Question

a solid is in the form of a right circular cylinder with a hemisphere shape at one end and a cone at the other end. their common diameter is 4.2 cm and heights of the cylindrical and conical portion are 12 cm and 7cm. respectively find the volume of the solid toy (π = 22/7)​

Answer 1

•SOLUTION:-

$\bf •The \: shape \: of \: the \: toy \: is \: attached \: above⇭$

$\bf➢ Radius \: of \: the \: hemisphere =2.1cm$

$\bf \: ➢ Radius \: of \: cylinder =2.1cm$

$\bf ➢Radius \: of \: the \: base \: of \: the \: cone =2.1 cm$

$\bf➢Height \: of \: the \: cylinder(H)=12cm$

$\bf➢Height \: of \: the \: cone(h)=7cm$

⇨Volume of the given toy =(volume of the hemisphere + volume of the cylinder + volume of the cone)

$\bf•Volume \: of \: hemisphere = \frac{2}{3} \pi r³$

$\bf•Volume \: of \: cylinder= \pi r²H$

$\bf • Volume \: of \: cone= \frac{1}{3} \pi r²h$

$\bf➦Now ,Volume \: of \: the \: toy=$

$\bf \: \: \: \: \: \: \: \: \: \: = \huge( \small \frac{2}{3} \pi r {}^{3} + \pi r {}^{2} H+ \frac{1}{3} \pi r²h \huge) \small cm {}^{3}$

$\: \: \: \: \: \: \: \: \: \: \bf = \huge[ \small \frac{2}{3} \pi×(2.1)³+ \pi×(2.1)²w×12+ \frac{1}{3} \pi×(2.1)² \huge] \small cm³$

$\: \: \: \: \: \: \: \: \: \: \bf = \frac{1}{3} \pi \times (2.1) {}^{2} \big [ \small2×(2.1)+3×12+7 \big ] \small \: cm {}^{3}$

$\: \: \: \: \: \: \: \: \: \: \bf = \huge[ \small \frac{1}{3} \pi×(2.1)²×47.2 \huge] \small \: cm³$

$\: \: \: \: \: \: \: \: \: \: = \bf \huge( \small \frac{1}{ \cancel{3}} × \frac{22}{ \cancel{7}} × \frac{ {\cancel{21 }\: \:\cancel{3 }}}{10} \times \frac{21}{10} \times \frac{472}{10} \huge) \small \: cm {}^{3}$

$\bf \: \: \: \: \: \: \: \: \: \: = 218.064 \: cm {}^{3}$

➠Hence,the volume of the given toy is 218.064 cm³

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Answer 2

Given :-

• Common diameter of solid shapes is 4.2cm
• Height of cylinderical portion(h)=12
• Height of conical portion(H) is 7cm

To Find:-

• We have to find the volume of toy

Solution :-

$\sf \Big[\purple{Volume\ of \ solid \ toy = V_{cylinder}+V_{cone}+V_{Hemisphere}}\Big]$

$\underline{\sf\ \ Volume s\ \ }\begin{cases}\sf{Cylinder= \pi r^2 h}\\\sf{Cone= \dfrac{1}{3}\pi r^2 h}\\\sf{Hemisphere= \dfrac{2}{3}\pi r^3}\end{cases}$

$:\implies\sf\ Volume_{toy}= \pi r^2 h+\dfrac{1}{3}\pi r^2 H+\dfrac{2}{3}\pi r^3\\ \\ \\ \\ :\implies\sf\ Volume_{toy}= \pi r^2\Bigg\lgroup\ h+ \dfrac{1}{3}H+ \dfrac{2}{3}r\Bigg\rgroup\\ \\ \\ \therefore\sf\ H= 7cm\ ;\ \ h= 12cm\ \ ;\ \ r= 4.2/2=2.1cn\\ \\ \\ \\ :\implies\sf\ Volume_{toy}= \pi r^2 \Bigg\lgroup\ 12+ \dfrac{1}{3}\times 7 + \dfrac{2}{3}\times 2.1\Bigg\rgroup\\ \\ \\ \\ :\implies\sf\ Volume_{toy}= \pi r^2 \bigg\lgroup \ \dfrac{36+7+4.2}{3}\bigg\rgroup\\ \\ \\ \\ :\implies\sf\ Volume_{toy}= \pi r^2 \bigg\lgroup \dfrac{47.2}{3}\bigg\rgroup\\ \\ \\ \\ :\implies\sf\ Volume_{toy}= \dfrac{22}{\cancel{7}}\times \cancel{2.1}\times \cancel{2.1}\times \dfrac{47.2}{\cancel{3}}\\ \\ \\ \\ :\implies\sf\ Volume_{toy}= 22\times 0.3\times 0.7\times 47.2\\ \\ \\ \\ :\implies\underline{\boxed{\red{\sf\ Volume_{toy}= 218.064cm^3}}}$