Question

A bus stop is barricadded from the remaining part of the road , by using 50 hollow cones made of recyled cardboard . Each cones has a base diameter of 40 cm and height 1 m . if the outer side of each of the cones is to be painted and the cost of painting is Rs 12 per m^2 , what will be the cost of painting all these cones ?​

Answer 1

Radius of the cone (r) = 40/2 cm = 20 cm = 0.2 m

Height of the cone (h) = 1 m

Let l  be the slant height of a cone.

∴ l = √h2 + r2 

⇒ l = √12 + 0.22 

⇒ l = √1.04 

⇒ l = 1.02 m

Rate of painting = ₹12 per m2 

Curved surface of 1 cone = πrl m2

                                         = (3.14 × 0.2 × 1.02) m2

                                         = 0.64056 m2

Curved surface of such 50 cones = (50 × 0.64056) m2

                                                                 = 32.028 m2

Cost of painting all these cones = ₹(32.028 × 12)

                                                   = ₹384.34

Answer 2

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☃ ƛƝƧƜЄƦ ☃

$\small \red{{Diameter \: of \: base \: = \: 40 \: cm}} \\ \small \blue{{ Radius , r = \: \frac{40}{2} = 20 \: cm }} \\ \small \green{{Height, h = 1m = 100 \: cm}}$

$\small \pink{{l ^{2} = h^{2} = (100)^{2} + (20)^{2} = 10000 + 400 = 10400 }} \\ \small \orange{{l = \sqrt{10400} = 102 \: cm \: nearly }}$

$\small \red{{Surface \: area \: of \: one \: cone =\pi \: r \: l = 3.14 \times 20 \times 102 = 6405.6 \:^{2} }} \\ \small \pink{{Surface \: area \: of \: 50 \: cones = \: 50 \times 6405.6 = 320280 \: cm^{2}}} \\ \small \orange{{Rate \: of \: painting= \: Rs12 \: per \: m^{2}}} \\ \small \blue{{Cost \: of \: Painting \: = \frac{12 \times 320280}{1000} = \: }} \small \blue{{ \: Rs }} \small \red{{\: 384.34}}$

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