Question

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

Answer 1

GIVEN:-

•Radius of the cylinder=5cm

•Height of the cylinder =12 cm

TO FIND OUT:--

• The volume and the surface area of the remaining solid

SOLUTION :-

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

$Now ,</p><p> \\ </p><p>\begin{gathered}\bf volume \: _{cylinder} \: = {\bigg (} \frac{22}{7} \times 5 \times 5 \times 12{ \bigg)cm {}^{3} } \\ \\ \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf \frac{6600}{7} cm {}^{3}\end{gathered}volumecylinder=(722×5×5×12)cm3=76600cm3</p><p></p><p></p><p>$

$volume \: cone=(31×722×5×5×12) \: cm3 \\ =72200cm3</p><p></p><p></p><p>$

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

$\begin{gathered}\bf \implies { \bigg(} \frac{6600}{7} - \frac{2200}{7}{ \bigg)} cm {}^{3} = \frac{4400}{7} cm {}^{3} = 628.57 \: cm {}^{3} \\ \\\end{gathered} </p><p>⟹( </p><p>7</p><p>6600</p><p> </p><p> − </p><p>7</p><p>2200</p><p> </p><p> )cm </p><p>3</p><p> = </p><p>7</p><p>4400</p><p> </p><p> cm </p><p>3</p><p> =628.57cm </p><p>3</p><p> </p><p>$

Slant height of the cone(l)

$\begin{gathered}\bf \: l = \sqrt{r {}^{2} + h {}^{2} } \\\end{gathered} </p><p>l= </p><p>r </p><p>2</p><p> +h </p><p>2</p><p> </p><p> </p><p> </p><p> </p><p> </p><p></p><p>\begin{gathered}\bf \implies l = \sqrt{5 {}^{2} + (12) {}^{2} } = \sqrt{169} = 13cm \\ \\\end{gathered} </p><p>⟹l= </p><p>5 </p><p>2</p><p> +(12) </p><p>2</p><p> </p><p> </p><p> = </p><p>169</p><p> </p><p> =13cm</p><p> </p><p> </p><p></p><p>\begin{gathered}\textsf{CSA} \bf_{cone}={\bigg (} \frac{22}{7} \times 5 \times 13 { \bigg)}cm² = \frac{1430}{7} cm² \\\end{gathered} </p><p>CSA </p><p>cone</p><p> </p><p> =( </p><p>7</p><p>22</p><p> </p><p> ×5×13)cm²= </p><p>7</p><p>1430</p><p> </p><p> cm²</p><p> </p><p> </p><p></p><p>\begin{gathered}\textsf{CSA} \bf_{cylinder}= {\bigg (} 2 \times \frac{22}{7} \times 5 \times 12{ \bigg)}= \frac{2640}{7} cm² \\ \\\end{gathered} </p><p>CSA </p><p>cylinder</p><p> </p><p> =(2× </p><p>7</p><p>22</p><p> </p><p> ×5×12)= </p><p>7</p><p>2640</p><p> </p><p> cm²</p><p> </p><p> </p><p></p><p></p><p> </p><p></p><p></p><p>$

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

$\begin{gathered}= \bf \pi r {}^{2} sq. \: unit \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \bf{ \bigg(} \frac{22}{7} \times 5 \times 5 { \bigg)}cm {}^{2} = \frac{550}{7} cm {}^{2}\end{gathered} </p><p>=πr </p><p>2</p><p> sq.unit</p><p> =( </p><p>7</p><p>22</p><p> </p><p> ×5×5)cm </p><p>2</p><p> = </p><p>7</p><p>550</p><p> </p><p> cm </p><p>2$

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

$\begin{gathered}\implies\bf { \bigg(} \frac{2640}{7} + \frac{1430}{7} + \frac{550}{7} { \bigg)}cm {}^{2} = 660cm {}^{2} \\\end{gathered} </p><p>⟹( </p><p>7</p><p>2640</p><p> </p><p> + </p><p>7</p><p>1430</p><p> </p><p> + </p><p>7</p><p>550</p><p> </p><p> )cm </p><p>2</p><p> =660cm </p><p>2</p><p> </p><p>$

Answer 2

GIVEN:-

•Radius of the cylinder=5cm

•Height of the cylinder =12 cm

TO FIND OUT:--

• The volume and the surface area of the remaining solid

SOLUTION :-

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

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

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