the key for solving the above problem is knowing how to find the perimeter of a square inscribed inside a circle!

part one
finding the circal circumference

circal r = 2

circumference whole circle 2*pi*r = 4*pi

finding 3/4 of circumference = 3/4*4*pi = 3*pi

part two
draw a square inscribed inside a circle (look at the attachment)

the diagonal of the square is r=2*2=4 (diameter of circle).

the square forms two 45-45-90 triangles.

sqrt(2)*x=4

x=4/sqrt(2)

so the ratio is 1:1:sqrt(2) = 4/sqrt(2):4/sqrt(2):4

meaning the line in the attachment marked (1) equal 4/sqrt(2) and half the line is [4/sqrt(2)]/2

now:

from the triangle with height 5 we can make two smaller right triangles with base [4/sqrt(2)]/2.

using the Pythagorean theorem will yield [[4/sqrt(2)]/2]^2+5^2=x^2

2+25 = x^2

x = sqrt(27)

so we can conclude that the sign circumference is pi*3+2*sqrt(27) or in other words pi*3+3*2sqrt(3) = pi*3+6sqrt(3)

and as was said before the answer is then (B)

Attachments

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