Geometry, cleverly finding area, cutting and supplementing.
(Method 1)
Take the midpoint $E$ of $B C$, connect $D E$; connect $B D$;
$S_{\triangle A B D}=A B \times B E \div 2=28 \times 14 \div 2=196$ square centimeters;
$S_{\text {sector } B E D}=\frac{1}{4} \times \pi \times B E^{2}=\frac{1}{4} \times \pi \times 14^{2}=49 \pi$ square centimeters;
$S_{\triangle B E D}=B E \times D E \div 2=14 \times 14 \div 2=98$ square centimeters;
$S_{\text {shape } B D}=S_{\text {sector } B E D}-S_{\triangle B E D}=49 \pi-98$ square centimeters;
$S_{\text {shaded }}=S_{\triangle A B D}+S_{\text {shape } B D}=196+(49 \pi-98)=49 \pi+98=49 \times \frac{22}{7}+98=252$ square centimeters.
(Method 2)
Take the midpoint $E$ of $B C$, connect $D E$; connect $B D$;
Since $\triangle A B C$ and $\triangle B D E$ are both isosceles right triangles;
Therefore, $\angle A C B=\angle D B E=45^{\circ}$;
Therefore, $A C / / B D$;
Therefore, $S_{\triangle A B D}=S_{\triangle C B D}$;
$S_{\text {sector } B E D}=\frac{1}{4} \times \pi \times B E^{2}=\frac{1}{4} \times \pi \times 14^{2}=49 \pi$ square centimeters;
$S_{\triangle C E D}=C E \times D E \div 2=14 \times 14 \div 2=98$ square centimeters;
$S_{\text {shaded }}=S_{\text {sector } B E D}+S_{\triangle C E D}=49 \pi+98=49 \times \frac{22}{7}+98=252$ square centimeters.
As shown in the figure, $\triangle A B C$ is an isosceles right triangle, $A B=28 \mathrm{~cm}$. A semicircle is drawn with $B C$ as the diameter, and point $D$ is the midpoint of the semicircle arc. Try to find the area of the shaded part. (Take $\pi=\frac{22}{7}$.)