Question

From a solid cylinder of height 2.8 cm and diameter 4.2 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid [take π=22/7]

Answer 1

 

The following figure shows the required cylinder and the conical cavity.

Given:
Height (h) of the conical part = Height (h) of the cylindrical part = 2.8 cm
Diameter of the cylindrical part = Diameter of the conical part = 4.2 cm
∴ Radius (r) of the cylindrical part = Radius (r) of the conical part = 2.1 cm
Slant height (l) of the conical part ${\displaystyle =\sqrt{r^2+h^2}}$

${\displaystyle =\sqrt{{\left(2.1\right)}^2+{\left(2.8\right)}^2}cm}$

${\displaystyle =\sqrt{4.41+7.84}cm}$

${\displaystyle =\sqrt{12.25}cm=3.5cm}$

Total surface area of the remaining solid = Curved surface area of the cylindrical part + Curved surface area of the conical part + Area of the cylindrical base

${\displaystyle =2\pi rh+\pi rl+\pi r^2}$

${\displaystyle =(2\times \frac{22}{7}\times 2.1\times 2.8+\frac{22}{7}\times 2.1\times 3.5+\frac{22}{7}\times 2.1\times 2.1)c{m}^2}$

${\displaystyle =(36.96+23.1+13.86)c{m}^2}$

${\displaystyle =73.92c{m}^2}$

Thus, the total surface area of the remaining solid is 73.92 cm².