Question

A milk container is made of metal sheet in the shape of frustum of a cone whose volume is 10459 3/7 cm³. 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 Rs. 1.40 per cm².

Answer 1

Cost of metal sheet of the container = Rs. 2745.60

Step-by-step explanation:

Bottom radius r = 8 cm

Top radius R = 20cm

Let Height of bucket be h  

Volume of the bucket = 1/3πh(R² + r² +Rr)

10459 (3/7)   = 1/3 * 22/7 * h (20² + 8² + 160)

10459.43 = 1/3 * 22/7 * h * 624

Therefore h = (10459.43 * 3 * 7) / (22 * 624) = 16 cm

  = 8800 cu.cm.

Slant height of frustum l = root of (h² + (R-r)²) = root (16² + (20-8)²)

   = root ( 256 + 144)  

   = root 400

   = 20 cm

Curved Surface area of bucket = π(R+r)l = 22/7 * 28 * 20 = 1760 sq.cm.

Area of base = πr² = 22/7 * 8 * 8 = 201.14 sq.cm.

Total surface area of the metal sheet =  1760 + 201.14 = 1961.14 sq.cm.

Cost of one cm² of metal sheet = Rs. 1.40

Therefore cost of metal sheet of the container = 1.4 * 1961.14 = Rs. 2745.60

Answer 2

Given:

Given:r1 = 20cm,

Given:r1 = 20cm,r2 = 80cm

Let h cm be the height of the milk container

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 =

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

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