Answer: (d) = 340π m2
Area of canvas required = Lateral surface area of the conic frustum + Curved surface area of the cone.
Height of conical part = 28 m – 24 m = 4 m
Slant height (l1) of the conical part = \(\sqrt{4^2+3^2}\) = \(\sqrt{25}\) = 5m.
Slant height (l2) of the frustum
= \(\sqrt{h^2+(R-r)^2}\) = \(\sqrt{24^2+(10-3)^2}\)
= \(\sqrt{576+49} = \sqrt{625}\) = 25 m.
∴ Area of canvas reqd = π(R + r)l2 + πrl1
= π× 13 × 25 + π × 3 × 5
= (325π + 15π) m2
= 340π m2.