1 Answer
Solution :-
Here AB be the chord which is subtending an angle 90° at the center O.
It is given that the radius (r) of the circle = 10 cm
(i) Area of minor sector = \(\dfrac{90}{360°}×πr^2 \)
= \( (\dfrac{1}{4})×(\dfrac{22}{7})×10^2\)
Or, Area of minor sector = 78.5 cm2
Also, area of ΔAOB = \( \dfrac{1}{2}\)×OB×OA
Here, OB and OA are the radii of the circle i.e. = 10 cm
So, area of ΔAOB = \( \dfrac{1}{2}×10×10\)
= 50 cm2
Now, area of minor segment = area of minor sector – area of ΔAOB
= 78.5 – 50
= 28.5 cm2
(ii) Area of major sector = Area of circle – Area of minor sector
= (3.14×102)-78.5
= 235.5 cm2
Answered by Pragya Singh | 1 year agoRelated Questions
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