document.write( "Question 847874: Find the area of a circle in which a chord AB is 48cm and the perpendicular bisector of the chord, originating at the centre of the circle, is 10cm. \n" ); document.write( "

Algebra.Com's Answer #510731 by hamsanash1981@gmail.com(151) $\"\"$ $\"About$

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Let the center of the circle be O and O be the point at which it bisects the chord AB such that OPA forms the right angled triangle and OA is radius of the circle then OP = 10cm and AP = 24
\n" ); document.write( "By Pythagoras theorem,\r
\n" ); document.write( "\n" ); document.write( " $\"OA%5E2+=+OP%5E2%2BAP%5E2\"$
\n" ); document.write( " $\"OA%5E2+=+10%5E2+%2B+24%5E2\"$
\n" ); document.write( " $\"OA+=+sqrt%28676%29\"$
\n" ); document.write( "OA = 26 \r
\n" ); document.write( "\n" ); document.write( "RADIUS = 26
\n" ); document.write( "AREA = 22/7*26*26
\n" ); document.write( "=2124.57 Sq cm
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