Question

Circumference of the base of a cylinder, open at the top, is 132 cm. The sum of radius and height is 41 cm. Find cost of polishing the outer surface area of cylinder at the rate of Rs. 10 per square decimeter. ( Take π = 22/7 )

Answer 1

Given : (i) Circumference is 132cm.
( ii ) radius + height = 41cm

Let the radius be r.
then ,
Circumference = 2 π r
132 = 2 × 22/7 × r
132 × 7 / 2 × 22 = r
21 cm = r

CASE 1 :
Radius + Height = 41 cm
let the height be h.
21 + h = 41
h = 41-21
h = 20 cm

CASE 2 :
Outer Surface = CSA of cylinder
CSA of cylinder = 2 π r h
= 2 × 22/7 × 21 × 20
= 2640 cm^2

CASE 3 :
Rate of polishing the outer surface =
Rs. 10 per dm

1 dm = 10 cm
10 dm = 100 cm
= Rs. 10 / 100 cm

Cost = 2640 × 10/100 = Rs. 264
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Hope it helps...!!!!!!!!!
Thanks...........!!!!!!!!!!!!!

Answer 2

Here, l = 1.5 m, b = 1.25 m

                 



                 ∵ It is open from the top.

                 ∵ Its surface area = [Lateral surface area] + [Base area]

                 = [2(l + b)h] + [l × b]

                 = [2(1.50 + 1.25)0.65 m2] + [1.50 × 1.25 m2]

                 = [2 × 2.75 × 0.65 m2] + [1.875 m2]

                 = 3.575 m2 + 1.875 m2 = 5.45 m2

                 ∵ The total surface area of the box = 5.45 m2

                 ∴ Area of the sheet required for making the box = 5.45 m2

           (ii) Rate of sheet = Rs. 20 per m2

                 Cost of 5.45 m2 = Rs. 20 × 5.45

                 

                 ⇒ Cost of the required sheet = Rs.109