document.write( "Question 1099501: In parallelogram ABCD, line EF is perpendicular to line AB and line GH is perpendicular to BD. Compute the area of the parallelogram, given AD=22, AE=13, EF=12, BG=17, and GH=8. \r
\n" ); document.write( "\n" ); document.write( "link for the problem https://gyazo.com/fc882226a5f618fc9931d6a27147f4ac \n" ); document.write( "

Algebra.Com's Answer #713934 by ikleyn(52787) $\"\"$ $\"About$

You can put this solution on YOUR website!
.
\n" ); document.write( "

\r\n" );
document.write( "1.  From the right-angled triangle AFE  sin( < FAE) = .\r\n" );
document.write( "\r\n" );
document.write( "    Hence, sin( < BAD) = sin( < FAE) = .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "2.  From the right-angled triangle BHG  sin( < GBH) = .\r\n" );
document.write( "\r\n" );
document.write( "    Hence,  sin( < CBD) = sin( < GBH) = .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "3.  Angles CBD and ADB are, obviously, congruent.\r\n" );
document.write( "\r\n" );
document.write( "    Therefore, sin( < ADB) = sin( < CBD) = .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "4.  Thus in triangle ADB we know\r\n" );
document.write( "\r\n" );
document.write( "    AD = 22,  sin( < BAD) =   and  sin( < ADB) = .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "    Having sines of two angles at the base of the triangle ADB, we can find sin( < ABD):\r\n" );
document.write( "\r\n" );
document.write( "    sin( < ABD) = sin(180 - < BAD - < ADB) = sin( < BAD + < ADB) = sin*cos + cos*sin =  = \r\n" );
document.write( "\r\n" );
document.write( "    =  = .\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "5.  Now apply the sine law theorem to find the side AB:\r\n" );
document.write( "\r\n" );
document.write( "     =   ====>  |AB| = 22* =  =  =  = 10.4.\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "\r\n" );
document.write( "6.  Now calculate the area of the parallelogram ABCD by applying the formula\r\n" );
document.write( "\r\n" );
document.write( "     = |AD|*|AB|*sin( < A) = 22*10.4* = 22*0.8*12 = 211.2.\r\n" );
document.write( "

\r
\n" ); document.write( "\n" ); document.write( "Answer. The area of the parallelogram ABCD is 211.2 square units.\r
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "

\n" );