Question

Form a solid cylinder whose height is 2.4cm ,diameter is 1.4cm a conical cavity of same height and diameter is hollowed out find the total surface area of the remaining solid to. The nearest cm2

Answer 1

total surface area=(pi*r*l)+(2*pi*r*h)+(pi*r*r)

l=sqroot(r^2+h^2)

Answer 2

For the cylindrical part :

$\sf{r = \frac{1.4}{2} cm = 0.7cm}$

$\sf{h=2.4cm}$

For the conical part:

$\sf{r=0.7cm}$

$\sf{h=2.4cm}$

Slant Height :

$\sf{l = \sqrt{ {r}^{2} + {h}^{2} }}$

$\sf{\sqrt{(0.7)}^{2} + {(2.4)}^{2}cm}$

$\sf{2.5cm}$

Total surface area of the remaining solid

= CSA of the cylinder + Top area of the cylinder + CSA of the cone

$\longrightarrow$$\sf{2\pi \: rh + \pi \: {r}^{2} + \pi \: rl = \pi \: r(2h + r + l)}$

$\longrightarrow$$\sf{\frac{22}{7} \times 0.7(4.8 + 0.7 + 2.5) {cm}^{2}}$

$\longrightarrow$$\sf{\frac{22}{7} \times 0.7 \times 8.0 { cm }^{2}}$

$\longrightarrow$${\boxed{\sf{\red{{17.6cm}^{2}}}}}$