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what is the surface area of a cylindrical ring where the diameter of th…

Question

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) 256.22 in²
b) 849.73 in²
c) 950.02 in²
d) 529.86 in²

Explanation:

Step1: Find the circumference of the cross - section

The formula for the circumference of a circle is \(C=\pi d\). Given \(d = 6.3\) in, then \(C=\pi\times6.3\approx3.14\times6.3 = 19.782\) in.

Step2: Calculate the surface area of the cylindrical ring

The surface area \(A\) of a cylindrical ring is given by \(A = C\times l\), where \(l\) is the length of the center - line. Given \(l = 48\) in and \(C\approx19.782\) in. Then \(A=19.782\times48=950.016\approx950.02\) \(in^{2}\).

Answer:

C. \(950.02\ in^{2}\)