document.write( "Question 1132209: O, is the centre of circle HKL. Side HK= 16cm, side HL = 10cm and the perpendicular from O to HK is 4cm. With the aid of a diagram what is the lenght of the perpendicular from O to HL \n" ); document.write( "

Algebra.Com's Answer #749171 by Edwin McCravy(20060) $\"\"$ $\"About$

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document.write( "Draw in radii OK, OH and OL and let their lengths be r.\r\n" );
document.write( "Let OA = 4, the perpendicular from O to HK.\r\n" );
document.write( "Let the red line OB = x, the perpendicular from O to HL, the length\r\n" );
document.write( "of which is what we are to find.\r\n" );
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document.write( "Triangle OHK is isosceles, so the perpendicular from O to HK, which \r\n" );
document.write( "is 4, bisects the base HK.  HK is given as 16, so AH = AK = 8.\r\n" );
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document.write( "Similarly, triangle OHL is isosceles, so the perpendicular from O to HL,\r\n" );
document.write( "which is our unknown x, bisects the base HL.  HL is given as 10, so \r\n" );
document.write( "BH = BL = 5.\r\n" );
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document.write( "We use the Pythagorean theorem on either right triangle OAH or OAK:\r\n" );
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document.write( "r² = 8²+4² = 64+16 = 80\r\n" );
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document.write( "We use the Pythagorean theorem on either right triangle OBH or OBL:\r\n" );
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document.write( "r² = x²+5², and we have already found r² = 80, so\r\n" );
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document.write( "80 = x²+25\r\n" );
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document.write( "55 = x²\r\n" );
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document.write( "So the solution is \r\n" );
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document.write( "Edwin

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