Question

the vertical height of the conical tent is14 m and the diameter of its base is 18 m. how many persons can it accumulateif each person is to be allowed 108 m^3 of space.​

Answer 1

AnswEr:-

❀ Your Answer Is 11 persons.

ExplanaTion:-

Given:-

• Vertical height of a cone = 14 m

• Diameter of base = 18 m.

• So its radius = 18/2 m = 9 m.

• Space allowed to one person = 108 m³.

See the diagram below:-

$\\ \setlength{\unitlength}{1 mm}\begin{picture}(5,5)\put(2,2){\line(1,-4){5}}\put(2,2){\line(-7,-4){5}} \qbezier(7,-18)(2,-13)(-3,-18)\qbezier(7,-18)(2,-20)(-3,-18)\put(9,-10){\vector( 0,-1){8}}\put(9,-6){\vector(0,1){8}}\put(6,-9){ \tt{14\:m} }\put(2,-18){\circle*{0.5}}\put( -3,-18){\line(5,0){10}}\put(1.5,-21){ \tt{18\: m} }\end{picture}$

To Find:-

• How many person can the tent accumulate.

Formula Used:-

$\\ \longrightarrow{\boxed{\boxed{\bf V = \dfrac{1}{3}\pi {r}^{2}h.}}}$

Where,

• V = Volume of a cone.

• $\tt\pi = \dfrac{22}{7}.$

• r = Radius of the Cone.

• h = Height of the cone.

So Here,

• r = 9 m.

• h = 14 m.

Therefore,

$\\ \bf\bigstar{\underline{ \underline{Volume \: \: Of \: \: The \: \: Tent, \: }}} \\ \\ \tt = \frac{1}{3} \pi {r}^{2} h. \\ \\ \tt = \frac{1}{3} \times \frac{22}{ \cancel7} \times (9 \: m) {}^{2} \times \cancel{14} \: m. \\ \\ \tt = \frac{1}{ \cancel3} \times 22 \times \cancel{81} \:m {}^{2} \times 2 \: m. \\ \\ \tt = 1 \times 22 \times 27 \: m {}^{2} \times 2 \: m. \\ \\ \tt = 22 \times 54 \: m {}^{3} \\ \\ \boxed{ \bf =1188 \: m {}^{3}.}$

Now,

It is given that one person occupy 108 m³ of space and we have to find out how many people can occupy 1188 m³ of space so lets find out:-

$\\ \tt\mapsto No. \: \: of \: \: persons = \cancel\dfrac{1188 \: m {}^{3} }{108 \: m {}^{3} } \\ \\ \bf\mapsto \boxed{ \bf No. \: \: of \: \: persons = 11.}$

$\bf\therefore$ 11 persons can accumulate in the given tent.

Answer 2

${ \blue{ \underline{ \red{ \sf{ \underline{ \huge{ \tt{question}}}}}}}}$





▪ The vertical height of the conical tent is 14 cm and the diameter of its base is 18 m. How many persons can it accumulate, if each person is to be allowed 108 m^3 of space??






${ \blue{ \underline{ \red{ \underline{ \tt{ \huge{solution}}}}}}}$





${ \bold{ \dagger{ \blue{ \: \: \: given}}}}$




${ \sf{ \pink{ \underline{ \underline{dimension \: of \: the \: conical \: tent}}}}}$



${ \star{ \orange{ \sf{ \: \: vertical \: height = 14 \: m}}}} \\ \\ { \star{ \orange{ \sf{ \: \: diameter \: of \: base = 18 \: m}}}}$



${ \orange{ \rightarrow{ \sf{radius = \frac{diameter}{2} }}}} \\ \\ { \implies{ \orange{ \sf{radius = \frac{18 \: m}{2} = 9 \: m}}}}$





${ \green{ \sf{ \underline{ \underline{volume \: of \: conical \: tent}}}}} \\ \\ { \boxed{ \boxed{ \green{ \sf{ \: \: v = \frac{1}{3} \: \pi \: {r}^{2} h \: \: }}}}} \\ \\ { \sf{where}} \\ \\ { \sf{ r = { \green{ radius \: of \: the \: cone}}}} \\ \\ { \sf{h = { \green{vertical \: height \: of \: the \: cone}}}}$




▪ putting the above given values in the formula.....





${ \blue{ \sf{volume = \frac{1}{3} \: \pi \: {(9 \: m)}^{2} \times 14m}}}$





${ \implies{ \blue{ \sf{volume = \frac{1}{3} \times \frac{22}{7} \times 81 \times 14 {m}^{3}}}}}$





${ \implies{ \blue{ \sf{volume = 22 \times 27 \times 2 {m}^{3}}}}}$




▪ it's given that the volume occupied by each person is 108 m^3





${ \red{ \sf{no. \: of \: persons = \frac{volume \: of \: conical \: tent}{volume \: occupied \: by \: 1 \: person}}}}$






${ \implies{ \red{ \sf{no. \: of \: persons = \frac{22 \times 27 \times 2 \: {m}^{3} }{108 \: \frac{ {m}^{3} }{person}}}}} } \\ \\ { \implies{ \red{ \sf{ no. \: of \: persons = \frac{22 \times 2 \: persons}{4}}}}} \\ \\ { \sf{ \implies{ \underline{\boxed{ \pink{no. \: of \: persons = 11 \: persons}}}}}}$