【ID】672 【题型】解答题 【类型】中考真题 【来源】2021年安徽省中考数学试卷

\begin{aligned} &\therefore A D / / E F, \angle E=\angle F=90^{\circ} \\ &\therefore \angle B A D=\angle E B A=53^{\circ} \end{aligned}

$\therefore \sin \angle E B A=\frac{A E}{A B} \approx 0.80, \quad \cos \angle E B A=\frac{B E}{A B} \approx 0.60$,
$$\therefore A E=8, \quad B E=6$$
$\angle A B C=90^{\circ}$,
\begin{aligned} &\therefore \angle F B C=90^{\circ}-\angle E B A=37^{\circ}, \\ &\therefore \angle B C F=90^{\circ}-\angle F B C=53^{\circ}, \end{aligned}

$\therefore \sin \angle B C F=\frac{B F}{B C} \approx 0.80, \cos \angle B C F=\frac{F C}{B C} \approx 0.60$
$$\therefore B F=\frac{24}{5}, \quad F C=\frac{18}{5}$$
$\therefore E F=6+\frac{24}{5}=\frac{54}{5}$
$$\therefore S_{\text {四边形EFD. }}=A E \cdot E F=8 \times \frac{54}{5}=\frac{432}{5},$$
\begin{aligned} &S_{\triangle A B E}=\frac{1}{2} \cdot A E \cdot B E=\frac{1}{2} \times 8 \times 6=24 \\ &S_{\triangle B C F}=\frac{1}{2} \cdot B F \cdot C F=\frac{1}{2} \times \frac{24}{5} \times \frac{18}{5}=\frac{216}{25}, \end{aligned}
$\therefore$ 截面的面积 $=S_{\text {四过形 } E F D A}-S_{\triangle A B E}-S_{\triangle B C F}=\frac{432}{5}-24-\frac{216}{25}=53 \frac{19}{25}\left(\mathrm{~cm}^{2}\right)$.