Question

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''​

Answer 1

Answer:

63.25cm2

Step-by-step explanation:

h=4cm

circumference=11cm

$2\pi r$=11

r=$\frac{7}{4\\}$

TSA of a cylinder=$2\pi r(r+h)$=2*22/7*7/4(7/4+4)

=63.25cm^2

Answer 2

Given :

• Dimensions of rectangular paper = 11 cm × 4 cm
• Height of cylinder = 4 cm

To find :

• Surface area of cylinder

Solution :

When the rectangle is folded into a cylinder , one of its side becomes the height while the other becomes the circumference of the base circle.

When the 4 cm side becomes height , the circumference of the base circle is 11 cm .

Therefore ,

2πr = 11

$\tt 2 \times \dfrac{22}{7} \times r = 11 \\ \\ \tt \implies r = \frac{11 \times 7}{2 \times 22} \\ \\ \implies \boxed{ \tt r = 1.75 \: cm }$

Now , we have the height of the cylinder as 4 cm and radius of the base circle as 1.75 cm .

The surface area of cylinder is given by :

$\boxed{ \sf surface \: area \: of \: cylinder = 2 \pi r(r + h)}$

$\tt = 2 \times \dfrac{22}{7} \times 1.75(1.75 + 4) \\ \\ \tt = 2 \times \frac{22}{7} \times \frac{175}{100} \left( \frac{575}{100} \right) \\ \\ \tt = 2 \times 22 \times \frac{25}{100} \times \frac{23}{4} \\ \\ \tt = 11 \times 23 \times \frac{1}{4} \\ \\ = \tt \frac{253}{4} \\ \\ \tt = 63.25 \: {cm}^{2}$