SOLUTION: Alex's dog, Fluffy, is tied witha 20 foot rope to the center of the back wall of the shed, which has dimensions of 14 ft by 18 ft. If trhe back wall is the longer wall, over what a

Algebra ->  Surface-area -> SOLUTION: Alex's dog, Fluffy, is tied witha 20 foot rope to the center of the back wall of the shed, which has dimensions of 14 ft by 18 ft. If trhe back wall is the longer wall, over what a      Log On


   



Question 149136: Alex's dog, Fluffy, is tied witha 20 foot rope to the center of the back wall of the shed, which has dimensions of 14 ft by 18 ft. If trhe back wall is the longer wall, over what area can Fluffy play, to the nearest square foot? Would your answer change if the back wall was 14 ft instead?
Thank you so much!

Answer by josmiceli(19441) About Me  (Show Source):
You can put this solution on YOUR website!
If the rope is stretched tight, Fluffy can reach everything at
a distance of 20' from the center of the 18' wall to the left and right
as long as Fluffy doesn't start to bend the rope around the corners
So far that is a semicircle with area of
pi%2Ar%5E2+%2F+2 where r+=+20ft
3.1416%2A400+%2F+2+=+628.32ft2
When the rope bends around the corner, it's radius becomes 20+-+%2818%2F2%29
20+-+9+=+11, and Fluffy has to stop when he gets to the side
wall. That is a quadrant (1/4th) of a circle on each side with total
area of pi%2A11%5E2+%2F+4+%2B+pi%2A11%5E2+%2F+4
pi%2A11%5E2+%2F+2+
3.1416%2A121+%2F+2
190.067 ft2
The sum of these areas is:
628.32+%2B+190.067+=+818.39 ft2
------------------------------
If the back wall is 14', the answer does change.
The big semicircle is still 628.32 ft2,
but the area of the side quadrants is bigger
the radius is 20+-+%2814%2F2%29
+20+-+7+=+13
The area is
pi%2A13%5E2+%2F+4+%2B+pi%2A13%5E2+%2F+4
pi%2A13%5E2+%2F+2
3.1416+%2A+169+%2F+2+=+265.46
So, the total area is
628.32+%2B+265.46+=+893.78 ft2