+0  
 
0
1236
3
avatar

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?


 Jun 13, 2017

Best Answer 

 #1
avatar+98061 
+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

 

 

 

cool cool cool

 Jun 14, 2017
edited by CPhill  Jun 14, 2017
 #1
avatar+98061 
+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

 

 

 

cool cool cool

CPhill Jun 14, 2017
edited by CPhill  Jun 14, 2017
 #2
avatar+7350 
+2

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

 

Here is how I saw it from your answer:

 

 Jun 14, 2017
 #3
avatar+98061 
+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!!!!

 

 

 

cool cool cool

 Jun 14, 2017

32 Online Users

avatar
avatar