Algebra.Com's Answer #94226 by Edwin McCravy(20055)![\"\"](\"/images/stars/stars-13.gif\")  
You can put this solution on YOUR website!
ABCD is a trapezoid with BC perpendicular to AB and BC perpendicular to CD, AB = 13, BC = 12, and CD = 8. A line segment is drawn from A to E, the midpoint of CD. a. Find the area of triangle AED. b. Find the perimeter of triangle AED.
\n" );
document.write( "\r\n" );
document.write( "Make the drawing below. We are given that BC = 12, and since\r\n" );
document.write( "E is given as the midpoint of CD, which is 8, then CE and ED are 4 each. \r\n" );
document.write( "\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( "Draw EF parallel to BC, then EF = BC = 12, and\r\n" );
document.write( "since BF = CE = 4 and AB = 13, then by subtraction,\r\n" );
document.write( "AF = AB - BF = 13 - 4 = 9:\r\n" );
document.write( "\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( "We can find the area of trapezoid AFED using the\r\n" );
document.write( "formula\r\n" );
document.write( "\r\n" );
document.write( " /b1 + b2\\r\n" );
document.write( "Area = |---------|h \r\n" );
document.write( " \ 2 /\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "Area of trapezoid AFED =  =  =  =  =  \r\n" );
document.write( "\r\n" );
document.write( "We can find the area of triangle AFE using the \r\n" );
document.write( "formula:\r\n" );
document.write( "\r\n" );
document.write( "Area = {{(bh)/2}}} =  =  =  =  \r\n" );
document.write( "\r\n" );
document.write( "Now since\r\n" );
document.write( "\r\n" );
document.write( "Area of trapezoid AFED = Area of triangle AFE + Area of triangle AED\r\n" );
document.write( "\r\n" );
document.write( " 78 = 54 + Area of triangle AED\r\n" );
document.write( "\r\n" );
document.write( "so Area of triangle AED = 78 - 54 = 24\r\n" );
document.write( "\r\n" );
document.write( "So that's the answer to part a.\r\n" );
document.write( "\r\n" );
document.write( "------------------------------------------\r\n" );
document.write( "\r\n" );
document.write( "Now we must find the perimeter of triangle AED.\r\n" );
document.write( "\r\n" );
document.write( "We have one side of it, DE, which is 4.\r\n" );
document.write( "\r\n" );
document.write( "We can find side AE for it is the hypotenuse of right triangle AFE.\r\n" );
document.write( "\r\n" );
document.write( "We use the Pythagorean theorem\r\n" );
document.write( "\r\n" );
document.write( "c² = a² + b²\r\n" );
document.write( "\r\n" );
document.write( "AE² = AF² + EF² = 9² + 12² = 81 + 144 = 225\r\n" );
document.write( "\r\n" );
document.write( "so  =  =  \r\n" );
document.write( "\r\n" );
document.write( "Now we only need side DA.\r\n" );
document.write( "\r\n" );
document.write( "To do this we need to draw DG parallel to BC and EF.\r\n" );
document.write( "\r\n" );
document.write( "then DG = EF = BC = 12, and\r\n" );
document.write( "since GF = DE = 4 and AF = 9, then by subtraction,\r\n" );
document.write( "AG = AF - GF = 9 - 4 = 5:\r\n" );
document.write( "\r\n" );
document.write( " \r\n" );
document.write( "\r\n" );
document.write( "We can find side AD for it is the hypotenuse of right triangle AGD.\r\n" );
document.write( "\r\n" );
document.write( "We use the Pythagorean theorem\r\n" );
document.write( "\r\n" );
document.write( "c² = a² + b²\r\n" );
document.write( "\r\n" );
document.write( "AD² = AG² + DG² = 5² + 12² = 25 + 139 = 169\r\n" );
document.write( "\r\n" );
document.write( "so  =  =  \r\n" );
document.write( "\r\n" );
document.write( "Now the perimeter of triangle AED is the sum of\r\n" );
document.write( "its three sides:\r\n" );
document.write( "\r\n" );
document.write( "Perimeter = AE + DE + AD = 15 + 4 + 13 = 32\r\n" );
document.write( "\r\n" );
document.write( "Edwin \n" );
document.write( " |