Area- Walkway - Solution/Explanation

[Guest]

A circular courtyard has a diameter of 80 ft. A cement walkway a yard in width borders the courtyard on the inside. Determine area of walkway.

Solution provided with problem:

Outer boundary of walkway is the circumference of a circle with 40 ft. radius - the inner boundary is the circumference of a circle with 37 ft. radius.

Area of walkway in sq. ft. - 77 ft. x 3 x 3.1416 = 725.7096.


Rule provided - Determine area of each circle, and subtract the area of the inner circle from the area of the outer circle - or multiply the sum of the radii by the difference and product by 22/7, or 3.1416 for larger accuracy.

I don't understand the solution with the rule.

[Guest]

You have just used the rule

[Guest]

[tex]\pi = 3.1416...[/tex] or roughly [tex]22/7[/tex]
[tex]R =[/tex] radius of courtyard
[tex]r =[/tex] radius of inner courtyard (part without a walkway)
Area of courtyard [tex]= \pi R^2[/tex]
Area of inner courtyard [tex]= \pi r^2[/tex]
Area of walkway = Area of courtyard - Area of inner courtyard
[tex]=\pi R^2-\pi r^2[/tex]
This is the first way the solution suggests.

The second way comes from realizing that [tex]\pi R^2-\pi r^2[/tex] can be factorized
[tex]\pi R^2-\pi r^2[/tex]
[tex]= \pi(R^2-r^2)[/tex]
[tex]=\pi(R+r)(R-r)[/tex] (See the difference of two squares https://en.wikipedia.org/wiki/Difference_of_two_squares)

Either formula is an acceptable way of calculating the area.

Hope this helped,

R. Baber.

[Guest]

That helped.

A couple of questions:

Are the courtyard and walkway also considered to be a ring ?

Can the length of the walkway be determined ?

Thanks again.

[Guest]

The entire courtyard is a disc/circle, the inner courtyard (the part that doesn't have the walkway) is another disc/circle (of smaller radius). The walkway is a ring or annulus. The length of the walkway is ambiguous, it is longer on the outside ([tex]2\pi R[/tex]) than on the inside [tex]2\pi r[/tex]. Usually you take the average and call that the length.

Hope this helped,

R. Baber.

[Guest]

That helped. Thanks.