Math, asked by madanraisampang, 9 months ago

From a solid cylinder of height 12 cm and the diameter of the base 10cm a conical cavity of the same height and same diameter is hollowed out find the surface area of the remaining solid.​

Answers

Answered by JessicaJha
1

Step-by-step explanation:

the answer is 172.85 cm2

Answered by Anonymous
3

\Large{\underline{\underline{\red{Given:-}}}}

•Radius of the cylinder=5cm

•Height of the cylinder =12 cm

\Large{\underline{\underline{\green{Find:-}}}}

• The volume and the surface area of the remaining solid

\Large{\underline{\underline{\red{solution:-}}}}

 \textsf{formula used}\begin{cases} \bf volume   \: _{cylinder} =  \pi  r {}^{2} h\\ \bf  CSA_{cylinder}  = 2 \pi rh\\ \bf CSA\: _{cone}=  \pi  rl\\ \bf \: volume   \: _{cone} =  \frac{1}{3}  \pi r {}^{2} h</p><p></p><p>\end{cases}

Now ,

 \bf volume   \: _{cylinder}  \:  =   {\bigg (} \frac{22}{7} \times 5 \times 5 \times 12{ \bigg)cm {}^{3} }  \\ \\   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \: =   \bf \frac{6600}{7}  cm {}^{3}

 \bf  \: volume \: _{cone}  \: ={  \bigg(} \frac{1}{3} \times  \frac{22}{7} \times 5 \times 5  \times 12 { \bigg)}cm {}^{3}    \\  \\ \:  \:  \:  \:  \:  \:  \:  \:   =  \bf \frac{2200}{7} cm {}^{3}

•Volume of the remaining solid =(volume of the cylinder) -(volume of the cone)

 \bf \implies { \bigg(} \frac{6600}{7}  -  \frac{2200}{7}{  \bigg)} cm {}^{3}  =  \frac{4400}{7} cm {}^{3}  = 628.57 \: cm {}^{3}  \\  \\

•Slant height of the cone(l)

 \bf \: l =  \sqrt{r {}^{2} + h {}^{2}  }  \\

 \bf  \implies l =  \sqrt{5 {}^{2} + (12) {}^{2}   }   =   \sqrt{169} = 13cm \\   \\

 \textsf{CSA} \bf_{cone}={\bigg (} \frac{22}{7}  \times 5 \times 13 { \bigg)}cm² =  \frac{1430}{7} cm² \\

 \textsf{CSA} \bf_{cylinder}= {\bigg (} 2 \times \frac{22}{7}  \times 5 \times 12{ \bigg)}= \frac{2640}{7} cm² \\  \\

•Now, area of upper circular base of base of cylinder =

  = \bf  \pi r {}^{2}  sq. \: unit \\  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \ \:  \:  \:   \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  =  \bf{ \bigg(} \frac{22}{7}  \times 5 \times 5 { \bigg)}cm {}^{2}  =  \frac{550}{7} cm {}^{2}

•Whole surface area of the remaining solid =CSA(cylinder)+CSA(cone)+area of upper base of cylinder

  \implies\bf { \bigg(} \frac{2640}{7}  +  \frac{1430}{7}  +  \frac{550}{7}  {  \bigg)}cm {}^{2} = 660cm {}^{2}   \\

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