotosh Khan 
Posted  04/30/2008 : 20:50:56 quote: Originally posted by raghu
But, How? sqrt(85) <> 17<<<< because (85) is not the correct answer

raghu 
Posted  04/30/2008 : 15:00:21 But, How? sqrt(85) <> 17 
Subhotosh Khan 
Posted  04/30/2008 : 13:26:30 17" is the correct answer. 
raghu 
Posted  04/30/2008 : 10:44:02 This is a problem that appeared in Georgia tech math competition. The answer suggested is 17 inches. I got sqrt(85) as the answer, 
Subhotosh Khan 
Posted  04/30/2008 : 09:17:24 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 ..."? 