A chord of a circle of radius 10 cm subtends a right angle at the centre. Find the area of the corresponding:

(i) minor segment

(ii) major sector. (Use π = 3.14) Asked by Abhisek | 1 year ago |  103

##### 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 ago

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