Grain is heaped on a threshing floor in the shape of a right circular cone of base radius 12 m and vertical height 5 m. What area of canvas, in square metres, is needed to cover the heap exactly? (Use π = 3.14.)

Canvas covers only the sloping surface, so first find the slant height: $l$ = √$(12^2+5^2)=$ √169 = 13 m. Curved surface = π$rl=3.14\times 12\times$ 13 = 489.84 sq m. The trap 188.4 uses πrh with the vertical height 5 m in place of the slant height 13 m; 942 adds the base circle 452.16 sq m, but a heap standing on the floor needs no canvas underneath; 753.6 is the heap's VOLUME (1/3)π$r^2$h, a cubic quantity offered where an area was asked.