Given,
The diameter of thecylinder (also the same for cone) = 24 m.
So, its radius (R)= 24/2 = 12 m
The height of theCylindrical part (H1) = 11m
So, Height of the conepart (H2) = 16 – 11 = 5 m
Now,
Vertex of the coneabove the ground = 11 + 5 = 16 m
Curved Surface area ofthe Cone (S1) = πRL = 22/7 × 12 × L
The slant height (L)is given by,
L = √(R2 +H22) = √(122 + 52) = √169
L = 13 m
So,
Curved Surface Area ofCone (S1) = 22/7 × 12 × 13
And,
Curved Surface Area ofCylinder (S2) = 2πRH1
S2 =2π(12)(11) m2
Thus, the area ofCanvas required for tent
S = S1 +S2 = (22/7 × 12 × 13) + (2 × 22/7 × 12 × 11)
S = 490 + 829.38
S = 1319.8 m2
S = 1320 m2
Therefore, the area ofcanvas required for the tent is 1320 m2