Area of Rectangle

Let $AD=BC=a$ , $AB=CD=b$ , $AP=x$ and $QC=y$ . Then we have:

$\begin{cases} \dfrac {ax}2 = 5 & \implies ax = 10 & ... (1) \\ \dfrac {(b-x)(a-y)}2 = 4 & \implies (b-x)(a-y) = 8 & ... (2) \\ \dfrac {by}2 = 3 & \implies by = 6 & ... (3) \end{cases}$

$\begin{aligned} (2): \quad (b-x)(a-y) & = 8 \\ ab-{\color{#3D99F6}ax}-{\color{#3D99F6}by}+xy & = 8 & \small \color{#3D99F6}\text{Note that }(1): ax = 10, \ (3): by = 6 \\ ab-{\color{#3D99F6}10}-{\color{#3D99F6}6}+xy & = 8 \\ ab + xy & = 24 & \small \color{#3D99F6}\text{Multiply both sides by }ab \\ (ab)^2 + \color{#3D99F6}axby & = 24ab \\ (ab)^2 + \color{#3D99F6}60 & = 24ab \\ (ab)^2 -24 ab + 60 & = 0 \end{aligned}$

$\begin{aligned} \implies ab & = \frac {24 \pm \sqrt{24^2-4(60)}} 2 \\ & = 12 \pm \sqrt{84} \end{aligned}$

Note that $ab = [ABCD]$ , the area of rectangle $ABCD$ . Then the area of $\triangle DPQ$ :

$\begin{aligned} [DPQ] & = [ABCD] - [APD] - [PBQ] - [QCD] \\ & = {\color{#3D99F6} 12 + \sqrt{84} } - 5 - 4 - 3 & \small \color{#3D99F6} [ABCD] = 12 - \sqrt{84} \text{ is too small.} \\ & = \boxed{ \sqrt{84}} \end{aligned}$

Let BC = AD= y and AB = DC= u and m, d be real numbers such that m* y = BQ and d* u = AP. We know that d u y =10, m* y u – m *y *d *u =8 or m *y u -10m =8. But y *u = E, where E is the area of the rectangle ABCD, so m = 8/(E-10) (1) . We know that (1-m) *y = 6 (2). Combining (1) and (2) we get: E-8E/(E-10) =6 or E^2 -24E + 60=0 (3). Solving equation (3) we get 12 + sqrt(84) (the other root is smaller than 12 and E must be greater than 12). That means that the area of triangle PQD is sqrt(84).