A tent in the form of cylinder of diameter 4.2 m and height 8 m surmou

1.	A tent in the form of cylinder of diameter 4.2 m and height 8 m surmounted by a cone of equal base and height 6 m find the volume of the air in the tent
Answer» $\large\underline\purple{\bold{Solution :- }}$ $\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{radius \: of \: cylinder \: be \: r \: m} \\ &\sf{<klux>HEIGHT</klux> \: of \: cylinder \: be \: h \: m} \end{cases}\end{gathered}\end{gathered}$ $\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{radius \: of \: cone \: be \: r \: m} \\ &\sf{height \: of \: cylinder \: be \: H \: m} \end{cases}\end{gathered}\end{gathered}$ Here, ● Dimensions of Cylinder ☆ Diameter of cylinder, = 4.2 m ☆ Radius of cylinder, r = 2.1 m ☆ Height of cylinder, h = 8 m ● Dimensions of cone ☆ Diameter of cone, = 4.2 m ☆ Radius of cone, r = 2.1 m ☆ Height of cone, H = 6 m $\red{ \bf \: Now, }$ $\boxed { \purple{ \bf \: V_{(air \: in \: tent)} = V_{(Cylinder)} + V_{(Cone)} }}$ $\tt \longrightarrow \: V_{(air \: in \: tent)} = \pi \: {r}^{2} h + \dfrac{1}{3} \pi \: {r}^{2} H$ $\tt \longrightarrow \: V_{(air \: in \: tent)} = \pi \: {r}^{2} (h + \dfrac{H}{3} )$ $\tt \longrightarrow \: V_{(air \: in \: tent)} = \dfrac{22}{7} \times 2.1 \times 2.1 \times (8 + \dfrac{6}{3})$ $\tt \longrightarrow \: V_{(air \: in \: tent)} = \dfrac{22}{7} \times \dfrac{21}{10} \times \dfrac{21}{10} \times 10$ $\tt\implies \: \boxed{ \blue{ \tt \: V_{(air \: in \: tent)} = 138.6 \: {m}^{3} }}$ ─━─━─━─━─━─━─━─━─━─━─━─━─ More information:- Perimeter of rectangle = 2(length× breadth) DIAGONAL of rectangle = √(length ²+breadth ²) 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

A tent in the form of cylinder of diameter 4.2 m and height 8 m surmounted by a cone of equal base and height 6 m find the volume of the air in the tent

Answer»

$\large\underline\purple{\bold{Solution :- }}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{radius \: of \: cylinder \: be \: r \: m} \\ &\sf{<klux>HEIGHT</klux> \: of \: cylinder \: be \: h \: m} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf Let = \begin{cases} &\sf{radius \: of \: cone \: be \: r \: m} \\ &\sf{height \: of \: cylinder \: be \: H \: m} \end{cases}\end{gathered}\end{gathered}$

Here,

● Dimensions of Cylinder

☆ Diameter of cylinder, = 4.2 m

☆ Radius of cylinder, r = 2.1 m

☆ Height of cylinder, h = 8 m

● Dimensions of cone

☆ Diameter of cone, = 4.2 m

☆ Radius of cone, r = 2.1 m

☆ Height of cone, H = 6 m

$\red{ \bf \: Now, }$

$\boxed { \purple{ \bf \: V_{(air \: in \: tent)} = V_{(Cylinder)} + V_{(Cone)} }}$

$\tt \longrightarrow \: V_{(air \: in \: tent)} = \pi \: {r}^{2} h + \dfrac{1}{3} \pi \: {r}^{2} H$

$\tt \longrightarrow \: V_{(air \: in \: tent)} = \pi \: {r}^{2} (h + \dfrac{H}{3} )$

$\tt \longrightarrow \: V_{(air \: in \: tent)} = \dfrac{22}{7} \times 2.1 \times 2.1 \times (8 + \dfrac{6}{3})$

$\tt \longrightarrow \: V_{(air \: in \: tent)} = \dfrac{22}{7} \times \dfrac{21}{10} \times \dfrac{21}{10} \times 10$

$\tt\implies \: \boxed{ \blue{ \tt \: V_{(air \: in \: tent)} = 138.6 \: {m}^{3} }}$

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More information:-

Perimeter of rectangle = 2(length× breadth)

DIAGONAL of rectangle = √(length ²+breadth ²)

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