The curved surface area of a cylindrical pillar is 264 m^2 and its volume is 924 m^3. The height of the pillar is
Share
The curved surface area of a cylindrical pillar is 264 m^2 and its volume is 924 m^3. The height of the pillar is
Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
[tex]\begin{gathered}\begin{gathered}\bf \:Given-\begin{cases} &\sf{Volume_{(cylinder)} = 924 \: {m}^{3} } \\ &\sf{CSA_{(cylinder)} = 264 \: {m}^{2} } \end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{height_{(cylinder)}}\end{cases}\end{gathered}\end{gathered}[/tex]
[tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered}[/tex]
[tex] \underline{ \boxed{ \bf \: Volume_{(cylinder)} = \: \pi \: {r}^{2}h}} [/tex]
[tex] \underline{ \boxed{ \bf \: CSA_{(cylinder)} = 2\pi \: rh}}[/tex]
where,
[tex]\: \: \: \: \: \: \: \: \: \bull \: \sf \:r \: is \: radius \: of \: cylinder[/tex]
[tex]\: \: \: \: \: \: \: \: \: \bull \: \sf \:h \: is \: height \: of \: cylinder[/tex]
[tex]\large\underline{\sf{Solution-}}[/tex]
[tex]\begin{gathered}\begin{gathered}\bf \:Let-\begin{cases} &\sf{radius \: of \: cylinder \: = \: r \: meter} \\ &\sf{height \: of \: cylinder \: = \: h \: meter} \end{cases}\end{gathered}\end{gathered}[/tex]
Given that,
[tex]\rm :\longmapsto\:Volume_{(cylinder)} = 924[/tex]
[tex]\rm :\longmapsto\:\pi \: {r}^{2}h = 924 - - - (1) [/tex]
and
[tex]\rm :\longmapsto\:CSA_{(cylinder)} = 264[/tex]
[tex]\rm :\longmapsto\:2\pi \: rh = 264 - - - (2)[/tex]
Now, Divide equation (1) by equation (2), we get
[tex]\rm :\longmapsto\:\dfrac{\pi \: {r}^{2}h}{2\pi \: rh} = \dfrac{924}{264} [/tex]
[tex]\bf\implies \:r \: = \: 7 \: m[/tex]
On substituting the value of 'r' in equation (2), we get
[tex]\rm :\longmapsto\:2 \times \dfrac{22}{7} \times 7 \times h = 264[/tex]
[tex]\bf\implies \:h \: = \: 6 \: m[/tex]
[tex]\overbrace{ \underline { \boxed { \bf \therefore \: The \: height_{(cylinder)} \: is \: 6 \: meter}}}[/tex]
Additional Information :-
Perimeter of rectangle = 2(length× breadth)
Diagonal of rectangle = √(length²+breadth²)
Area of square = side²
Perimeter of square = 4× side
Volume of cylinder = πr²h
T.S.A of cylinder = 2πrh + 2πr²
Volume of cone = ⅓ πr²h
C.S.A of cone = πrl
T.S.A of cone = πrl + πr²
Volume of cuboid = l × b × h
C.S.A of cuboid = 2(l + b)h
T.S.A of cuboid = 2(lb + bh + lh)
C.S.A of cube = 4a²
T.S.A of cube = 6a²
Volume of cube = a³
Volume of sphere = 4/3πr³
Surface area of sphere = 4πr²
Volume of hemisphere = ⅔ πr³
C.S.A of hemisphere = 2πr²
T.S.A of hemisphere = 3πr²
[tex]\begin{gathered}\begin{gathered}\bf \: To \: Find - \begin{cases} &\sf{height_{(cylinder)}}\end{cases}\end{gathered}\end{gathered} [/tex]
[tex]\begin{gathered}\Large{\bold{{\underline{Formula \: Used - }}}} \end{gathered} [/tex]
[tex]\underline{ \boxed{ \bf \: Volume_{(cylinder)} = \: \pi \: {r}^{2}h}} [/tex]
[tex]\underline{ \boxed{ \bf \: CSA_{(cylinder)} = 2\pi \: rh}} [/tex]
where,
[tex]\: \: \: \: \: \: \: \: \: \bull \: \sf \:r \: is \: radius \: of \: cylinder[/tex]
[tex]\: \: \: \: \: \: \: \: \: \bull \: \sf \:h \: is \: height \: of \: cylinder[/tex]
[tex]\large\underline{\sf{Solution-}} [/tex]
Solution−
[tex]\begin{gathered}\begin{gathered}\begin{gathered}\bf \:Let-\begin{cases} &\sf{radius \: of \: cylinder \: = \: r \: meter} \\ &\sf{height \: of \: cylinder \: = \: h \: meter} \end{cases}\end{gathered}\end{gathered}\end{gathered} [/tex]
Given that,
[tex]\rm :\longmapsto\:Volume_{(cylinder)} = 924[/tex]
[tex]:\longmapsto\:\pi \: {r}^{2}h = 924 - - - (1)[/tex]
and
[tex]\rm :\longmapsto\:CSA_{(cylinder)} = 264[/tex]
[tex]\rm :\longmapsto\:2\pi \: rh = 264 - - - (2)[/tex]
Now, Divide equation (1) by equation (2), we get
[tex]\rm :\longmapsto\:\dfrac{\pi \: {r}^{2}h}{2\pi \: rh} = \dfrac{924}{264}[/tex]
[tex] \bf\implies \:r \: = \: 7 \: m[/tex]
On substituting the value of 'r' in equation (2), we get
[tex]\rm :\longmapsto\:2 \times \dfrac{22}{7} \times 7 \times h = 264[/tex]
[tex]\bf\implies \:h \: = \: 6 \: m [/tex]
[tex]\overbrace{ \underline { \boxed { \bf \therefore \: The \: height_{(cylinder)} \: is \: 6 \: meter}}} [/tex]
[tex] \large{\red{ \tt \: Additional \: \: \: Information :-}}[/tex]
Perimeter of rectangle = 2(length× breadth)
Diagonal of rectangle = √(length²+breadth²)
Area of square = side²
Perimeter of square = 4× side
Volume of cylinder = πr²h
T.S.A of cylinder = 2πrh + 2πr²
Volume of cone = ⅓ πr²h
C.S.A of cone = πrl
T.S.A of cone = πrl + πr²
Volume of cuboid = l × b × h
C.S.A of cuboid = 2(l + b)h
T.S.A of cuboid = 2(lb + bh + lh)
C.S.A of cube = 4a²
T.S.A of cube = 6a²
Volume of cube = a³
Volume of sphere = 4/3πr³
Surface area of sphere = 4πr²
Volume of hemisphere = ⅔ πr³
C.S.A of hemisphere = 2πr²
T.S.A of hemisphere = 3πr²