Question

a solid toy is in the form of a right circular cylinder with a hemispherical shape at one end and a cone at other and their diameter is 4.2 the height of a cylinderical and conical porons. are 24 cm and 14 cm respectively find the volume of toy

Answer 1

Diameter = 4.2cm

Radius = 4.2/2 = 2.1cm

Height of CYLINDRICAL portion = 24cm

Height of Conical portion = 14cm

height of hemispherical portion = 2.1cm

VOLUME OF TOY = VOLUME OF CYLINDER PORTION+ VOLUME OF CONICAL PORTION + VOLUME OF HEMISPHERICAL PORTION.

= πR²H + 1/3πR³+ 4/3πR³.

PUT ALL VALUES AND SOLVE..U WILL GET THE VOLUME OF TOY

Answer 2

$\bf\huge\color{blue}\mathcal{HEYA\:\:MATE}\mathbb{\::)}\\\bf\large\color{green}{Here\:is\:the\:answer}$
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$\bf{Given}\:-\:radius \: of \: cone \: \\ = \bf{radius \: of \: cylinder }\: \\ = \bf\: radius \: of \: hemisphere \: \\ = \: \frac{4.2}{2} = 2.1 \: cm$
$=> height \: of \: cylinderical \: portion \\ (h) \: = 24 \: cm \\ \\=> height \: of \: conical \: portion \: \\ (H) \: = 14 \: cm$
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$\\\bf{To \: find\:- \: \: volume \: of \: the \: toy}$
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$Solution\:- \: volume \: of \: toy \: = \\ volume \: of \: \\ (cylinder + cone + hemisphere)$
$= > \pi {r}^{2} h \: + \: \frac{1}{3} \pi {r}^{2} H \: + \frac{2}{3}\pi {r}^{3}$

$= > \: \pi {r}^{2}(h \: + \: \frac{1}{3}H \: + \: \frac{2}{3}r)$
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Just plug-in the values to get the answer
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Hope this helps ❤️❤️❤️❤️❤️