Question

Circumference of the the base of a cylinder open at the top is 132 cm. the sum of radius and height is 41 cm.find the cost of polishing the outer surface area of cylinder at the rate of rs 10 per square decimeter

Answer 1

Here is your answer :-

Circumference of the base of cylinder =
➡ 2πr = 132

➡ 2* 22/7 * r = 132

➡ r = 132* (1/2) * (7/22)

➡ r = 21cm

It is given that sum of radius & height is 41cm

➡ r + h = 41

➡ 21 + h = 41

➡ h = 41 - 21

➡ h = 20cm

Now, we have to cost of polishing the outer surface area, so

➡ Curved surface area of cylinder= 2πrh

➡ 2 * (22/7) *21 *20

➡ 2640 cm²

We have to polish an area of 2640 cm²

➡ It is given that,

cost of polishing 1 dm² = Rs 10

[ 1 dm² = ( 10 * 10 ) cm² = 100cm² ]

➡ Cost of polishing 100cm² = Rs 10

➡ cost o polishing 2640 cm² =
( 10/100 ) * 2640

➡ 2640/10 = Rs 264.

So, required answer is Rs 264.

Hope it will help you !!!

Answer 2

Circumference =2×pie×r=132cm

2×22/7×r=132cm

22/7×r=132/2

22/7×r=66

22r=66×7

22r=462

r=21cm

r+h=41cm

21+h=41cm

h=41-21cm

h=20cm

It is given that it is open at the top. Which means that we should include the base.

Therefore,

CSA+area of the circle (base)

2×pie×r×h + pie×r^2

=2×22/7×21×20 + 22/7×21^2

=44×60+22/7×441

=2640+1386=4026cm^2

=0.4026m^2

Cost of polishing 1m^2= rupees 10

Therefore,

Cost of polishing 0.4026m^2= ?

=0.4026×10

= rupees 4.026

I hope that my answer is correct and it would help you

Thank you