Q. 384.4( 8 Votes )

# A milk container

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, ℓ2 = h2 + (R – r)2

2 = 162 + (20 – 8)2

2 = 256 + (12)2

2 = 256 + 144

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 + πR2 + πr2

= π(20 + 8) × 20 + π(20)2 + π(8)2

= 28 × 20π + 400π + 64π

= 560π + 464π

= 1024π eqn1

Area of upper circular end = πR2

= 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 cm2

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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