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 $\pi =\frac{22}{7}$]

Answer 1

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

Given
Height (b) 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
$\therefore$ Radius(r) of the cylindrical part = Radius (R) of the conical part = 2.1 cm
Slant height (l) of the conical part $=\sqrt{r^2+h^2}$
$=\sqrt{(2.1{)}^2+(2.8{)}^2}cm$
$=\sqrt{4.41+7.81}cm$
$=\sqrt{12.25}cm$
$=3.5cm$

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

= 2πrh +πrl +π $r^2$

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

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

=73.92 $c{m}^2$

Thus, the total surface area of the remaining solid is 73.92 $c{m}^2$