The dimensions of a rectangle $ABCD$ are $51 \, cm \times 25 \, cm$. $A$ trapezium $PQCD$ with its parallel sides $QC$ and $PD$ in the ratio $9:8$ is cut off from the rectangle as shown in the figure. If the area of the trapezium $PQCD$ is $\frac{5}{6}$ th part of the area of the rectangle,find the lengths $QC$ and $PD$.

(N/A) $ABCD$ is a rectangle in which $AB = 51 \, cm$ and $BC = 25 \, cm$.
Since the parallel sides $QC$ and $PD$ are in the ratio $9:8$,let $QC = 9x$ and $PD = 8x$.
The height of the trapezium $PQCD$ is the same as the width of the rectangle,which is $25 \, cm$.
The area of the trapezium $PQCD = \frac{1}{2} \times (QC + PD) \times \text{height} = \frac{1}{2} \times (9x + 8x) \times 25 = \frac{1}{2} \times 17x \times 25$.
The area of the rectangle $ABCD = \text{length} \times \text{width} = 51 \times 25$.
It is given that the area of the trapezium $PQCD = \frac{5}{6} \times \text{Area of rectangle } ABCD$.
Therefore,$\frac{1}{2} \times 17x \times 25 = \frac{5}{6} \times 51 \times 25$.
Dividing both sides by $25$ and multiplying by $2$,we get $17x = \frac{5}{6} \times 51 \times 2 = \frac{5}{3} \times 51 = 5 \times 17 = 85$.
Thus,$x = \frac{85}{17} = 5$.
Hence,the length $QC = 9x = 9 \times 5 = 45 \, cm$.
And the length $PD = 8x = 8 \times 5 = 40 \, cm$.