How i solve the ncert class 10 chapter areas related to circle of exercise 12.2 question no. 4 . Find the best and easiest way to solve the question of this chapter . Give me the simplest solution of this question.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)
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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) Q.4(1),(2)
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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 = (90/360°)×πr2
= (¼)×(22/7)×102
Or, Area of minor sector = 78.5 cm2
Also, area of ΔAOB = ½×OB×OA
Here, OB and OA are the radii of the circle i.e. = 10 cm
So, area of ΔAOB = ½×10×10
= 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