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# A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in figure. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take $\pi = 3.14$)

Last updated date: 12th Aug 2024
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Hint: We can identify the value of various dimensions of both the cone and cylindrical figures constituting the toy rocket. Then applying formulas for the areas of respective figures, we can find the required areas to be painted by orange and yellow colors respectively.
Formulas to be used:
$l = \sqrt {{r^2} + {h^2}}$ where, l, r and h are the slant height, radius and height of the cone respectively.
Curved surface area of cone = $\pi rl$
Curved surface area of cylinder = $2\pi rh$
Area of circle = $\pi {r^2}$

We have been given:
Height of entire rocket = 26 cm
Height of conical part = 6 cm
Diameter of conical part = 5 cm
ğŸ¡ª Radius of conical part = $\dfrac{5}{2} = 2.5cm$
(As diameter is equal to twice the value of radius)
Diameter of cylindrical part = 3 cm
Radius of cylindrical part = $\dfrac{3}{2} = 1.5cm$
(As diameter is equal to twice the value of radius)
Slant height (l) of the cone can be calculated as:
$l = \sqrt {{r^2} + {h^2}}$ , here,
Height (h) = 6 cm
$\Rightarrow l = \sqrt {{{\left( {2.5} \right)}^2} + {6^2}} \\ \Rightarrow l = \sqrt {6.25 + 36} \\ \Rightarrow l = \sqrt {42.25} \\ \therefore l = 6.5 cm \;$
Now, we have to paint conical portion as orange and the cylindrical portion as yellow:

Area to be painted as orange:
Curved surface area of cone + Base area of cone â€“ Base area of cylinder _____ (1)
Area to be painted as yellow:
Curved surface area of cylinder + One base area of cylinder _______ (2)
Finding the respective values:
i) Curved surface area of cone (CSA) = $\pi rl$
Slant height (l) = 6.5 cm
$\Rightarrow CSA = 3.14 \times 2.5 \times 6.5\left( {\because \pi = 3.14} \right) \\ \therefore CSA = 51.025\;c{m^2} \\$
ii) Base area of cone = Area of circle (as base is circular in shape)
Base area of cone (B.C)= $\pi {r^2}$
$\Rightarrow BC = \pi {\left( {2.5} \right)^2} \\ \Rightarrow BC = 3.14 \times 6.25\left( {\because \pi = 3.14} \right) \\ \therefore BC = 19.625\;c{m^2} \\$
iii) Curved surface area of cylinder (CSA) = $2\pi rh$
Height (h) = height of entire rocket â€“ height of circular portion
$\Rightarrow h = 26 - 20 \\ \therefore h = 20cm \\$
$\Rightarrow CSA = 2 \times 3.14 \times 1.5 \times 20 \\ \therefore CSA = 188.4 c{m^2} \\$
ii) Base area of cylinder = Area of circle (as base is circular in shape)
Base area of cone (B.C)= $\pi {r^2}$
$\Rightarrow BC = \pi {\left( {1.5} \right)^2} \\ \Rightarrow BC = 3.14 \times 2.25\left( {\because \pi = 3.14} \right) \\ \therefore BC = 7.065\; c{m^2} \;$
$\Rightarrow \left( {51.025 + 19.625 - 7.0625} \right)c{m^2} \\ \Rightarrow \left( {70.65 - 7.0625} \right)c{m^2} \\ \therefore 63.58\; c{m^2} \;$
$\Rightarrow \left( {188.4 + 7.065} \right)c{m^2} \\ \therefore 195.465\; c{m^2} \;$
Therefore, the area of the rocket painted with each orange and yellow colours are $63.58\;c{m^2}$ and $195.465\;c{m^2}$ respectively.
Note: As all the quantities were given in cm, we obtained the respective areas in $c{m^2}$. In such questions we carefully observe which areas are required to be added and subtracted respectively to get the required area. In the curved surface areas of the bodies, we do not include their bases, it's like cutting the figure parallel to the base.