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A circle is placed in a 90 degree corner, and that circle is touching both walls of that corner. a point on the circumference of the circle is 18 inches from one wall and 25 inches from the other wall. what is the radius of the circle? if some work could be shown it would be very nice to know for the future.

 Apr 15, 2021
 #1
avatar+118586 
+1

See  the  following  image :

 

Let  the  corner be defined  by FGB

Let A  be  the  point  on  the circumference

Let AC  =  18

Let FD  =18 - r

Let AD = DE = FB  =  r

Let AF  =  25 - r

 

Triangle  AFD  is  right  such  that

 

FD^2  + AF^2  =  AD^2

 

(18 - r)^2  +  ( 25 - r)^2  =  r^2   simplify

 

r^2  - 36r + 324  + r^2  - 50r + 625  =  r^2

 

r^2  - 86r  +  949  =  0     we can factor  this  as

 

(r - 73) (r - 13)  =  0

 

Setting  both factors  to 0  and solving for  r  we  have

 

r = 73    (reject.......r cannot  be >18)

 

r  =13 inches

 

cool cool cool

 Apr 15, 2021
 #2
avatar+31412 
+1

A slightly different approach....see diagram below:

the angle between the inscribed angle legs of 18 and 25 is  45 degrees

   find c using cosine law

       c^2 = 18^2 + 25^2 - 2 (18)(25)(cos 45) = 312.604

 

Now  c  is also the hypotenuse of the isosceles triangle formed by the two radii

  use pythagorean theorem

     r^2 + r^2 = c^2

       2 r^2 = 312.604

          r = 12.50 inches         Hmmmmmm.....not the same as CPhill's answer....but close !       ~ EP

                    OK...I see what I did differently....I went to the points of contact at the wall.....not directly to the walls !!!

                                 SEE CPhill's answer ......it is best !!!!!!!!!

 

 

 

 Apr 15, 2021
edited by ElectricPavlov  Apr 15, 2021
edited by ElectricPavlov  Apr 15, 2021

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