Question

An open metallic bucket is in the shape of a frustum of a cone mounted on hollow cylindrical base made of metallic sheet. If the diameters of the two circular ends of the bucket are $45cm$ and $25cm$, the total vertical height of the bucket is $30cm$ and that of the cylindrical portion is $6cm$, find the area of the metallic sheet used to make the bucket. Also, find the volume of the water it can hold. (Take $\pi =\frac{22}{7}$)

Answer 1

We have,

Height of the frustum of the cone$h=(30-6)=24cm$

Radii of the circular ends are $r_1=225$ and $r_2=12.5cm$

$l=$Slant height of the frustum

$l=\sqrt{h^2+(r_1-r_2{)}^2}$

$l=\sqrt{{24}^2+(22.5-12.5{)}^2}=26cm$

Area of metallic sheet used=Curved surface area of the frustum of cone+area of circular base+curved surface area of cylinder

$=\pi (r_1-r_2)l+\pi r_2^2+2\pi r_2{h}_2$

$=\pi \left[\right(22.5-12.5)\times 26+{12.5}^2+2\times 12.5\times 6]$

$=\frac{22}{7}\times 1216.25=3822.5c{m}^2$

Volume of water that the bucket can hold

$=\frac{1}{3}\pi (r_1^2+r_2^2+r_1{r}_2)h$

$=\frac{1}{3}\times \frac{22}{7}\times ({22.5}^2+{12.5}^2+22.5\times 12.5)\times 24$

$=23719.02c{m}^3$

$=23.719liters$