# Circular park

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• Aug 19th 2008, 11:53 PM
devi
Circular park
A circular park of 20m diameter has a circular path just inside the boundary with a width of 1m.The are of the path is
15pi
17pi
19pi
25pi
• Aug 20th 2008, 12:00 AM
Chris L T521
Quote:

Originally Posted by devi
A circular park of 20m diameter has a circular path just inside the boundary with a width of 1m.The are of the path is
15pi
17pi
19pi
25pi

Find the area within the park, and then subtract from that the area within the path.

So you should get $400\pi-361\pi=\dots$

However, I don't see the answer as one of the choices you gave...hmmm..I'm probably overlooking something pretty obvious... (Wondering)

--Chris
• Aug 20th 2008, 12:03 AM
devi
Quote:

Originally Posted by Chris L T521
Find the area within the park, and then subtract from that the area within the path.

So you should get $400\pi-361\pi=\dots$

However, I don't see the answer as one of the choices you gave...hmmm..I'm probably overlooking something pretty obvious... (Wondering)

--Chris

The answer was given as 19Pi
• Aug 20th 2008, 12:09 AM
Chris L T521
Quote:

Originally Posted by devi
The answer was given as 19Pi

Quote:

Originally Posted by Chris L T521
Find the area within the park, and then subtract from that the area within the path.

So you should get $400\pi-361\pi=\dots$

However, I don't see the answer as one of the choices you gave...hmmm..I'm probably overlooking something pretty obvious... (Wondering)

--Chris

I caught my error...It had a 20 m diameter...

so the area of the park would be $(10~m)^2\pi=100\pi~m^2$

The area of the park enclosed by the path would be $(10~m-1~m)^2\pi=(9~m)^2\pi=81\pi~m^2$

So then we see that the difference between the two will give us the area of the path.

$100\pi~m^2-81\pi~m^2=\color{red}\boxed{19\pi~m^2}$

Does this make sense? Sorry about that; it was a mistake on my part...misread the problem.

--Chris