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

GIVEN :-

• Diameter of cylinder , hemisphere and cone = 4.2 cm
• Height of the cylindrical portion , $\sf h_1$ = 12 cm
• Height of the conical portions , $\sf h_2$ = 7 cm

TO FIND :-

• Volume of the solid toy

SOLUTION :-

   Figure ,

                 

               $\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\multiput(0,0)(2.5,0){2}{\line(0,1){3}}\qbezier(0,0)(1,1)(2.5,0)\qbezier(0,0)(1,-1)(2.5,0)\qbezier(0,0)(1.2,-3)(2.5,0)\qbezier(0,3)(1,4)(2.5,3)\qbezier(0,3)(1,2)(2.5,3)\qbezier(0,3)(0,3)(1.25,6)\qbezier(2.5,3)(2.5,3)(1.25,6)\multiput(0,3)(0.2,0){12}{\line(1,0){0.1}}\multiput(1.25,0)(0,-0.2){8}{\line(0,-1){0.1}}\put(0,0){\line(1,0){2.5}}\put(1.25,0){\line(0,1){6}}\put(3,-0.5){\vector(0,-1){1}}\put(3,-1.5){\vector(0,1){1.5}}\put(3.1,-0.8){ \bf 2.1\ cm }\put(3,1.5){\vector(0,-1){1.5}}\put(2.7,1.6){ \bf 12\ cm }\put(3,2){\vector(0,1){1}}\put(3,4){\vector(0,-1){1}}\put(2.7,4.1){ \bf 7\ cm }\put(3,4.5){\vector(0,1){1.5}}\end{picture}$

                           

   Formulas to be used ,

        $\longrightarrow\sf Volume \ of \ cone = \dfrac{1}{3} \pi r^2 h\\\\\longrightarrow Volume \ of \ cylinder = \pi r^2h\\\\\longrightarrow Volume \ of \ hemisphere = \dfrac{2}{3} \pi r^3$

 → Radius of cylinder , cone and hemisphere = 4.2/2 = 2.1 cm

    We know that ,

    Volume of the solid toy = volume of the conical portion + volume of

                                             cylindrical portion + volume of the hemispherical

                                              portion

       $\hookrightarrow \sf Volume \ of \ cone = \dfrac{1}{3}\pi r^2 h_2+\pi r^2 h_1+\dfrac{2}{3}\pi r^3$

                                   $=\sf \dfrac{1}{3} \pi (2.1)^2 \times 7 + \pi (2.1)^2 \times 12 +\dfrac{2}{3} \pi (2.1)^3\\\\= \dfrac{1}{3} \pi \times (2.1)^2 [7+3\times12 + 2 \times 2.1 ]\\\\=\dfrac{1}{3} \pi \times (2.1)^2 [7+36+4.2]\\\\= \dfrac{1}{3} \pi \times(2.1)^2 \times 47.2\\\\= \dfrac{1}{3}\times \dfrac{22}{7} \times 2.1 \times 2.1 \times 47.2\\\\=22 \times 0.3 \times 0.7 \times 47.2 \\\\= \bf 218.064 \ cm^3$

                                                   

Answer 2

•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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