Assuming the centre of the road passes through the centre of the circle, the sides of the road will form chords on either side of the centre 15' from the centre, thus bisecting the radii perpendicular to the sides of the road.
The angle of the sector AOB is 120º. The area of the sector, A, is given by A/πr^2=120/360=1/3 where r=30', the radius of the circle, OA, OB. So A=(1/3)*900π=300π sq ft. The height of each right-angled triangle forming the triangle AOB is given by Pythagoras: √(30^2-15^2)=15√3. The area of the right-angled triangle is (1/2)15*15√3=112.5√3, so the total area of the 4 such triangles=450√3 sq ft. The area of the two sectors=600π and the total area to the sides of the road = 600π-450√3=1105.53 sq ft (552.77 sq ft on each side).