Match the following with the area of shaded regions in cm2.
A
250π3
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B
50π3−25√3
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C
50π3
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Solution
Here, r=10 cm and θ=60∘.
For the given figure: Areaofminorsector =θ360×πr2 =60360×π×10×10 =50π3cm2
For the given figure: Areaofmajorsector=Areaofcircle−Areaofminorsector
Areaofcircle =πr2 =π×102 =100πcm2
⇒Areaofmajorsector =100π−50π3 =250π3cm2
For the given figure: Areaofthesegment=Areaofminorsector−Areaof△OAB Areaof△OAB =√34×10×10 =25√3cm2
[△OAB is an equilateral triangle with side 10 cm.] ⇒ Area of the segment =50π3−25√3cm2