ABCD is a rectangle. P is a point on the side AB as shown in the given figure. If DP = 13, CP = 10 and BP = 6, then what is the value of AP?

Solution (By Examveda Team)

$$\eqalign{ & {\text{In }}\Delta PCB \cr & BC = \sqrt {{{10}^2} - {6^2}} = 8\,{\text{cm}} \cr & BC = AD = 8\,{\text{cm}} \cr & {\text{In }}\Delta ADP \cr & AP = \sqrt {{{13}^2} - {8^2}} \cr & AP = \sqrt {169 - 64} = \sqrt {105} \cr} $$