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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 minor segment.**

Answer:

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

Area of minor sector = (90/360°)×πr^{2}

= (¼)×(22/7)×10^{2}

Or, Area of minor sector = 78.5 cm^{2}

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 cm^{2}

Now, area of minor segment = area of minor sector – area of ΔAOB

= 78.5 – 50

= 28.5 cm^{2}

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