document.write( "Question 877351: an altitude of a triangle is 2 cm longer than its base. what are the dimensions of the altitude and the base if the area of the triangle is 40 cm^2 \n" ); document.write( "
Algebra.Com's Answer #529303 by nerdybill(7384)\"\" \"About 
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an altitude of a triangle is 2 cm longer than its base. what are the dimensions of the altitude and the base if the area of the triangle is 40 cm^2
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\n" ); document.write( "Let b = base
\n" ); document.write( "then
\n" ); document.write( "b+2 = altitude
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\n" ); document.write( "area of any triangle = (1/2)bh
\n" ); document.write( "40 = (1/2)b(b+2)
\n" ); document.write( "80 = b(b+2)
\n" ); document.write( "80 = b^2+2b
\n" ); document.write( "0 = b^2+2b-80
\n" ); document.write( "0 = (b+10)(b-8)
\n" ); document.write( "b = {-10, 8}
\n" ); document.write( "can't be negative so:
\n" ); document.write( "b = 8 cm (base)
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\n" ); document.write( "altitude:
\n" ); document.write( "b+2=8+2= 10 cm\r
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