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

Step-by-step explanation:

$\underline{\underline{\maltese\: \: \textbf{\textsf{Question}}}}$

• 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.

$\underline{\underline{\maltese\: \: \textbf{\textsf{Answer}}}}$

• Area of the canvas used for making the tent = 44 m².

• Cost of the canvas of the tent at the rate of Rs. 500 m² = Rs. 22,000.

$\underline{\underline{\maltese\: \: \textbf{\textsf{Given}}}}$

• 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 m.

$\underline{\underline{\maltese\: \: \textbf{\textsf{To\ Find}}}}$

• 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 m².

$\underline{\underline{\maltese\: \: \textbf{\textsf{Basic\ Terms}}}}$

• Diameter : A straight line passing from side to side through the centre of a body or figure.

• Radius : A straight line from the centre to the circumference of a figure, especially in a circle.

$\underline{\underline{\maltese\: \: \textbf{\textsf{Formula\ Used}}}}$

• CSA of cylinder = 2πrh.
• CSA of top = πrl.
• Total area = CSA cylindrical part + CSA of top.
• Total cost = Total are of canvas × Given amount.
• Radius = Diameter/2.

$\underline{\underline{\maltese\: \: \textbf{\textsf{Diagrams}}}}$

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

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

$\rule{300}{2}$

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

The slant height is given. Hence, it is at the slant part not in the middle part.

$\underline{\underline{\maltese\: \: \textbf{\textsf{Calculations}}}}$

~ As through the given data, we know that the diameter of the cylindrical part is given. Then, to find the CSA of the cylindrical part it is important to have it's radius. So, to find it's radius, our required formula is as follows :- Radius = Diameter/2.

⇒ Radius = Diameter/2

⇒ Radius = 4/2

⇒ Radius = 2 m.

~ Now we have the radius of the cylindrical part. Then, by using the formula of CSA of cylinder we will find out the CSA of the cylindrical part. Formula :- CSA of cylinder = 2πrh.

• Here, π = 3.14, r = 2, h = 2.1

⇒ CSA of cylinder = 2 × 3.14 × 2 × 2.1

⇒ CSA of cylinder = 2 × 6.28 × 2.1

⇒ CSA of cylinder = 2 × 13.188

⇒ CSA of cylinder = 26.376

⇒ CSA of cylinder = 26.4 m² (approx).

~ Hence, the CSA of the cylindrical part is 26.4m².

~ Now we will find out the CSA of the top. Formula :- CSA of the top = πrl.

• Here, π = 3.14, r = 2, l = 2.8

⇒ CSA of the top = 3.14 × 2 × 2.8

⇒ CSA of the top = 6.28 × 2.8

⇒ CSA of the top = 17.6m².

~ Hence, CSA of the top is 17.6m².

~ Now we have the CSA of the cylindrical part and the CSA of the top. So now, we will find out the total area of the canvas. In simple words :- Total area = CSA of cylindrical part + CSA of the top.

⇒ Total area = 26.4 + 17.6

⇒ Total area = 44m².

~ Hence, the total area of the canvas = 44m².

~ Now, we have the total area of the canvas. So at last, we will find out the cost of the canvas of the tent at the rate of Rs 500 per m².

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

⇒ Total cost = 44 × 500

⇒ Total cost = Rs. 22,000.

~ Hence, the cost of the canvas of the tent at the rate of Rs 500 per m² is Rs. 22,000.

$\underline{\underline{\maltese\: \: \textbf{\textsf{Request}}}}$

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$\underline{\underline{\maltese\: \: \textbf{\textsf{See\ this\ answer\ at\ :}}}}$

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