Q. In the figure, ABCD is a square of side 1 dm and Angle PAQ = 45°.

1.	Q. In the figure, ABCD is a square of side 1 dm and Angle PAQ = 45°. The perimeter (in dm) of the triangle PQC is(NTSE STAGE-II 2014)PLS ANSWER IT FAST.DONT ANSWER IF DONT KNOW.
Answer» To FIND: The PERIMETER of $\Delta PQC$ . SOLUTION: Given, $\angle PAQ=45^{\circ}$ and $AB=BC=CD=DA=1\:dm$ Let, $\angle DAQ=x^{\circ}$ Then, $\angle BAP=90^{\circ}-(45^{\circ}+x^{\circ})=45^{\circ}-x^{\circ}$ For the triangle $\Delta ADQ$ . Here $\Delta ADQ$ is a right-angled triangle whose HEIGHT is $DQ$ and base is $AD$ . Then, $\quad tanx=\frac{DQ}{AD}$ $\Rightarrow tanx=\frac{DQ}{1}$ $\Rightarrow DQ=tanx$ $\Rightarrow 1-QC=tanx\quad[\because CD=DQ+QC]$ $\Rightarrow \boxed{QC=1-tanx}$ For the triangle $\Delta ABP$ . Here $\Delta ABP$ is a right-angled triangle whose height is $BP$ and base is $AB$ . Then, $\quad tan(45^{\circ}-x)=\frac{BP}{AB}$ $\Rightarrow tan(45^{\circ}-x)=\frac{BP}{1}$ $\Rightarrow BP=\frac{tan45^{\circ}-tanx}{1+tan45^{\circ}\:tanx}$ $\Rightarrow 1-PC=\frac{1-tanx}{1+tanx}\quad[\because BC=BP+PC]$ $\Rightarrow PC=1-\frac{1-tanx}{1+tanx}$ $\Rightarrow \boxed{PC=\frac{2\:tanx}{1+tanx}}$ For the triangle $\Delta PQC$ . Here $\Delta PQC$ is a right-angled triangle whose height is $PC$ and base is $QC$ . Then $\quad {PQ}^{2}={PC}^{2}+{QC}^{2}$ $\Rightarrow {PQ}^{2}=\big(\frac{2\:tanx}{1+tanx}\big)^{2}+(1-tanx)^{2}$ $\Rightarrow {PQ}^{2}=\frac{4\:tan^{2}x+1-2\:tan^{2}x+tan^{4}x}{(1+tanx)^{2}}$ $\Rightarrow {PQ}^{2}=\big(\frac{1+tan^{2}x}{1+tanx}\big)^{2}$ $\Rightarrow \boxed{PQ=\frac{1+tan^{2}x}{1+tanx}}$ Finding the perimeter of the triangle $\Delta PQC$ . $\therefore$ the perimeter of the triangle $\Delta$ is $=PQ+QC+PC$ $=\frac{1+tan^{2}x}{1+tanx}+(1-tanx)+\frac{2\:tanx}{1+tanx}$ $=\frac{1+tan^{2}x+1-tan^{2}x+2tanx}{1+tanx}$ $=\frac{2+2\:tanx}{1+tanx}$ $=\frac{2\:(1+tanx)}{1+tanx}$ $=\bold{2}$ Answer: Therefore, the perimeter of the triangle $\Delta PQC$ is 2 dm.