【Question 20】
As shown, $A B C D$ is a square, and points $E$ and $F$ are the midpoints of sides $A B$ and $B C$, respectively. $D E$ and $D F$ intersect the diagonal $A C$ at points $M$ and $N$. If the area of square $A B C D$ is $48 \mathrm{~cm}^{2}$, find the area of pentagon $E B F N M$.
【Analysis and Solution】
Geometry, Area, Equal Area Transformation.
Since point $E$ is the midpoint of side $AB$;
Therefore, $S_{\triangle ADE} = S_{\text{square } ABCD} \times \frac{1}{4} = 48 \times \frac{1}{4} = 12$ square centimeters;
In square $ABCD$, $AE \parallel CD$;
According to the hourglass model, $EM: DM = AE: CD = 1: 2$;
Therefore, $S_{\triangle AEM}: S_{\triangle ADM} = 1: 2$;
Therefore, $S_{\triangle AEM} = S_{\triangle ADE} \times \frac{1}{3} = 12 \times \frac{1}{3} = 4$ square centimeters;
Similarly, $S_{\triangle CFN} = 4$ square centimeters;
Since $S_{\triangle ABC} = S_{\text{square } ABCD} \times \frac{1}{2} = 48 \times \frac{1}{2} = 24$ square centimeters;
Therefore, $S_{\text{pentagon } EBENM} = S_{\triangle ABC} - S_{\triangle AEM} - S_{\triangle CFN} = 24 - 4 - 4 = 16$ square centimeters.