document.write( "Question 1137384: The area of triangle ABC is 231 square inches, and point P is marked on side AB so that AP:PB = 3:4. What are the areas of triangles APC and BPC? \n" ); document.write( "
Algebra.Com's Answer #755257 by ikleyn(52797)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Draw the perpendicular CD from the vertex C of the triangle ABC to its side AB (so the point D lies on the straight line AB).\r\n" );
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document.write( "Then CD is the common height in triangles ABC, APC and BPC.\r\n" );
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document.write( "Since the triangle area is half the product of the base length and the height drawn to the base, it is clear that\r\n" );
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document.write( "    - the area of the triangle ABC is the sum of areas of triangles APC and BPC,  and\r\n" );
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document.write( "    - the ratio of the area of triangle APC to the area of triangle BPC  is  the same as the ratio of the segment \r\n" );
document.write( "      lengths |AP| to |BP|, i.e. 3:4.\r\n" );
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document.write( "Then it is clear that area of the triangle APC = 3x and the area of the triangle BPC = 4x, where 3x + 4x = 231.\r\n" );
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document.write( "Hence,  7x = 231,  x = \"231%2F7\" = 33,  \r\n" );
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document.write( "area of the triangle APC = 3x = 3*33 = 99 square inches and area of the triangle BPC = 4*33 = 132 square inches.     ANSWER\r\n" );
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\n" ); document.write( "\n" ); document.write( "In the post by the other tutor,  I see the words  \"the solution is attached\".\r
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\n" ); document.write( "\n" ); document.write( "I do not understand their meaning,  because I do not see  \"the attached solution\"  there.\r
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