area and volume

(a) Add the small right circular cone VAB to a frustum ACDB. Then the big right circular cone VCD has height 3+4=7 cm.

So, the ratio of the curved surface area of the small circular cone to that of the big circular cone is 3^2 : 7^2 = 9 : 49.

Now the curved surface area of frustum ACDB is equal to curved surface area of big circular cone - curved surface area of small circular cone. So the ratio of curved surface area of the small circular cone to that of the frustum is 9 : (49-9) = 9 : 40

(b) 2 : ( cans of green paint for frustum) = 9 : 40
40 * 2 = 9(cans of green paint for frustum)
cans of green paint for frustum = 80/9

sorry about that...where is your graph