Question

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, 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 cm². [Use \(\pi=\frac{22}{ 7}\)]

Answer 1

Given that, 
Height (h) of the conical part = Height (h) of the cylindrical part = 2.4 cm 
The diameter of the cylindrical part = 1.4 cm 
Therefore, the radius (r) of the cylindrical part = 0.7 cm 

Slant height \((l)\)of conical part = \(\sqrt{r^2+h^2}\)
\(=\sqrt{(0.7)^2+(2.4)^2}\)
\(=\sqrt{0.49+5.76}\)
\(=\sqrt{6.25}\)
\(=2.5\)

Total surface area of the remaining solid will be = surface area of conical cavity + TSA of the cylinder
\(= \pi rl+(2\pi rh+\pi r^2)\)
\(= \pi r(l+2h+r)\)
\(= (\frac{22}{7})× 0.7\times(2.5+4.8+0.7)\)
\(= 2.2×8 = 17.6\) cm²

The total surface area of the remaining solid to the nearest cm² is 18 cm²