SOLUTION: 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

Algebra ->  Rectangles -> SOLUTION: 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      Log On


   

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
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
There are 3 angles at P , whose measures add up to 180%5Eo .
DPA%2BCPB%2BAPQ=180%5Eo
DPA%2BCPB%2B90%5Eo=180%5Eo-->DPA%2BCPB=180%5Eo-90%5Eo-->DPA%2BCPB=90%5Eo
In right triangle ADP , DPA%2BDAP=90%5Eo .
system%28DPA%2BCPB=90%5Eo%2CDPA%2BDAP=90%5Eo%29--->CPB=DAP .
In right triangle PCQ , CQP%2BCPB=90%5Eo.
system%28DPA%2BCPB=90%5Eo%2CCBP%2BCPB=90%5Eo%29--->DPA=CQP .
So, triangles ADP and PCQ are similar, because they have congruent pairs of angles.
Since they are similar corresponding sides are proportional:
DP%2FDA=CQ%2FCP
x%2F9=CQ%2F%2812-x%29%2FCQ--->CQ=x%2812-x%29%2F9
BQ=BC-CQ--->BQ=9-CQ
system%28BQ=9-CQ%2CCQ=x%2812-x%29%2F9%29--->highlight%28BQ=9-x%2812-x%29%2F9%29