A tent in the shape of cylinder surmounted by a conical top. If the he

1.	A tent in the shape of cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 3.5 m and 6 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 500 per m2
Answer» Given - Height of cylindrical part = 3.5m Radius of Cylindrical part = 6/2 = 3m Slant Height of conical part = 2.8m To find - Area of canvas and RATE. Solution - First we will find the curved surface area of cylindrical and conical parts. So, Curved surface area of cylinder = 2πrh where, π = $\dfrac{<klux>22</klux>}{7}$ r = radius h = height On substituting the VALUES - CSA = 2πrh CSA = 2 × $\dfrac{22}{7}$ × 3 × 3.5m CSA = 2 × 22 × 3 × 0.5m $\longmapsto$ CSA = 66m² Similarly, Curved surface area of cone = πrl where, π = $\dfrac{22}{7}$ r = Radius l = slant Height On substituting the values - CSA = πrl CSA = $\dfrac{22}{7}$ × 3 × 2.8 CSA = 22 × 3 × 0.4 $\longmapsto$ CSA = 226.4m² Area used for making canvas = CSA of cylindrical part + CSA of conical part $\longmapsto$ 66m + 226.4m $\longmapsto$ 92.4m² Cost of canvas of tent = Rate given × total area $\longmapsto$ 500 × 92.4 $\longmapsto$ ₹46200 $\therefore$ Area for making canvas is 92.4m³ and cost is ₹46200 _____________________________________________

A tent in the shape of cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 3.5 m and 6 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 500 per m2

Answer»

Given -

To find -

Solution -

First we will find the curved surface area of cylindrical and conical parts.

So,

Curved surface area of cylinder = 2πrh

where,

π = $\dfrac{<klux>22</klux>}{7}$

r = radius

h = height

On substituting the VALUES -

CSA = 2πrh

CSA = 2 × $\dfrac{22}{7}$ × 3 × 3.5m

CSA = 2 × 22 × 3 × 0.5m

$\longmapsto$ CSA = 66m²

Similarly,

Curved surface area of cone = πrl

where,

π = $\dfrac{22}{7}$

r = Radius

l = slant Height

On substituting the values -

CSA = πrl

CSA = $\dfrac{22}{7}$ × 3 × 2.8

CSA = 22 × 3 × 0.4

$\longmapsto$ CSA = 226.4m²

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

$\longmapsto$ 66m + 226.4m

$\longmapsto$ 92.4m²

Cost of canvas of tent = Rate given × total area

$\longmapsto$ 500 × 92.4

$\longmapsto$ ₹46200

$\therefore$ Area for making canvas is 92.4m³ and cost is ₹46200

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