Question

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.5 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 600 per m².

Answer 1

Step-by-step explanation:

${\large{\underline{\underline{\pmb{\frak{Let's\ understand\ the\ concept:-}}}}}}$

☀️ Concept : As per the information, we know that the diameter of the cylindrical part is given. In order to find the CSA of the cylindrical part, it is important to have it's radius. Then, firstly we will find out the radius of the cylindrical part. After that, we will find out the CSA of the cylindrical part.

❍ Then, we will find out the CSA of the top. After that, we will add the CSA of the cylindrical part with the CSA of the top to find the total area of the canvas used for marking the tent.

${\large{\underline{\underline{\pmb{\frak{Given:-}}}}}}$

★ A tent is in the shape of a cylinder surmounted by a conical top.

★ The height of the cylindrical part is 2.5 m.

★ Diameter of the cylindrical part is 4 m.

★ Slant height of the top is 2.8 m.

★ The cost of the rate of the canvas can be find out at the rate of Rs. 600 per m².

${\large{\underline{\underline{\pmb{\frak{To\ find:-}}}}}}$

★ The cost of the canvas of the tent at the rate of Rs 600 per m².

${\large{\underline{\underline{\pmb{\frak{Using\ concepts:-}}}}}}$

★ Formula to find the radius of the cylindrical part.

★ Formula to find the CSA of cylinder.

★ Formula to find the CSA of top.

★ Formula to find total area.

★ Formula to find total cost.

${\large{\underline{\underline{\pmb{\frak{Using\ formulas:-}}}}}}$

★ Radius = Diameter/2.

★ CSA of cylinder = 2πrh.

★ CSA of top = πrl.

★ Total area = CSA of cylindrical part + CSA of top.

★ Total cost = Total area of canvas × Given rate.

${\large{\underline{\underline{\pmb{\frak{Solution:-}}}}}}$

★ Total area of the canvas used for making the tent is 49 m².

★ Total cost of the canvas of the tent at the rate of Rs. 600 per m² is Rs. 29400.

${\large{\underline{\underline{\pmb{\frak{Diagrams:-}}}}}}$

~ The figure given below tell us about the cylindrical part and the cone shape part of the tent.

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{2 \: m}}\put(9,17.5){\sf{2.5 \: m}}\end{picture}$

$\rule{300}{2}$

$\setlength{\unitlength}{1.2mm}\begin{picture}(5,5)\thicklines\put(0,0){\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\put(-0.5,-1){\line(1,2){13}}\put(25.5,-1){\line(-1,2){13}}\multiput(12.5,-1)(2,0){7}{\line(1,0){1}}\multiput(12.5,-1)(0,4){7}{\line(0,1){2}}\put(18,1.6){\sf{r}}\put(9.5,10){\sf{h}}\end{picture}$

The slant height is given. Hence, it is at the slant part nor in the middle part.

${\large{\underline{\underline{\pmb{\frak{Full\ solution:-}}}}}}$

~ Let us find out the radius of the cylindrical part by dividing the diameter of the cylindrical part with 2.

➟ Radius = Diameter/2

➟ Radius = 4/2

➟ Radius = 2 m.

• Henceforth, the radius of the cylindrical part is 2 m.

~ Now let us find out the CSA of the cylindrical part by using the formula of CSA of cylinder.

➟ CSA of cylinder = 2πrh

➟ CSA of cylinder = 2 × 3.14 × 2 × 2.5

➟ CSA of cylinder = 2 × 6.28 × 2.5

➟ CSA of cylinder = 2 × 15.7

➟ CSA of cylinder = 31.4 m².

• Henceforth, the CSA of the cylindrical part is 31.4 m².

~ Now let us find out the CSA of the top by using the required formula.

➟ CSA of the top = πrl

➟ CSA of the top = 3.14 × 2 × 2.8

➟ CSA of the top = 6.28 × 2.8

➟ CSA of the top = 17.6 m².

• Henceforth, the CSA of the top is 17.6 m².

~ Now we will find out the total area of the canvas by adding the CSA of the cylindrical part with the CSA of the top.

➟ Total area = CSA of cylindrical part + CSA of the top

➟ Total area = 31.4 + 17.6

➟ Total area = 49 m².

• Henceforth, the total area of the canvas is 49 m².

~ Now, finally we will find out the total cost of the canvas of the tent at the rate of Rs 600 per m².

➟ Total cost = Total area of the canvas × Given rate

➟ Total cost = 49 × 600

➟ Total cost = Rs. 29400.

• Henceforth, the total canvas of the tent at the rate of Rs 600 per m² is Rs. 29400.

${\large{\underline{\underline{\pmb{\frak{Request:-}}}}}}$

• If there is any difficulty viewing this answer in app, kindly see this answer at web (https://brainly.in/question/) for clear steps and understanding.

${\large{\underline{\underline{\pmb{\frak{See\ this\ answer\ at:-}}}}}}$

• https://brainly.in/question/39543193.

K N O W M O R E :-

Q. A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of Rs 500 per m².

https://brainly.in/question/37485386.

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Answer 2

$\bf \: \underline{Given} :$ ↴

$\small \sf \implies A \: tent \: is \: in \: the \: shape \: of \: a \: cylinder \: surmounted \: by \: a \: conical \: top.$

$\small \sf \implies Height \: of \: cylindrical \: part = 2.5 \: m$

$\small \sf \implies Diameter \: of \: cylindrical \: part = 4 \: m$

$\small \sf \implies Slant \: height \: of \: the \: top = 2.8 \: m$

$\bf \: \underline{To \: find} :$ ↴

$\small \sf \implies Area \: of \: canvas \: used \: for \: making \: the \: tent.$

$\small \sf \implies Cost \: of \: the \: canvas \: of \: the \: tent \: at \: the \: rate \: of \: Rs. \: 600 \: per \: m ^{2}$

$\bf \: \underline{\underline{Solution} }$

$\sf \: Area \: of \: canvas : CSA \: of \: cone + CSA \: of \: cylinder$

$\sf \: Finding \: the \: curved \: surface \: area \: of \: cone.$

$\sf \: Diameter \: of \: cone = Diameter \: of \: cylinder = 4$

$\sf \implies Radius = \dfrac{Diameter}{2} = \dfrac{4}{2} = 2 \: m$

$\sf \implies Slant \: height \: (L) = 2.8 \: m$

$\red{ \sf \implies Curved \: surface \: area \: of \: cone = \pi r L }$

$\sf \implies \dfrac{22}{7} \times 2 \times 2.8$

$\sf \implies \dfrac{44}{7} \times 2.8$

$\sf \implies \dfrac{123.2}{7}$

$\sf \implies 17.6 \: m^{2}$

$\boxed{ \pink{ \sf \: Curved \: surface \: area \: of \: cone = 17.6 \: m^{2}}}$

$\sf \: Now, finding \: the \: curved \: surface \: area \: of \: cylinder.$

$\sf Diameter \: of \: cylinder = 4$

$\sf \implies Radius = \dfrac{Diameter}{2} = \dfrac{4}{2} = 2 \: m$

$\sf \implies Height = 2.5 \: m$

$\red{ \sf \implies Curved \: surface \: area \: of \: cylinder = 2\pi r h }$

$\sf \implies 2 \times \dfrac{22}{7} \times 2 \times 2.5$

$\sf \implies \dfrac{44}{7} \times 2 \times 2.5$

$\sf \implies \dfrac{44}{7} \times 5$

$\sf \implies \dfrac{220}{7}$

$\sf \implies 31.4 \: m^{2}$

$\boxed{ \pink{ \sf \: Curved \: surface \: area \: of \: cylinder = 31.4 \: m^{2}}}$

$\sf Now,$

$\sf \: Area \: of \: canvas : CSA \: of \: cone + CSA \: of \: cylinder$

$\sf \: Area \: of \: canvas \: : 17.6 \: m^{2} + 31.4\:m^{2}$

$\underline{ \underline{\boxed{\purple{\sf \: Area \: of \: canvas \: : 49\:m^{2}}}}}$

$\sf Now, we \: will \:find :$ ↴

$\small \sf The\: cost \: of \: the \: canvas \: of \: the \: tent \: at \: the \: rate \: of \: Rs. \: 600 \: per \: m ^{2}$

$\sf Total \: cost = Total \: area \: of \: the \: canvas \times Given \: rate$

$\sf Total \: cost = 49 \times 600$

$\underline{\underline{\boxed{\purple{\sf Total \: cost = Rs. \: 29400.}}}}$