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What is the surface area of a cylindrical ring where the diameter of the cross section is 6.3 in and the center line has a length of 48 in? A. 529.86 in2 B. 256.22 in2 C. 950.02 in2 D. 849.73 in2

Respuesta :

djsdoodle
The surface area of the cylindrical ring is given by
Ï€dh
where d is the diameter and h is the height:
π•6.3•48 = 950.02
The answer is C. 950.02 in^2

Answer: Option 'C' is correct.

Step-by-step explanation:

Since we have given that

Diameter of cylinder = 6.3 inches

Radius of cylinder is given by

[tex]\frac{6.3}{2}\ in[/tex]

Height of cylinder = 48 in.

As we know the "Surface area of cylinder":

[tex]Area=2\pi rh\\\\Area=2\times \frac{22}{7}\times \frac{6.3}{2}\times 48\\\\Area=950.02\ in^2[/tex]

Hence, Option 'C' is correct.

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