A wooden toy rocket is in the shape of a cone | vedclass

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Question

(C) Let the radius of the cone be $r$, slant height of the cone be $l$, height of the cone be $h$, radius of the cylinder be $r^{\prime}$, and height

Vedclass · Accepted Answer

$195.465$

Vedclass · Answer

(C) Let the radius of the cone be $r$, slant height of the cone be $l$, height of the cone be $h$, radius of the cylinder be $r^{\prime}$, and height of the cylinder be $h^{\prime}$.Given: $r = 2.5\, cm$, $h = 6\, cm$, $r^{\prime} = 1.5\, cm$, and $h^{\prime} = 26 - 6 = 20\, cm$.The slant height $l = \sqrt{r^2 + h^2} = \sqrt{2.5^2 + 6^2} = \sqrt{6.25 + 36} = \sqrt{42.25} = 6.5\, cm$.The conical portion has its circular base resting on the base of the cylinder. Since the base of the cone is larger than the base of the cylinder, a ring-shaped area at the base of the cone is also to be painted orange.Area to be painted orange = ($CSA$ of the cone) + (Base area of the cone - Base area of the cylinder)$= \pi r l + (\pi r^2 - \pi (r^{\prime})^2)$$= 3.14 	imes 2.5 	imes 6.5 + 3.14 	imes (2.5^2 - 1.5^2)$$= 51.025 + 3.14 	imes (6.25 - 2.25)$$= 51.025 + 3.14 	imes 4 = 51.025 + 12.56 = 63.585\, cm^2$.Area to be painted yellow = ($CSA$ of the cylinder) + (Area of the bottom base of the cylinder)$= 2 \pi r^{\prime} h^{\prime} + \pi (r^{\prime})^2$$= \pi r^{\prime} (2h^{\prime} + r^{\prime})$$= 3.14 	imes 1.5 	imes (2 	imes 20 + 1.5)$$= 4.71 	imes 41.5 = 195.465\, cm^2$.Thus, the area painted orange is $63.585\, cm^2$ and the area painted yellow is $195.465\, cm^2$.

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