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The circular table in the diagram is pushed tangent to two perpendicular walls. Let P be a point on the circumference of the table, such that P is on the minor arc between the two points of tangency (like in the diagram). The distances from P to the two walls are 27 and 6. What is the radius of the table?

Guest Jun 13, 2017

#1**+2 **

Let the center of the circle be O

And we can construct right triangle OQP with angle OQP being = 90°

Let the hypotenuse of this triangle be OP = the radius, R

Let one of the legs of this triangle = PQ = (R - 27)

And let the other leg = OQ = (R - 6)

And by the Pythagorean Theorem, we have that

PQ^2 + OQ^2 = OP^2

(R - 27)^2 + (R - 6)^2 = R^2 simplify

R^2 - 54R + 729 + R^2 - 12R + 36 = R^2

2R^2 - 66R + 765 = R^2

R^2 - 66R + 765 = 0

This can be factored as

(R - 15) (R - 51) = 0

Setting each factor to 0 we have that R = 15 or R = 51

We must reject R = 15 because one of the legs, PQ, would have a negative length

So the radius = 51

CPhill Jun 14, 2017

#1**+2 **

Best Answer

Let the center of the circle be O

And we can construct right triangle OQP with angle OQP being = 90°

Let the hypotenuse of this triangle be OP = the radius, R

Let one of the legs of this triangle = PQ = (R - 27)

And let the other leg = OQ = (R - 6)

And by the Pythagorean Theorem, we have that

PQ^2 + OQ^2 = OP^2

(R - 27)^2 + (R - 6)^2 = R^2 simplify

R^2 - 54R + 729 + R^2 - 12R + 36 = R^2

2R^2 - 66R + 765 = R^2

R^2 - 66R + 765 = 0

This can be factored as

(R - 15) (R - 51) = 0

Setting each factor to 0 we have that R = 15 or R = 51

We must reject R = 15 because one of the legs, PQ, would have a negative length

So the radius = 51

CPhill Jun 14, 2017