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14 Sep 2018, 01:23
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B
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Difficulty:
25%
(medium)
Question Stats:
78% (01:28) correct
22%
(01:43)
wrong
based on 34
sessions
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What is the ratio of the inner circle's circumference to the outer circle's circumference, if the inner circle's area is 80% of the outer circle's area?
A. 4 : 5
B. \(2 : \sqrt{5}\)
C. \(\sqrt{5}: 2\)
D. \(4 \pi: \sqrt{5}\)
E. It cannot be determined from the information given.
Attachment:
image018.jpg [ 3.86 KiB | Viewed 2229 times ]
Area Of Inner circle with radius "r" = 80/100 * Area of outer circle with radius "R"
\(\pi * r^2 = \frac{80}{100} * \pi * R^2\)
cancel pi and take square root on both sides
\(r = \sqrt\frac{8}{10} * R\)
\(\frac{r}{R} = \frac{2}{\sqrt5}\)
Multiply \(2*\pi\) in numerator and denominator of LHS to get the circumference
\(\frac{2*\pi*r}{2*\pi*R} = \frac{2}{\sqrt5}\)
Answer: B
\(\pi * r^2 = \frac{80}{100} * \pi * R^2\)
cancel pi and take square root on both sides
\(r = \sqrt\frac{8}{10} * R\)
\(\frac{r}{R} = \frac{2}{\sqrt5}\)
Multiply \(2*\pi\) in numerator and denominator of LHS to get the circumference
\(\frac{2*\pi*r}{2*\pi*R} = \frac{2}{\sqrt5}\)
Answer: B
14 Sep 2018, 10:30
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