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If the radius of a cylinder was shrunk down to a quarter of its orignal size and the height was reduced to a third of its original size, what would be the formula to find the modified surface area?

Respuesta :

ivycoveredwalls

Surface area, S, is

[tex] S=2\pi r^2 + 2\pi rh [/tex]


Replace the radius with (1/4)r and the height with (1/3)h and you get the new surface area


[tex] S^\prime = 2\pi(\frac{1}{4}r)^2 + 2\pi(\frac{1}{4}r)(\frac{1}{3}h) \\
= 2\pi(\frac{1}{16})r^2 + 2\pi(\frac{1}{12})rh \\
=\frac{1}{8}\pi r^2 + \frac{1}{6} \pi rh [/tex]

lublana

Answer:

[tex]\frac{\pi r^2}{8}+\frac{1}{6}\pi rh[/tex]

Step-by-step explanation:

Let h be the original height of cylinder and r be the original radius of cylinder

According to question

[tex]r'=\frac{1}{4}r[/tex]

[tex]h'=\frac{1}{3}h[/tex]

Surface area of cylinder=[tex]2\pi r(r+h)[/tex]

Using the formula

Surface area of cylinder=[tex]2\pi \times \frac{1}{4}r(\frac{1}{4}r+\frac{1}{3}h)[/tex]

Surface area of cylinder=[tex]\frac{\pi r^2}{8}+\frac{1}{6}\pi rh[/tex]

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