document.write( "Question 1191970: In an isosceles triangle ABC, side AB is twice as long as AC. The perimeter of the triangle is 200cm. Find the length of AC.
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Algebra.Com's Answer #823839 by ikleyn(52781) $\"\"$ $\"About$

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document.write( "Since the triangle is isosceles (given) and since side AB is twice as long as AC,\r\n" );
document.write( "it means that AC is the base of the triangle, while AB and BC are the lateral sides.\r\n" );
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document.write( "It is only one possible configuration for the triangle to exist.\r\n" );
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document.write( "So, if x is the length of the base AC, then the lateral sides AB and BC have the lengths 2x, each.\r\n" );
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document.write( "Then for the perimeter we have this equation\r\n" );
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document.write( "    AC + AB + BC = 200 cm,\r\n" );
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document.write( "or\r\n" );
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document.write( "    x  + 2x + 2x = 200,\r\n" );
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document.write( "         5x      = 200\r\n" );
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document.write( "          x      = 200/5 = 40.\r\n" );
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document.write( "ANSWER.  The base AC is 40 cm long.  The lateral sides are  2*40 = 80 cm long: they are congruent.\r\n" );
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