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

Volume of a frustum cone=volume of milk container.
1/3πh(R²+r²+Rr)=10459.42857
1/3×22/7×h((20²)+8²+20×8)=10459.42857
1/3×22/7×h(400+64+160)=10459.42857
1/3×22/7×h×624=10459.42857
h=10459.42857×21/22×624=16cm
h=16cm
Slant height(l)=√h²(R-r)²
=√16²(20-8)²
√256+144
=√400
=20cm
Slant height(l)=20cm
Cost of a metal sheet used in a container @ 1.40 per cm²=3218.285714×1.40=Rs.4505.6

Answer 2

Answer:

Step-by-step explanation:

For frustum of a cone,

R= 20 cm

r= 8 cm

Volume = 10459 3/7 cm^3

1/3 Pi h (R^2 +r^2 +Rr) =10459 3/7

1/3×22/7×h(400 +64 + 160) = 73216/7

1/3×22×h(624) = 73216

1/3×22×h(78) =9152

1/3×11×h(39) = 2288

13×h =208

h= 16 cm

Slant height(l) = root(R-r)^2 +h^2

=root(20-8)^2 +16^2

= root 12^2 +16^2

=root 20^2 - by pythgorian triplet

L = 20 cm

Costof metal sheet = Rate×[CSA of frustum + Base area]

= 1.4 ×[pi(R+r)L + piR^2]

=1.4×pi[20(20+8) + 20^2]

=1.4×22/7 ×20 [28+20]

=88×48

=4224

Cost of metal sheet= Rs. 4224