Question

A milk container is made of metal sheet in the shape of frustum of cone where its volume is 10459 3/7 cm^3 . The radii of its lower and upper circular ends are 8 and 30 cm

Answer 1

Given:

Given:r1 = 20cm,

Given:r1 = 20cm,r2 = 80cm

Given:r1 = 20cm,r2 = 80cmLet h cm be the height of the milk container

Given:r1 = 20cm,r2 = 80cmLet 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 =

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

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