A strange quadrilateral

Let's do some name calling: $AP=x, PB=y$ . Using the information given to us, we can also write $AD = x; AB=x+y; DQ=AQ-AD=(x+y)-x=y$ .

Now let's write the expression for the diagonal BD, applying the Pythagorean theorem to, say, $\Delta ADB$ : $BD^2=AB^2+AD^2$ $BD^2=(x+y)^2+x^2=2x^2+2xy+y^2=36$

We might need it somewhere, who knows? Anyway, thinking about the area of the polygon we want to find, the first urge is to break it into two parts: trapezoid $APCD$ and right triangle $DQC$ . Their corresponding areas are $A_{APCD}=\frac{x+(x+y)}{2}x=(x+\frac{y}{2})x=x^2+\frac{xy}{2}$ and $A_{\Delta DQC}=\frac{1}{2}y(x+y)=\frac{1}{2}y^2+\frac{xy}{2}$

Let's sum them up: $A_{APCQ}=x^2+\frac{xy}{2}+\frac{xy}{2}+\frac{y^2}{2}=\frac{1}{2}(2x^2+2xy+y^2)$

Look at you! You're just half of $BD^2$ , or simply $\frac{36}{2}=\boxed{18}$

Please upvote It if you like it

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First of all seeing the conditions in the questions I assumed that $ABCD$ is a square and $Q$ coincides $D$ and $P$ coincides $B$

Now as $BP$ is a diagonal $\therefore BP=\sqrt{2}\cdot AB$

i.e $6=\sqrt{2}\cdot AB \Rightarrow AB=3\sqrt 2$

As area of square is $AB^2$ where $AB$ is side of a square

$\Rightarrow \therefore$ Area= $(2\sqrt 3)^2 \Rightarrow 18cm^2$

Hence the answer is $\Huge\Rightarrow\color{royalblue}{\boxed{18}}$

Because BD=6 Then, $AD^2$ + $AB^2$ = $6^2$ = 36

From above figure total area is= $AD^2$ + $\frac{1}{2}$ PX CX + $\frac{1}{2}$ DQ CD

$\rightarrow$ $AD^2$ + $\frac{1}{2}$ AD (AB-AD) + $\frac{1}{2}$ (AB-AD) AB

$\rightarrow$ $AD^2$ + $\frac{1}{2}$ (AB AD - $AD^2$ + $AB^2$ - AB AD)

$\rightarrow$ $AD^2$ + $\frac{1}{2}$ ( $AB^2$ - $AD^2$ )

$\rightarrow$ $\frac{1}{2}$ (2 $AD^2$ + $AB^2$ - $AD^2$ )

$\rightarrow$ $\frac{1}{2}$ $AD^2$ + $AB^2$

$\rightarrow$ $\frac{36}{2}$

$\rightarrow$ 18

So, Total area is 18.

Lets create straight line AC and straight line PQ.

AC is perpendicular to PQ. $Area_{APCQ}=\frac{AC\times PQ}{2}=\frac{6\times6}{2}=18unit^{2}$

Area of $APQC=\triangle ACQ+\triangle APC$ .

As $AP=AD=BC \triangle APC$ area is $\frac{1}{2}(AD)(AD)=\frac{1}{2}AD^{2}$ .

As $AB=DC=AQ \triangle AQC$ area is $\frac{1}{2}(AB)(AB)=\frac{1}{2}AB^{2}$ .

Thus $APQC$ area is $(\frac{1}{2})(AB^{2}+AD^{2})$

We know that $AB^{2}+AD^2=6^2=36$

So area $APQC=(\frac{1}{2})(AB^{2}+AD^{2})=(\frac{1}{2})(36)=\boxed{18}$

$\text{According to }\color{#0000ff}{Ilya}\text{ }\color{#0000ff}{Andreev}\text{, }2x^2+2xy+y^2=36.$

$\text{The assumption of one }\color{#0000ff}{Mehul}\text{ }\color{#0000ff}{Chaturvedi}\text{ can be explained by similarity. }$

$AP=AD\text{ and }AB=AQ.\text{ All are sides of two squares. All squares are similar.}$

$\text{Refer to the figure above. }ABRQ\text{ is a square. Its area is }A_{ABRQ}=(x+y)^2$

$\color{#ffffff}{\LaTeX}$

$\text{The area of }APCQ\text{ can be solve by subtracting the areas of }\Delta{PBC}\text{ and }\Delta{CRQ}.$

$A_{\Delta{PBC}}=\frac{1}{2}xy$

$A_{\Delta{CRQ}}=\frac{1}{2}y(x+y)$

$\color{#ffffff}{\LaTeX}$

$A_{APCQ}=A_{ABRQ}-A_{PBC}-A_{CRQ}$

$A_{APCQ}=(x+y)^2-\frac{1}{2}xy-\frac{1}{2}y(x+y)$

$A_{APCQ}=x^2+2xy+y^2-\frac{1}{2}xy-\frac{1}{2}xy-\frac{1}{2}y^2$

$\boxed{A_{APCQ}=x^2+xy+\frac{1}{2}y^2}$

$\color{#ffffff}{\LaTeX}$

$\text{Solve for }y^2\text{ in }2x^2+2xy+y^2=36$

$y^2=36-2x^2-2xy\leftarrow\text{ Plug this to the other equation.}$

$A_{APCQ}=x^2+xy+\frac{1}{2}(36-2x^2-2xy)$

$A_{APCQ}=x^2+xy+18-x^2-xy$

$\color{#20A900}{\boxed{\huge{A_{APCQ}=18}}}$

$\color{#FFFFFF}{\LaTeX}$

$\color{#EC7300}{\LaTeX}$

$\text{PS. Help me know how to tag! lol Please upvote. :)}$