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