Question

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 3/7cm3. The radii of its lower and upper ends are 8cm and 20 cm respectively. Find the cost of metal sheet used in making the container at the rate of Rs 1.40 per square centimeter.

Answer 1

your answer in attachment

hope this helps you

Answer 2

Given:

r1 = 20cm,

r2 = 80cm

Let h cm be the height of the milk container

Volume 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 =

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

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