Question

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 conical heap of sand is formed, if the height of the conical heap is 24 cm, find the radius and slant height of the heap.

Answer 1

Given,

Radius of cylinder $\left(R\right)=18cm$

Height of cylinder $\left(H\right)=32cm$

Height of cone $\left(h\right)=24cm$

To find out,

Radius and slant height of the cone.

We know that, volume of a cylinder $=\pi r^{2}h$

$=\pi \times {\left(18\right)}^2\times 32$

We also know that, volume of a cone $=\frac{1}{3}\pi r^{2}h$

$=\frac{1}{3}\times \pi \times r^2\times 24$

Since sand of bucket is emptied on the ground and a conical heap of sand is formed.

Hence, volume of cone $=$ Volume of cylinder

$\pi \times {\left(18\right)}^2\times 32=\frac{1}{3}\times \pi \times r^2\times 24$

$\Rightarrow r^2=\frac{(18{)}^2\times 32\times 3}{24}$

$\Rightarrow r^2=(18{)}^2\times 4$

$\Rightarrow r^2=(18{)}^2\times (2{)}^2$

$\Rightarrow r^2=(18\times 2{)}^2$

$\Rightarrow r^2=(36{)}^2$

$\therefore r=36cm$

We know that, slant height of a cone $l=\sqrt{r^2+h^2}$

$\Rightarrow l=\sqrt{{\left(24\right)}^2+{\left(36\right)}^2}$

$=\sqrt{(12\times 12\times 2\times 2)+(12\times 12\times 3\times 3)}$

$=\sqrt{(12\times 12)(4+9)}$

$=12\sqrt{13}cm$

Hence, the radius and slant height of the conical heap is $36cm$ and $12\sqrt{13}cm$ respectively.