Question

From a solid cylinder whose height is $2.4cm$ and diameter $1.4cm$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $c{m}^2$.

Answer 1

Step 1: Solve for slant height of cone

Given,

1. Diameter of cylinder $=$ Diameter of cone i.e. $2r=1.4cm$
2. Height of cylinder $=$ Height of cone i.e. $h=2.4cm$

Let $l$ be the slant height of the cone.

From the figure,

$l^2=h^2+r^2$

$$\begin{gathered} ={\left(2.4\right)}^2+{\left(0.7\right)}^2 \\ \Rightarrow l=\sqrt{5.76+0.49} \\ =\sqrt{6.25} \\ \Rightarrow l=2.5cm \end{gathered}$$

Step 2: Solve for curved surface area of cone

Curved surface area of cone$=\mathrm{\pi r}l$

$$\begin{gathered} =\frac{22}{7}\times 0.7\times 2.5 \\ =5.5{\mathrm{cm}}^2 \end{gathered}$$

Step 3: Solve for curved surface area of cylinder

Curved surface area of cylinder$=2\mathrm{\pi rh}$

$$\begin{gathered} =2\times \frac{22}{7}\times 0.7\times 2.4 \\ =10.56{\mathrm{cm}}^2 \end{gathered}$$

Step 4: Solve for area of circular base of cylinder

Area of circular base$={\mathrm{\pi r}}^2$

$=\frac{22}{7}\times {\left(0.7\right)}^2=1.54{\mathrm{cm}}^2$

Step 5: Solve for Total surface area of remaining solid

Total surface area of the remaining solid $=$Curved surface area of cylinder $+$ area of circular base of cylinder $+$ curved surface area of cone

$$\begin{gathered} =10.56+1.54+5.5 \\ =17.6c{m}^2 \\ \approx 18c{m}^2 \end{gathered}$$

Hence, total surface area of remaining solid is $18c{m}^2$