From a solid cylinder of height 7 cm and base diameter 12 cm, a conical cavity of same height and same base diameter is hollowed out. Find the total surface area of the remaining solid.

Or

A cylindrical bucket, 32 cm high and with the radius of base 18 cm, is filled with sand. This bucket is emptied on the ground and a conical heap of sand is formed. If the height of the concial heap is 24 cm, then find the radius and slant height of the heap.

We have

We can see that, total surface area of the remaining solid in the figure can be obtained by adding:

(a). The curved surface area of the cylinder.

(b). The curved surface area of the hollowed-out cone.

(c). The area of solid top of the cylinder.

Let Total Surface Area be given by TSA, Curved Surface Area as CSA.

Let radius of cylinder as well as cone = r and height of cylinder as well as cone = h

We need to find slant height (l) =

So, the formula of the total surface area of the remaining solid can be given as

TSA of remaining solid = CSA of the cylinder + CSA of the cone + Area of circular top

⇒ TSA of remaining solid = 2πrh + πrl + πr²

⇒ TSA of remaining solid = πr(2h + l + r)

⇒ TSA of remaining solid = [∵, √85 = 9.219]

⇒ TSA of remaining solid =

Hence, total surface area of remaining solid is 551.01 cm².

Or

Let radius of the conical heap be x and radius of cylindrical bucket be r

Let height of the conical heap be h^’ and height of cylindrical bucket be h

Since sand volume is going to be equal to volume of the cylindrical bucket before it adjusts itself to conical shape, we can write

Volume of conical heap = Volume of cylindrical bucket

…(i)

[∵, Volume of cone = and Volume of cylinder = πr²h]

Given are: r = 18 cm, h = 32 cm and h^’ = 24 cm

Substituting r and h in the equation (i), we get

⇒ x² = 18 × 18 × 2 × 2

⇒ x = 18 × 2 = 36

Thus, radius of the conical heap is 36 cm.

∴, Slant height of heap is given by

⇒

⇒ l=√(1296+576)=√1872=√(12×12×13)=12√13

⇒ l = 43.27 cm

Thus, slant height of the conical heap is 43.27 cm.