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Consider the centre of circle to be O, ends of chord as A, B and their mid-point as M.
We have, radius, r = 5√2 cm and chord length, l = 10 cm.
Now you can observe, ∆OAM & ∆OBM are isosceles & right angled at M.
Here, angle AOM = angle BOM = 45°, then area of ∆ AOB = 25 cm²
and the area of sector AOB = ¼(circle area) = (12.5)π cm², since angle AOB = 90°.
Then area of segments are 50(π-2)/4 and 50(3π+2)/4.
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we have r=√50cm
 we know  of the segment make angle Ф at centre
       we have sinФ/2=5/√50=45 degree
                                     ⇒ Ф=90
   Area of segment=(Фπ/180-sinФ)r²/2
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Actually, sin45°=1/√2
i agrre answer is 14.26cm2
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