document.write( "Question 706561: Can someone assist with this problem? PLEASE?\r
\n" ); document.write( "\n" ); document.write( "I have a triangle, divided into two triangles. I have to figure out the area of the entire one triangle as a whole. The left side of the triangle has a length of 10 meters, and it gives two base lengths, one is 8 meters of the one triangle on the left, and the other base length is 15 meters, with no additional info. given, please help solve.\r
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Algebra.Com's Answer #435274 by Edwin Parker(36)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "It can be worked if the triangle on the left is isosceles,\r\n" );
document.write( "like this:\r\n" );
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document.write( " \"matrix%286%2C1%2C%0D%0A%0D%0Aor%2C+a%2C+right%2Ctriangle%2Clike%2C%22this--%3E%22%29\" \r\n" );
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document.write( "If it's the first case, we draw the height, h, or altitude (in green), and\r\n" );
document.write( "dividing the 8 base into two segments 4 each\r\n" );
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document.write( "In either case we use the Pythagorean theorem to get the height h\r\n" );
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document.write( "4˛ + h˛ = 10˛                                 8˛ + h˛ = 10˛\r\n" );
document.write( "16 + h˛ = 100                                 64 + h˛ = 100\r\n" );
document.write( "     h˛ = 84                                       h˛ = 36\r\n" );
document.write( "      h = √84                                       h = √36\r\n" );
document.write( "      h = √4*21                                     h = 6\r\n" );
document.write( "      h = 2√21\r\n" );
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document.write( "The formula for the area of a triangle is A = \"1%2F2\"base×height\r\n" );
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document.write( "We have the base in both cases as 8+15 or 23.\r\n" );
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document.write( "The case on the left is \r\n" );
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document.write( "A = \"1%2F2\"(23)2√21\r\n" );
document.write( "  = 23√21\r\n" );
document.write( "  = 105.4 square meters, approximately\r\n" );
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document.write( "On the right,\r\n" );
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document.write( "A = \"1%2F2\"(23)(6) \r\n" );
document.write( "  = 69 square meters\r\n" );
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document.write( "Edwin
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