第65页

信息发布者：
C
5
99
101
解：连接​$BD$​
∵​$∠A = 90°，$​∴​$AB^2+AD^2=BD^2$​
∵​$AB = 3，$​​$AD = 4$​
∴​$BD^2=3^2+4^2=9 + 16=25，$​∴​$BD = 5$​
∵​$BD^2+CD^2=5^2+12^2=25 + 144=169，$​​$BC^2=13^2=169$​
∴​$BD^2+CD^2=BC^2$​
∴​$∠BDC = 90°$​
∴​$S_{四边形ABCD}=S_{\triangle BAD}+S_{\triangle BDC}=\frac 12×3×4+\frac 12×5×12=36$​

解：连接$PP'。$
$\because \triangle P'AB\cong\triangle PAC，$
$\therefore \angle BAP'=\angle CAP，$$AP' = AP = 6，$$P'B = PC = 10。$
$\because \triangle ABC$是等边三角形，
$\therefore \angle BAC=\angle CAP+\angle PAB = 60^{\circ}，$
$\therefore \angle P'AP=\angle BAP'+\angle PAB = 60^{\circ}，$
$\therefore \triangle P'AP$是等边三角形，
$\therefore AP = AP'=P'P = 6，$$\angle APP' = 60^{\circ}，$点$P$与点$P'$之间的距离为$6。$
$\because P'P^{2}+PB^{2}=6^{2}+8^{2}=36 + 64=100，$$P'B^{2}=10^{2}=100，$
$\therefore P'P^{2}+PB^{2}=P'B^{2}，$
$\therefore \angle P'PB = 90^{\circ}，$
$\therefore \angle APB=\angle P'PB+\angle APP'=90^{\circ}+60^{\circ}=150^{\circ}。$
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