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 2.1 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.

Answer 1

${\pmb{\sf{\underline{Question...}}}}$

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 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 m sq.

${\pmb{\sf{\underline{Given \; that}}}}$

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

★ The height of the cylindrical part = 2.1 m

★ The diameter of the cylindrical part = 4 metres.

Henceforth, radius is 2 metres.

${\small{\underline{\boxed{\sf{\because Radius \: = \dfrac{D}{2}}}}}}$

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

${\sf{:\implies Radius \: = 2 \: m}}$

★ The slant height of the top = 2.8 m

${\pmb{\sf{\underline{To \; find}}}}$

★ The area of the canvas used for making the tent.

★ The cost of the canvas of the tent at the rate of Rs 500 per metre sq.

${\pmb{\sf{\underline{Solution}}}}$

★ The area of the canvas used for making the tent = 44m²

★ The cost of the canvas of the tent at the rate of Rs 500 per metre sq. = Rupees 22,000

${\pmb{\sf{\underline{Diagrams}}}}$

Figure that tell us about the cylindrical part and the cone shape part of the given tent. These are looking something like the below respectively :

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{2 \: m}}\put(9,17.5){\sf{2.1 \: m}}\end{picture}$

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

But as the slant height is given henceforth, it is at the slant part nor in the middle part!

• Request: Kindly see the both diagrams from web. Thank you.

${\pmb{\sf{\underline{Using \; concepts}}}}$

★ Formula to find curved surface area of the cylinder / cylindrical part.

★ Formula to find curved surface area of the top.

★ Formula to find total area of canvas.

★ Formula to find total cost(according to the question)

${\pmb{\sf{\underline{Using \; formulas}}}}$

${\small{\underline{\boxed{\sf{CSA \: of \: cylinder = 2 \pi rh}}}}}$

${\small{\underline{\boxed{\sf{CSA \: of \: top \: = \pi rl}}}}}$

${\small{\underline{\boxed{\sf{Total \: area \: = CSA \: cylinderal \: part \: + CSA \: of \: top}}}}}$

${\small{\underline{\boxed{\sf{Total \: cost \: = Total \: area \: of \: canvas \times Given \: Amount}}}}}$

${\pmb{\sf{\underline{Full \; Solution}}}}$

~ Firstly let us find the CSA of the cylindrical part by using the given formula:

${\small{\underline{\boxed{\sf{CSA \: of \: cylinder = 2 \pi rh}}}}}$

${\sf{:\implies CSA \: of \: cylinder = 2 \pi rh}}$

${\sf{:\implies CSA \: of \: cylinder = 2 \times 3.14 \times 2 \times 2.1}}$

${\sf{:\implies CSA \: of \: cylinder = 2 \times 6.28 \times 2.1}}$

${\sf{:\implies CSA \: of \: cylinder = 2 \times 13.188}}$

${\sf{:\implies CSA \: of \: cylinder = 26.376}}$

${\sf{:\implies CSA \: of \: cylinder = 26.4 \: m \: sq. \: (approx)}}$

Henceforth, 26.4 m sq. is the curved surface area of the cylindrical part.

~ Now by using the given formula let us find the CSA of the top:

${\small{\underline{\boxed{\sf{CSA \: of \: top \: = \pi rl}}}}}$

${\sf{:\implies \pi rl}}$

${\sf{:\implies 3.14 \times 2 \times 2.8}}$

${\sf{:\implies 6.28 \times 2.8}}$

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

Henceforth, curved surface area of the top is 17.6 m sq.

~ Now let's find the total area of the canvas.

${\small{\underline{\boxed{\sf{Total \: area \: = CSA \: cylinderal \: part \: + CSA \: of \: top}}}}}$

${\sf{:\implies 26.4 + 17.6}}$

${\sf{:\implies 44 \: m \: sq.}}$

Henceforth, 44 m sq. is the total area of the canvas.

~ Now at last let us find the total cost of the canvas of the tent at the rate of Rupees 500 per m sq.

${\small{\underline{\boxed{\sf{Total \: cost \: = Total \: area \: of \: canvas \times Given \: Amount}}}}}$

${\sf{:\implies 44 \times 500}}$

${\sf{:\implies Rs. \: 22,000}}$

Answer 2

$\bf \large \underline{Given: }$

$\text{Height (h) of the cylindrical part = 2.1 m} \\\text{Diameter of the cylindrical part = 4 m} \\ \text{Radius of the cylindrical part = 2 m} \\\text{Slant height (l) of conical part = 2.8 m}$

$\textbf{\large\underline {Formula Used : }}$

$\boxed {\leadsto \sf \pink{CSA \: of \: cone = πrl}} \\\boxed{\leadsto \sf \green{CSA \: of \: Cylinder = 2πrh}}$

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

Area of canvas used = CSA of conical part + CSA of cylindrical part

$\text \purple{ Area of canvas used=πrl+2πrh}$

$\text{=π×2×2.8+2π×2×2.1} \\\text{=2π[2.8+4.2] }\\ \: \: \: \: \: \: \: \: \: \: \: \: {=2× \frac{22}{7} ×7 }\\ \boxed{ \large \orange{=44m^{2} }}$

Cost of 1 m² canvas = ₹ 500

Cost of 44 m² canvas = 44 × 500= ₹ 22000

Therefore, it will cost ₹ 22000 for making such a tent.

Diagrams:

$\setlength{\unitlength}{1mm}\begin{picture}(5,5)\thicklines\multiput(-0.5,-1)(26,0){2}{\line(0,1){40}}\multiput(12.5,-1)(0,3.2){13}{\line(0,1){1.6}}\multiput(12.5,-1)(0,40){2}{\multiput(0,0)(2,0){7}{\line(1,0){1}}}\multiput(0,0)(0,40){2}{\qbezier(1,0)(12,3)(24,0)\qbezier(1,0)(-2,-1)(1,-2)\qbezier(24,0)(27,-1)(24,-2)\qbezier(1,-2)(12,-5)(24,-2)}\multiput(18,2)(0,32){2}{\sf{2 \: m}}\put(9,17.5){\sf{2.1 \: m}}\end{picture}$

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