document.write( "Question 1178014: A trapezium in which AD is parallel to BC and ∠ADC = ∠BCD = 90 °, The points A,B and C are (a,18), (12,-2) and (2,-7) respectively. Given that AB = 2BC, Find the value of a. Find the equation of AD. Find the equation of CD.Find the coordinates of D.Find the area of trapezium. \n" ); document.write( "
Algebra.Com's Answer #807183 by mananth(16946)\"\" \"About 
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\n" ); document.write( "D(BC )\r
\n" ); document.write( "\n" ); document.write( "Distance Equation Solution:
\n" ); document.write( "d=sqrt((2−12)2+(−7−(−2)))\r
\n" ); document.write( "\n" ); document.write( "d=sqrt((−10)2+(−5)\r
\n" ); document.write( "\n" ); document.write( "d=sqrt((100+25))\r
\n" ); document.write( "\n" ); document.write( "d(BC)=√125\r
\n" ); document.write( "\n" ); document.write( "d(AB) \r
\n" ); document.write( "\n" ); document.write( "Distance Equation Solution:
\n" ); document.write( "d=sqrt((a−12)^2+(18−(−2)^2))\r
\n" ); document.write( "\n" ); document.write( "d=sqrt((a^2-24a+144 +400))\r
\n" ); document.write( "\n" ); document.write( "d=sqrt( a^2-24a+544)\r
\n" ); document.write( "\n" ); document.write( "AB = 2BC\r
\n" ); document.write( "\n" ); document.write( "sqrt( a^2-24a+544) =10sqrt(5)\r
\n" ); document.write( "\n" ); document.write( "a^2-24a +544 =100*5\r
\n" ); document.write( "\n" ); document.write( "a^2-24a+44=0\r
\n" ); document.write( "\n" ); document.write( "a^2-22a-2a+44=0\r
\n" ); document.write( "\n" ); document.write( "(a-2)(a-22)=0\r
\n" ); document.write( "\n" ); document.write( "a=2 Or 22\r
\n" ); document.write( "\n" ); document.write( "slope of BC\r
\n" ); document.write( "\n" ); document.write( "Slope =y2−y1/x2−x
\n" ); document.write( " of AD\r
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\n" ); document.write( "\n" ); document.write( "−7−−2/2−12\r
\n" ); document.write( "\n" ); document.write( "−5/−10\r
\n" ); document.write( "\n" ); document.write( "1/2= slope of BC = slope of AD\r
\n" ); document.write( "\n" ); document.write( "equation of AD \r
\n" ); document.write( "\n" ); document.write( "use y = mx+c\r
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