document.write( "Question 1004239: The Perimeters of two triangles is in the ratio of 2:4. The sum of their areas is 100 cm squared. Find the area of each triangle? \n" ); document.write( "
Algebra.Com's Answer #620753 by josgarithmetic(39618)\"\" \"About 
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That ratio is also 1:2.
\n" ); document.write( "Let x,y,z be the lengths of each side for the small triangle. The altitude is also in the same ratio between the two triangles (because linear measures).\r
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\n" ); document.write( "\n" ); document.write( "Small perimeter, x+y+z
\n" ); document.write( "Large perimeter, 2(x+y+z)
\n" ); document.write( "If h is the altitude of the small triangle, the 2h is altitude for large triangle.\r
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\n" ); document.write( "\n" ); document.write( "Small area, \"%281%2F2%29zh\"
\n" ); document.write( "large area, \"%281%2F2%292z%2A2h=%281%2F2%29%2A4%2Azh\"\r
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\n" ); document.write( "\n" ); document.write( "Their summed areas is \"100=%281%2F2%29zh%2B%281%2F2%29%2A4zh\"
\n" ); document.write( "\"200=zh%2B4zh=5zh\"
\n" ); document.write( "\"zh=200%2F5\"
\n" ); document.write( "\"zh=40\"
\n" ); document.write( "This can now be used for finding the value of each area.\r
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\n" ); document.write( "\n" ); document.write( "Small Triangle, \"%281%2F2%29%2A40=highlight%2820%29\"
\n" ); document.write( "Large Triangle, \"%281%2F2%29%2A4%2Azh=2%2A40=highlight%2880%29\"
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