Question

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)

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Answer 1

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.

Answer 2

Answer:

In square ABCD, ∠DAQ=x

o

∴ tanx=

AD

DQ

=DQ

QC=1−DQ=1−tanx

∠PAB=45

o

−x

o

∴ tan(45−x)=

AB

BP

=

1

BP

⟹

cos(45−x)

sin(45−x)

=BP

⟹

cosx+sinx

cosx−sinx

=BP

⟹

1+tanx

1−tanx

=BP

∵ PC=1−BP

⟹ PC=1−(

1+tanx

1−tanx

)

⟹ PC=

1+tanx

2tanx

∵ △PCQ=90

o

⟹ PQ

2

=PC

2

+QC

2

⟹ PQ

2

=(

1+tanx

2tanx

)

2

+(1−tanx)

2

⟹ PQ=(

1+tanx

1+tan

2

x

)

Perimeter of △PQC=PQ+QC+PC

=(

1+tanx

1+tan

2

x

)+(1−tanx)+(

1+tanx

2tanx

)

=(

1+tanx

1+tan

2

x+2tanx

)+(1−tanx)

=

1+tanx

(1+tanx)

2

+(1−tanx)

=1+tanx+1−tanx=2

Hence, Perimeter of △PQC is 2.