Correct Answer: $\frac{(17\sqrt{21})}{5}$

Solution :  

Given: QD = 2 cm, QC = 5 cm, CR = 3 cm, BR = 4 cm, PB = 6 cm, PA = 5 cm and AD = 3 cm. PQR is a triangle and quadrilateral ABCD is inscribed in it.
From the figure,
We see that PQ = 10 cm, PR = 10 cm, and QR = 8 cm.
Since PQ = PR. Then the triangle PQR is isosceles.
We can draw a perpendicular PS on QR. Such that QS = RS = 4 cm.
The height of an isosceles triangle = $\sqrt{10^2-4^2}=2\sqrt{21}$ cm

Let $\angle$QPS = $\angle$RPS = $\theta$ and $\angle$QPR = $2\theta$
$\sin \theta = \frac{2}{5}$; $\cos \theta = \frac{\sqrt21}{5}$
Since $\sin 2\theta = 2\sin \theta\cos \theta $
⇒ $\sin 2\theta = \frac{4\sqrt{21}}{25} $
Area of triangle PQR = $\frac{1}{2}×10×10×\frac{4\sqrt{21}}{25}=\frac{1}{2}×8×2\sqrt{21}=8\sqrt{21}=\frac{40\sqrt{21}}{5}$ cm²
Area of triangle BPA = $\frac{1}{2}×5×6×\sin 2\theta=\frac{12\sqrt{21}}{5}$ cm²
Area of triangle DQC = $\frac{1}{2}×5×2×\sin (90^{\circ}-\theta)=\frac{1}{2}×5×2×\cos \theta=\sqrt{21}=\frac{5\sqrt{21}}{5}$ cm²
Area of triangle BPA = $\frac{1}{2}×3×4×\sin (90^{\circ}-\theta)=\frac{1}{2}×3×4×\cos \theta=\frac{6\sqrt{21}}{5}=$ cm²
The area of the quadrilateral ABCD = Area of triangle PQR – Area of triangle BPA – Area of triangle DQC – Area of triangle BPA
The area of the quadrilateral ABCD = $\frac{40\sqrt{21}}{5}-\frac{12\sqrt{21}}{5}-\frac{5\sqrt{21}}{5}-\frac{6\sqrt{21}}{5}=\frac{17\sqrt{21}}{5}$ cm²
Hence, the correct answer is $\frac{17\sqrt{21}}{5}$.