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.

maximum Sep 25, 2023

#2**+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).

Tiggsy Sep 25, 2023

#1**+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

CPhill Sep 25, 2023

#2**+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