Q. 384.4( 8 Votes )

Answer :

Upper end radius of container = R = 20 cm

Lower end radius of container =r = 8 cm

Height of the container be ‘h’ cm

As container is in shape of frustum

Volume of container = Volume of Frustum

Where R = Radius of larger or upper end and r = Radius of smaller or lower end and h = height of frustum π = 22/7

(putting the values)

⇒ 22 × h × 208 = 73216

⇒ 4576 × h = 73216

⇒ h = 73216/4576

∴ h = 16 cm

Let the slant height of the frustum be ‘ℓ’ cm

So, ℓ^{2} = h^{2} + (R – r)^{2}

⇒ ℓ^{2} = 16^{2} + (20 – 8)^{2}

⇒ ℓ^{2} = 256 + (12)^{2}

⇒ ℓ^{2} = 256 + 144

⇒ ℓ^{2} = 400

⇒ ℓ = √(400)

∴ ℓ = 20 cm

Now the Area of sheet used in making the container can be calculated by simply subtracting the area of the upper circular end from the T.S.A of frustum

⇒ T.S.A of frustum = π(R + r)ℓ + πR^{2} + πr^{2}

= π(20 + 8) × 20 + π(20)^{2} + π(8)^{2}

= 28 × 20π + 400π + 64π

= 560π + 464π

= 1024π →eqn1

Area of upper circular end = πR^{2}

= π(20)^{2}

= 400π →eqn2

T.S.A of container = 1024π – 400π (from eqn2 and 1)

⇒ T.S.A of container = 624π

⇒ Cost of metal sheet used = T.S.A of container × Rate per cm^{2}

⇒ Cost of mtal sheet used

= 13728 × 0.2

= Rs. 2745.6

__The cost of the metal sheet used is Rs. 2745.6__

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