Weird Circle

[[email protected]]

A cow is tethered to the corner of a rectangular shed. If the length of the rope is 5, and the shed has length 4 and width 3, what is the maximum area that is accessible to the cow? (The cow cannot enter the shed).

[ShortyII]

Nifty question.

You essentially have three circles with different radii.

One part has a radius of 5 - the length of the tether and this circle stretches through 270 degrees (3/4 area).

When it runs along the edge of a wall - at its full length it can wrap around the wall providing two small circles with radii of 1 and 2 that stretch for 90 degrees (1/4 area) - see below.



Ok so now we can break it down into three parts.

I have 3 quarters of the radius 5 circle = 0.75 x π x (5x5) = 18.75π

1 quarter of the radius 2 circle = 0.25 x π x (2x2) = π

1 quarter of the radius 1 circle = 0.25 x π x (1x1) = 0.25π

Therefore in total we have 20Ï€.

[GMATGuruNY]

As the cow moves around the shed, the rope tethering the cow will WRAP AROUND the shed, as shown in the figure above.
As the cow moves COUNTERCLOCKWISE, the 5-unit rope will wrap 4 UNITS TO THE LEFT (along the top of the shed) and 1 UNIT DOWN (along the left side of the shed).
As the cow moves CLOCKWISE, the 5-unit rope will wrap 3 UNITS DOWN (along the right side of the shed) and 2 UNITS TO THE LEFT (along the base of the shed).
The result is that the cow can access the 3 shaded regions.

Blue region:
This region is equal to 3/4 of a circle with a center at A and a radius of 5.
Area = (3/4)(π)(5²) = (75/4)π.

Yellow region:
This region is equal to 1/4 of a circle with a center at B and a radius of 2.
Area = (1/4)(π)(2²) = π.

Red region:
This region is equal to 1/4 of a circle with a center at C and a radius of 1.
Area = (1/4)(π)(1²) = (1/4)π.

Total accessible area = blue region + red region + yellow region = (75/4)π + (1/4)π + π = (76/4)π + π = 20π.

The correct answer is E.