document.write( "Question 1150800: Land in the shape of an isosceles triangle has a base of 130 m. An altitude from
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Algebra.Com's Answer #772266 by Edwin McCravy(20060)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "The green line BD is the altitude drawn from vertex of base angle B, so BD is\r\n" );
document.write( "perpendicular to AC. So ADB is a right triangle, so by the Pythagorean theorem,\r\n" );
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document.write( "\"AD%5E2%2BBD%5E2=AB%5E2\"\r\n" );
document.write( "\"AD%5E2%2B120%5E2=130%5E2\"\r\n" );
document.write( "\"AD%5E2=130%5E2-120%5E2\"\r\n" );
document.write( "\"AD%5E2=16900-14400\"\r\n" );
document.write( "\"AD%5E2=2500\"\r\n" );
document.write( "\"AD=sqrt%282500%29\"\r\n" );
document.write( "\"AD=50\"\r\n" );
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document.write( "CDB is also a right triangle, so also by the Pythagorean theorem,\r\n" );
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document.write( "\"BD%5E2%2BCD%5E2=BC%5E2\"\r\n" );
document.write( "\"120%5E2%2BCD%5E2=BC%5E2\"\r\n" );
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document.write( "But since BC = AD + CD\r\n" );
document.write( "          BC = 50 + CD\r\n" );
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document.write( "\"120%5E2%2BCD%5E2=%2850%2BCD%29%5E2\"\r\n" );
document.write( "\"120%5E2%2BCD%5E2=50%5E2%2B100CD%2BCD%5E2\"\r\n" );
document.write( "\"120%5E2=50%5E2%2B100CD\"\r\n" );
document.write( "\"14400-2500=100CD\"\r\n" );
document.write( "\"11900=100CD\"\r\n" );
document.write( "\"119=CD\"\r\n" );
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document.write( "\"AC+=+AD+%2B+CD\"\r\n" );
document.write( "\"AC+=+50+%2B+119\"\r\n" );
document.write( "\"AC+=+169\"\r\n" );
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document.write( "\"Area=expr%281%2F2%29base%2Aaltitude\"\r\n" );
document.write( "\"Area=expr%281%2F2%29AC%2ABD\"\r\n" );
document.write( "\"Area=expr%281%2F2%29169%2A120\"\r\n" );
document.write( "\"Area=matrix%281%2C2%2C10140%2Cm%5E2%29\"\r\n" );
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document.write( "Edwin

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