Mental Stretch

[J Sim]

"Merry-go-round" area solution
As Bagwa pointed out this is a limit problem and the area of the "merry-go-round" deck is 100 pi regardless of the size of the circles given the 20 foot chord.

Trigonometry proof follows:


The radius of the small circle is r2.
The radius of the large circle is d1 which is also equal to r1+r2.
Since the radius of the large circle serves as the hypotenuse of a right triangle
formed by the sides r2 and the 10 ft measurement, the square of the hypotenuse (d1)
is equal to the sum of the squared leg lengths. (d1) squared = (r2) squared + (10 ft) squared.
Pythagorean theorem - If a and b are the legs of a right triangle and c is its hypotenuse, then a2+b2=c2.

Area of a circle equals pi*(radius) squared.
The area of the large circle is pi * (d1) squared also equal to pi * [(r2) squared + (10 ft) squared]
by the substiution above. This is equal to pi*100 + pi*(r2) squared.

The area of the little circle is pi* (r2) squared.

Subtracting the area of the little circle from the area of the large circle is:
[pi*100 + pi*(r2) squared] - pi*(r2) squared = pi*100 or about 314 square feet.
So yes, one gallon of paint covers that area.

The amazing part is that it doesn't matter how big the circles are its just half of the bisecting segment length squared * pi.
If that holds true the really easy approach [as Bagwa proposed] is that the inner circle has a radius of zero (a point), then the bisecting
segment is actually the diameter. (1/2*20ft) squared * pi = 100 * pi