A circular grass plot 12 feet in diameter is cut by a straight gravel path 3 feet wide, one edge of which passes through the center of the plot. The number of square feet in the remaining grassy area is:

- 36 – 34
- 30 – 15
- 36 – 33
- 35 – 9√3
- 30 – 9√3

Area I = (6^{2})/2 = 18.

Area II = Sector AOB – △AOB.

We see that △AOC is a 30°-60° right △ since OC = 3 and

OA = 6. Thus ∠AOC = 60° and ∠AOB = 120°.

So the area of the sector = 120/360 · ((6^{2})) = 36/3 = 12.

△AOB has base AB = 6√3 and altitude OC = 3. Its area is

(6√3 · 3)/2 = 9√3.

So the grassy area = I + II = 18 + (12 – 9√3) = 30 – 9√3. The answer is (e).