+0  
 
-2
14
2
avatar+819 

A circular table is pushed into the corner of a square room so that a point $P$ on the edge of the table is $4''$ from one wall and $5''$ from the other wall as shown. Find the radius of the table in inches.

 

 Sep 25, 2023

Best Answer 

 #2
avatar+397 
+1

Run a horizontal through P to meet the circle again at A.

Drop a vertical through P to meet the circle again at B.

Angle APB is a rt-angle so AB will be a diameter of the circle.

To calculate the length of PA, drop a vertical from the point of contact with the wall onto PA and show that half the length of PA is           r - 4, so  that the length of PA is 2r - 8, where r is the radius of the circle.

Similarly the length of PB is 2r - 10. 

Now, by Pythagoras in APB, (2r - 8)^2 + (2r - 10)^2 = (2r)^2,

leading to

r^2 - 18r +41 = 0,

and r = 15.325 (3 dp).

 Sep 25, 2023
 #1
avatar+128826 
+1

The diagonal of the rectangle  = sqrt [ 5^2 + 4^2] = sqrt [ 41] in

 

Call the radius of the table  = r

 

By the Pythagorean Theorem

 

r^2 + r^2  =  [ r + sqrt 41]^2

 

2r^2  = r^2 + 2rsqrt 41 + 41

 

r^2 - 2sqrt 41 r  - 41  =  0

 

Solving for r gives that  r =  sqrt (41) + sqrt (82) =   sqrt (41) ( 1 + sqrt 2)  in  ≈ 15.46 in

 

cool cool cool 

 Sep 25, 2023
 #2
avatar+397 
+1
Best Answer

Run a horizontal through P to meet the circle again at A.

Drop a vertical through P to meet the circle again at B.

Angle APB is a rt-angle so AB will be a diameter of the circle.

To calculate the length of PA, drop a vertical from the point of contact with the wall onto PA and show that half the length of PA is           r - 4, so  that the length of PA is 2r - 8, where r is the radius of the circle.

Similarly the length of PB is 2r - 10. 

Now, by Pythagoras in APB, (2r - 8)^2 + (2r - 10)^2 = (2r)^2,

leading to

r^2 - 18r +41 = 0,

and r = 15.325 (3 dp).

Tiggsy Sep 25, 2023

2 Online Users

avatar