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# circle

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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

#2
+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,

r^2 - 18r +41 = 0,

and r = 15.325 (3 dp).

Sep 25, 2023

#1
+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

Sep 25, 2023
#2
+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,