SOLUTION: I have no clue how to solve, please help. I need to solve for tonight in order to get extra credit before we take finals on Monday. The problem is : A circular table is placed into

Question 35067: I have no clue how to solve, please help. I need to solve for tonight in order to get extra credit before we take finals on Monday. The problem is : A circular table is placed into the corner of a room, touching both walls which are perpendicular to each other. A point on the edge (circumference) of the table is one foot from one wall and six inches from the other wall. What is the diameter of the table?
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
A circular table is placed into the corner of a room, touching both walls which are perpendicular to each other. A point on the edge (circumference) of the table is one foot from one wall and six inches from the other wall. What is the diameter of the table?
TAKE CORNER AS ORIGIN AND THE 2 PERPENDICULAR WALLS AS X AND Y AXES.SO IF THE RADIUS OF THE CIRCLE IS R ,THEN SINCE THE 2 WALLS/AXES ARE TOUCHIG THE CIRCLE ,THE CNTRE OF THE CIRCLE IS (R,R)
HENCE EN OF CIRCLE IS
(X-R)^2+(Y-R)^2=R^2.....THE COORDINATES OF A POINT ON THE CIRCLE IS (12,6).SO
(12-R)^2+(6-R)^2=R^2
144-24R+R^2+36-12R+R^=R^2
R^2-36+180=0
(R-30)(R-6)=0
HENCE R=30"...OR...6"
SO DIAMETER IS 60"=5'...OR......12"=1'
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