Question

area of sector of a circle is 1/6 of area of circle find degree measure of its minor arc

Answer 1

Answer:

$degree\:measure \: of \\minor \: arc = 60 \degree$

Step-by-step explanation:

$Let \: radius \: of \: a \\ circle \:be\: r\: and \\ degree\:measure \: of \\minor \: arc = \theta$

According to the problem given,

area of sector of a circle is 1/6 of area of circle

$\frac{\theta}{360}\times \pi r^{2}=\frac{1}{6}\times \pi r^{2}$

$\implies \theta= \frac{360\times \pi r^{2}}{6\times \pi r^{2}}$

After cancellation, we get

$\theta = 60 \degree$

Therefore,

$degree\:measure \: of \\minor \: arc = 60 \degree$

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Answer 2

Answer:

degree of minor arc = 60°

Step-by-step explanation:

let the degree of minor arc be x

area of sector = 1/6 area of circle

   x/360°×π×r² = 1/6×π×r²    (cancel πr² in both sides)

            x/360° = 1/6

                     x = 360°/6

                     x = 60°

∴ degree of minor arc is 60°

Hope this answer was useful !!!