A container, open at the top and made up of metal sheet is in the form of a frustum of a
cone of height 16 cm with diameters of its lower and upper ends as 16 cm and 40 cm respectively. Find the cost of metal sheet used to make the container, if it costs 10 per 100cm2. [Take a = 3.14]
pls help me to solve this
Answers
Answer:
Radius r
1
of upper end of the container=20cm
Radius r
2
of lower end of the container=8cm
Height h of the container=16cm
Slant height l of the frustrum=
(r
1
−r
2
)
2
+h
2
=
(20−8)
2
+16
2
=
144+256
=20cm
Capacity of container=Volume of the frustrum
=
3
1
πh(r
1
2
+r
2
2
+r
1
r
2
)
=
3
1
×3.14×16(20
2
+8
2
+20×8)
=
3
1
×3.14×16(400+64+160)
=
3
1
×3.14×16×624
=10449.92 cubic.cm
=10.45litres
Cost of 1 litre of milk=Rs20
Cost of 10.45 litre of milk=Rs20×10.45=Rs209
Area of metal sheet used to make the container
=π(r
1
+r
2
)l+πr
2
2
=π(20+8)20+π8
2
=560π+64π=624πsq.cm
Cost of 100sq.cm metal sheet= Rs.8
Cost of 624πsq.cm metal sheet =
100
624×3.14×8
=156.75
∴ the cost of the milk which can completely fill the container is Rs209
and the cost of metal sheet used to make the container is Rs.156.75
Answer:
Step-by-step explanation: We have to find CSA and lower base area
R=20cm
r=8cm
slant height= √(20-8)²+16²=20cm
SA=πl(R+r)+πr²=π*20*28+π*8*8
=>π(560+64)
=>1959.3cm²
The cost per 100cm² is Rs 10
so cost per cm²=1/10
Hence the cost = 1959.3/10=195.93
=Rs 196(approx)
hope it helps you...