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Geometry

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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
+75302
+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

CPhill  Jun 14, 2017
edited by CPhill  Jun 14, 2017
Sort:

#1
+75302
+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

CPhill  Jun 14, 2017
edited by CPhill  Jun 14, 2017
#2
+4154
+2

Ohhh !!!    I was stumped on this one...!

Here is how I saw it from your answer:

hectictar  Jun 14, 2017
#3
+75302
+1

Thanks for the diagram, hectictar....!!!!  That really helps.....

To be honest....it was a little bit of a head-scratcher for me, too......I first tried to "over-complicate" things.....then, as usual......the "easy way" just kind of  "fell out ".....LOL!!!!

CPhill  Jun 14, 2017

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