Question

From a solid cylinder of height 7n 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. [Use $=\frac{22}{7}$]
OR
A cylindrical bucket, 32 cm high and with 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 conical heap is 24 cm, then find the radius and slant height of the heap.

Answer 1

It is given that, height (h) of cylindrical part = height (h) of the conical part = 7 cm
Diameter of the cylindrical part = 12 cm
$\therefore$ Radius (r) of the cylindrical part =$\frac{12}{2}cm=6cm$
$\therefore$ Radius of conical part = 6cm
Slant height (I) of conical part =$\sqrt{r^2+h^2}cm$
$=\sqrt{6^2+7^2}cm=\sqrt{36+49}cm=\sqrt{85}cm$
= 9.22 cm (approx.)
Total surface area of the remaining solid

= CSA of cylindrical part + CSA of conical part + Base area of the circular part
$=2\pi rh+\pi rl+\pi r^2=2\times \frac{22}{7}\times 6\times 7c{m}^2+\frac{22}{7}\times 6\times 9.22c{m}^2+\frac{22}{7}\times 6\times 6c{m}^2=264c{m}^2+173.86c{m}^2+113.14c{m}^2=551c{m}^2$
OR

Height $\left(h_1\right)$ of cylindrical bucket = 32 cm
Radius $\left(r_1\right)$ of circular end of bucket = 18 cm
Height $\left(h_2\right)$ of conical heap = 24 cm
Let the radius of the circular end of conical heap be r₂.
The volume of sand in the cylindrical bucket will be equal to the volume of sand in the conical heap.
Volume of sand in the cylindrical bucket = Volume of sand in conical heap
$\pi \times r_1^2\times h_1=\frac{1}{3}\pi \times r_2^2\times h_2\pi \times {18}^2\times 32\pi \frac{1}{3}\times 24\pi \times {18}^2\times 32=\frac{1}{3}\pi \times r_2^2\times 21r_2^2=\frac{3\times {18}^2\times 32}{24}={18}^2\times 4$
$r_2=18\times 2=36cm$
slant height =$\sqrt{{36}^2+{24}^2}{12}^2+(3^2+2^2)=12\sqrt{13}cm$
therefore, the radius and slant height of the conical heap are 36 cm and $12\sqrt{13}$ respectively.