Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class

Surface Area of Cylinder

Question 1
Deleted for CBSE Board 2024 Exams

Question 2 Deleted for CBSE Board 2024 Exams

Question 3 Important Deleted for CBSE Board 2024 Exams

Question 4 Important Deleted for CBSE Board 2024 Exams

Question 5 Deleted for CBSE Board 2024 Exams

Question 6 Important Deleted for CBSE Board 2024 Exams

Question 7 Deleted for CBSE Board 2024 Exams

Question 8 Deleted for CBSE Board 2024 Exams

Question 9 Important Deleted for CBSE Board 2024 Exams

Question 10 Deleted for CBSE Board 2024 Exams

Question 11 Important Deleted for CBSE Board 2024 Exams You are here

Chapter 11 Class 9 Surface Areas and Volumes

Serial order wise

Last updated at May 29, 2023 by Teachoo

Question 11 The students of a Vidyalaya were asked to participate in a competition for making and decorating penholders in the shape of a cylinder with a base, using cardboard. Each penholder was to be of radius 3 cm and height 10.5 cm. The Vidyalaya was to supply the competitors with cardboard. If there were 35 competitors, how much cardboard was required to be bought for the competition? [ = 22/7] Here the pen holder is open at the top, but covered at the bottom Surface area of 1 penholder = Curved Surface area of cylinder with + Area of bottom circle with radius 3 cm Radius (r) of the circular end of cylindrical penholder = 3 cm Height (h) of penholder = 10.5 cm Surface area of 1 penholder = Curved Surface area of cylinder with + Area of circle with radius 3 cm = 2 rh + r2 = (2 22/7 3 10.5+22/7 (3)2) cm2 = (22/7 (2 3 10.5+(3)2)) cm2 = (22/7 (63+9)) cm2 = (22/7(72)) cm2 Area of cardboard required = Surface area of 35 penholder = 35 (22/7 " " 72) cm2 = 5 (22" " 72) cm2 = 7920 cm2 Therefore, 7920 cm2 cardboard sheet will be bought.