Question

In the figure, ABCD is a square of side 1 dm and $\angle PAQ={45}^o$ .
The perimeter (in dm) of the triangle PQC is

• A. $2$
• B. $1+$ $\sqrt{2}$
• C. $2\sqrt{2}-1$
• D. $1+$ $\sqrt{3}$

Answer 1

The correct option is A $2$

In square ABCD, $\angle DAQ=x^o$

$\therefore \tan x=\frac{DQ}{AD}=DQ$

$QC=1-DQ=1-\tan x$

$\angle PAB={45}^o-x^o$

$\therefore \tan (45-x)=\frac{BP}{AB}=\frac{BP}{1}$

$⟹\frac{\sin (45-x)}{\cos (45-x)}=BP$

$⟹\frac{\cos x-\sin x}{\cos x+\sin x}=BP$

$⟹\frac{1-\tan x}{1+\tan x}=BP$

$\because PC=1-BP$

$⟹PC=1-\left(\frac{1-\tan x}{1+\tan x}\right)$

$⟹PC=\frac{2\tan x}{1+\tan x}$

$\because △PCQ={90}^o$

$⟹P{Q}^2=P{C}^2+Q{C}^2$

$⟹P{Q}^2={\left(\frac{2\tan x}{1+\tan x}\right)}^2+(1-\tan x{)}^2$

$⟹PQ=\left(\frac{1+{\tan }^{2}x}{1+\tan x}\right)$

Perimeter of $△PQC=PQ+QC+PC$

$=\left(\frac{1+{\tan }^{2}x}{1+\tan x}\right)+(1-\tan x)+\left(\frac{2\tan x}{1+\tan x}\right)$

$=\left(\frac{1+{\tan }^{2}x+2\tan x}{1+\tan x}\right)+(1-\tan x)$

$=\frac{(1+\tan x{)}^2}{1+\tan x}+(1-\tan x)$

$=1+\tan x+1-\tan x=2$

Hence, Perimeter of $△PQC$ is $2$.