5、In the figure, $\mathrm{ABCD}$ is a right trapezoid $\left(\angle \mathrm{DAB}=\angle \mathrm{ABC} C=90^{\circ}\right)$, and a rectangle $\mathrm{ADEF}$ is constructed outward with $\mathrm{AD}$ as one side, with an area of 6.36 square centimeters. Connecting $\mathrm{BE}$ intersects $\mathrm{AD}$ at $\mathrm{P}$, and then connecting $\mathrm{PC}$. The area of the shaded part in the figure is () square centimeters.
(A) 6.36
(B) 3.18
(C) 2.12
(D) 1.59
5. Choose B
Solution: As shown in the figure, connect $\mathrm{AE}, \mathrm{BD}$. Since $\mathrm{AD} / / \mathrm{BC}$, then: $S_{\triangle \mathrm{PDc}}=S_{\triangle \mathrm{PDB}}$
Also, since $A B / / E D$, then: $S_{\triangle E A D}=S_{\triangle E R D}$
$=3.18$ (square centimeters)
Explanation: The answer is independent of the shape of the right trapezoid. We can let the side $\mathrm{BC}$ approach the side $\mathrm{AD}$ until they coincide. At this point, $\mathrm{P}$ coincides with $\mathrm{A}$, and $\mathrm{PE}$ is the diagonal of $\mathrm{ADEF}$. Therefore, the area of the shaded part is half the area of $\mathrm{ADEF}$, which is 3.18 square centimeters.