A tent is made in the form of a frustum A of a right circular cone surmounted by another right circular cone B. The diameter of the ends of the frustum A are 8 m and 4 m, its height is 3 m and the height of the cone B is 2 m. What is the area of the canvas required for the tent?

85.77 m²

Slant height of the frustum

$=\sqrt{(2-4)^2+3^2}=\sqrt{13}$

slant height of cone

$=\sqrt{2^2+2^2}=2\sqrt{2}$

Tent canvas required = CSA of frustum + CSA of cone

$=π×\sqrt{13}(4+2)+π×2\sqrt{2}×2=85.77m^2$