a rectangular piece of paper 11cm*4cm is folded without overlapping to make a cylinder of height 4cm . find the surface area of the closed cylinder.
Class 10 please help fast we ith correct answer
AND I WANT AREA OF CLOSED SURFACE AREA. I DON'T WANT THE VOLUME PLEASE
Answers
Answered by
1
Answer:
Circumference of base of cylinder = 11cm
2*pi*r = 11
So r = 11/(2*pi). → retain as such for easy calculation
Volume of cylinder = pi*(r^2)*h
Substituting for r,
pi*((11/(2*pi))^2)*4
= 11*11*4/(4*pi)
= 11*11*7/22 (using pi=22/7)
= 77/2
= 38.5 cu cm
Step-by-step explanation:
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Answered by
4
Answer:
Given that,
Length of the rectangiular paper= 11cm
Breadth of the rectangular paper= 4cm
Circumfernce of circular part of cyclinder = 2∏r
also given that cyclinder is rolled along its length
HenceCircumfernce of circular part of cyclinder =Length of the rectangiular paper
2∏r =11cm
2 X 22 / 7 X r = 11
44 / 7 X r = 11
r = 7 / 4 cm
Height of cylinder =Breadth of the rectangular paper = h =4cm
Volume of cyclinder =∏r2h
=22 / 7 x (7/4)2X 4
= 38.5cm3
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