Since all circles are similar, a proportion can be set up using the circumference and diameter of each circle. Substitute the va
lues d1 = 1, C1 = π, and d2 = 2r into the proportion.
StartFraction C 1 Over d 1 EndFraction = StartFraction C 2 Over d 2 EndFraction
Which shows how to correctly solve for C2, the circumference of any circle with radius r?
A. Because StartFraction pi Over 1 EndFraction = StartFraction C 2 Over 2 r EndFraction , C 2 = 2 r pi
B. Because StartFraction 1 Over pi EndFraction = StartFraction C 2 Over 2 r EndFraction ,
2 = StartFraction 2 r Over pi EndFraction
C. Because StartFraction pi Over 2 r EndFraction = StartFraction C 2 Over 1 EndFraction , C 2 = StartFraction pi Over 2 r EndFraction
D. Because StartFraction pi Over 1 EndFraction = StartFraction C 2 Over 4 r EndFraction , C 2 = 4 r pi
StartFraction C 1 Over d 1 EndFraction = StartFraction C 2 Over d 2 EndFraction
Which shows how to correctly solve for C2, the circumference of any circle with radius r?
A. Because StartFraction pi Over 1 EndFraction = StartFraction C 2 Over 2 r EndFraction , C 2 = 2 r pi
B. Because StartFraction 1 Over pi EndFraction = StartFraction C 2 Over 2 r EndFraction ,
2 = StartFraction 2 r Over pi EndFraction
C. Because StartFraction pi Over 2 r EndFraction = StartFraction C 2 Over 1 EndFraction , C 2 = StartFraction pi Over 2 r EndFraction
D. Because StartFraction pi Over 1 EndFraction = StartFraction C 2 Over 4 r EndFraction , C 2 = 4 r pi
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