document.write( "Question 1012750: The side AB of a rectangular ABCD is 12cm and the side AD is 9cm. P is a point in the side CD and Q is a point in the side BC such that angle APQ is a right angle. Taking DP as xcm, express BQ in terms of x \n" ); document.write( "

Algebra.Com's Answer #628931 by KMST(5328) $\"\"$ $\"About$

You can put this solution on YOUR website!

There are 3 angles at $\"P\"$ , whose measures add up to $\"180%5Eo\"$ .
\n" ); document.write( " $\"DPA%2BCPB%2BAPQ=180%5Eo\"$
\n" ); document.write( " $\"DPA%2BCPB%2B90%5Eo=180%5Eo\"$ --> $\"DPA%2BCPB=180%5Eo-90%5Eo\"$ --> $\"DPA%2BCPB=90%5Eo\"$
\n" ); document.write( "In right triangle $\"ADP\"$ , $\"DPA%2BDAP=90%5Eo\"$ .
\n" ); document.write( " $\"system%28DPA%2BCPB=90%5Eo%2CDPA%2BDAP=90%5Eo%29\"$ ---> $\"CPB=DAP\"$ .
\n" ); document.write( "In right triangle $\"PCQ\"$ , $\"CQP%2BCPB=90%5Eo\"$ .
\n" ); document.write( " $\"system%28DPA%2BCPB=90%5Eo%2CCBP%2BCPB=90%5Eo%29\"$ ---> $\"DPA=CQP\"$ .
\n" ); document.write( "So, triangles $\"ADP\"$ and $\"PCQ\"$ are similar, because they have congruent pairs of angles.
\n" ); document.write( "Since they are similar corresponding sides are proportional:
\n" ); document.write( " $\"DP%2FDA=CQ%2FCP\"$
\n" ); document.write( " $\"x%2F9=CQ%2F%2812-x%29%2FCQ\"$ ---> $\"CQ=x%2812-x%29%2F9\"$
\n" ); document.write( " $\"BQ=BC-CQ\"$ ---> $\"BQ=9-CQ\"$
\n" ); document.write( " $\"system%28BQ=9-CQ%2CCQ=x%2812-x%29%2F9%29\"$ ---> $\"highlight%28BQ=9-x%2812-x%29%2F9%29\"$ \n" ); document.write( "

\n" );