Question

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 1

Answer:

Given:

Height (h) of the cylindrical part = 2.1 m

Diameter of the cylindrical part = 4 m

Radius of the cylindrical part = 2 m

Slant height (l) of conical part = 2.8 m

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

=πrl+2πrh

=π×2×2.8+2π×2×2.1

=2π[2.8+4.2]

=2×

7

22

×7

=44m

2

Cost of 1 m

2

canvas =500 rupees

Cost of 44 m

2

canvas =44×500=22000 rupees.

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

Answer 2

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 = $\dfrac{22}{7}$

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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