jmc

To find the area of quadrilateral $DRQC,$ we subtract the area of $△PRQ$ from the area of $△PDC.$

First, we calculate the area of $△PDC.$ We know that $DC=AB=5\text{ cm}$ and that $\angle DCP=90^{\circ }.$ When the paper is first folded, $PC$ is parallel to $AB$ and lies across the entire width of the paper, so $PC=AB=5\text{ cm}.$ Therefore, the area of $△PDC$ is $$\frac{1}{2}\times 5\times 5=\frac{25}{2}=12.5\text{mbox}{cm}^2.$$ Next, we calculate the area of $△PRQ.$ We know that $△PDC$ has $PC=5\text{ cm},$ $\angle PCD=90^{\circ },$ and is isosceles with $PC=CD.$ Thus, $\angle DPC=45^{\circ }.$ Similarly, $△ABQ$ has $AB=BQ=5\text{ cm}$ and $\angle BQA=45^{\circ }.$ Therefore, since $BC=8\text{ cm}$ and $PB=BC-PC,$ we have $PB=3\text{ cm}.$ Similarly, $QC=3\text{ cm}.$ Since $$PQ=BC-BP-QC,$$ we get $PQ=2\text{ cm}.$ Also, $$\angle RPQ=\angle DPC=45^{\circ }$$ and $$\angle RQP=\angle BQA=45^{\circ }.$$



Using four of these triangles, we can create a square of side length $2\text{ cm}$ (thus area $4\text{mbox}{cm}^2$).



The area of one of these triangles (for example, $△PRQ$) is $\frac{1}{4}$ of the area of the square, or $1\text{mbox}{cm}^2.$ So the area of quadrilateral $DRQC$ is therefore $12.5-1=\boxed{11.5}\text{mbox}{cm}^2.$