Question

A milk container is made of metal sheet in the
shape of frustum of a cone whose volume is
10459 cm². The radii of its lower and upper
circular ends are 8 cm and 20 cm respectively.
Find the cost of metal sheet used in making the
container at the rate of * 1.40 per square
centimetre.
291​

Answer 1

$\mathbb\\\boxed{ANSWER:}$

GIVEN:

R = 20 cm

r = 8 cm

h = Height in cm

Volume = (1/3) π h (R² + r² + R.r)

=> 10459 3/7 = (1/3) (22/7) h ( 20² + 8² + 20*8)

=> 73216/7 = (22h/21) (400 + 64 + 160)

=> 9984 = h (624)

=> h = 16

Height = 16 cm

Surface Area =  π(R + r)( √(R-r)² + h²)   + πR² + πr²

= (22/7) (20 + 8) (√(20-8)² + 16²)  + (22/7)20²  + (22/7)8²

=(22/7) (28) (20) + (22/7)( 400 + 64)

= 1760 + 10208/7

= 22528/7  cm²

Cost = 1.4 * 22528/7 = Rs 4505.6

If it is open from top then surface Area

Surface Area =  π(R + r)( √(R-r)² + h²)  + πR²

= (22/7) (28) (20) + (22/7)( 400)

= 1760 + 8800/7

= 21120/7 cm²

Cost = 1.4 * 21120/7 =  Rs 4224

$\mathcal{THANK \;\;YOU !!}$

Answer 2

Given:

r1 = 20cm,

r2 = 80cm

Let h cm be the height of the milk containerVolume of the milk container = Volume of frustum of cone

= $\sf\green{\frac{1}{3} \pi \: h( {r1}^{2} + {r2}^{2} + r1 + r2) = 10459 \frac{3}{7} {cm}^{3} }$

= $\sf\green{\frac{1}{3} \frac{22}{7} \times h(( {20)}^{2} + {8}^{2} + 20 \times 8)}$

= $\sf\green{ \frac{73216}{7} {cm}^{3}}$

= $\sf\green{ \frac{1}{3} \times \frac{22}{7} \times h \times 624 = \frac{73216}{7}}$

$\therefore$ $\sf\green{h = \frac{73216 \times 3}{22 \times 624} = 16cm}$

Slant height of the frustum =

$\longrightarrow$$\sf\red{l = \sqrt{ {h}^{2} + {(r1 - r2)}^{2} } }$

$\longrightarrow$$\sf\red{ \sqrt{ {(16)}^{2} - {(20 - 8)}^{2} } cm}$

$\longrightarrow$$\sf\red{ \sqrt{256 + 144} cm}$

$\longrightarrow$$\sf\red{ \sqrt{400} cm}$

$\longrightarrow$$\sf\red{20cm}$

Area of the metal sheet used in the container =

= CSA of the frustum + Area of the base

= $\sf\blue{\pi(r1 + r2)l + \pi {r2}^{2} }$

= $\sf\blue{( \frac{22}{7} (20 + 8) \times 20 + \frac{22}{7} \times 8 \times 8) {cm}^{2} }$

= $\sf\blue{\frac{22}{7} (560 + 64) {cm}^{2}}$

= $\sf\blue{ \frac{22}{7} \times 624 {cm}^{2}}$

Cost of the metal at the rate of Rs. 1.40 per cm square =

= $\sf\orange{Rs.\frac{22}{7} \times 624 \times 1.40}$

= ${\boxed{\sf{\orange{Rs.2745.60}}}}$