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?
+128407 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
+128407 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
I was stumped on this one...!
