Square Yards, Area

[Guest]

A park has a circular lake 240' in diameter. A walkway 6' wide is around the lake. Determine 1) Number of square yards in walkway, 2) Area of walkway if width is doubled.

My partial solution:

Area of ring - pi(R + r) (R - r)

A = pi (120 + 3) (120 - 3)

pi (123) (117)

pi (14391)

3.1416 (14391)

45120.7656 square yards.

How do I solve from here ?

[Guest]

Please respond to my question. Thanks.

[Guest]

Area of ring - pi(R + r) (R - r)

A = pi (120 + 3) (120 - 3)

I haven't thought about the question yet but do you mean.....

Area of ring - pi(R + r) (R - r)

A = pi (120 + 6) (120 - 6)

Because is the walkway not 6 feet wide..?????

[Guest]

Because is the walkway not 6 feet wide..?????



Because is the walkway not 6 feet wide..?????


"Because is the walkway not 6 feet wide..?????"

Yes. I was not sure. I thought the "r" would be the radius of the smaller diameter.

[Guest]

Is the smaller diameter not the lake.... = 240 feet....so 120 feet radius
and 1st walkway is 6 feet wide..........

Area = pi[(r+6)^2 - r^2]

= pi[ r^2 + 12r + 36 - r^2]

= pi[ 12r + 36 ]

= 3.142 [(12 x 120 ) + 36]

= 4637.59 sq feet

If walkway width is doubled to 12 feet

Area = pi[(r + 12)^2 - r^2]

= pi[ r^2 + 24r + 144 - r^2]

= pi[ 24r + 144]

= 3.142[(24 x 120) + 144]

= 9501.41 sq feet

[Guest]

Sorry for duplicated reply to question ... computer problem.

"pi[ r^2 + 12r + 36 - r^2]"

"pi[ r^2 + 24r + 144 - r^2]"

How was the 12r and 24r obtained ?

[Guest]

The last post is same as doing it your way.... as an expression of difference of 2 squares.......
Area = pi[(r+6)^2 - r^2]

= pi [ (r + 6 + r)(r + 6 - r) ]

the r's cancel in 2nd bracket so left with

= pi [ (2r + 6) x 6

= pi [ 12r + 36 ] as before............

and for the double width..............

= pi [ (r + 12 + r)(r + 12 - r) ]

the r's cancel in 2nd bracket so left with

= pi [ (2r + 12) x 12

= pi [ 24r + 144 ] as before............

[Guest]

"pi[ r^2 + 12r + 36 - r^2]"
"pi[ r^2 + 24r + 144 - r^2]"

How was the 12r and 24r obtained ?

The large radius was (r+6) and we have to square it and multiply by pi
The smaller radius is r and we have to square it and subtract it.
So that gives Area = pi[(r+6)^2 - r^2]

(r+6)^2 = (r+6)(r+6) multiply together

Multiply each term in one bracket on each term in the other bracket

r(r+6) + 6(r+6)

r^2 + 6r + 6r + 36

Gives r^2 + 12r + 36

Then you subtract off the other -r^2

Leaves 12 + 36

and work the same way for the double width path.........

[Guest]

I understand now. Thanks again.

[Guest]

Gives r^2 + 12r + 36

Then you subtract off the other -r^2

Leaves 12 + 36 .........Typo error should be 12r + 36 ...........xxxxxxxx

and work the same way for the double width path........