解:連接?$ A C $?
∵?$∠A B C=90°, A B=90 \mathrm{m}, B C=120 \mathrm{m} $?
∴?$A C=\sqrt{A B^{2}+B C^{2}} =\sqrt{90^{2}+120^{2}}=150 \mathrm{m} $?
?$\text { 作 } A E \perp C D, C E=130-D E ,$?
?$A C=150 \mathrm{m}, A E \perp D C $?
∴?$∠A E D=90°， ∠A E C=90°$?
?$\text { 在 Rt } \triangle A E D \text { 中,由勾股定理可得, } A E^{2}+D E^{2}=A D^{2}$?
在?$ Rt \triangle A E C $?中,由勾股定理可得，
?$A E^{2}+E C^{2}=A C^{2}, \text { 即 } $?
?$A E^2+(C D-D E)^2=A C^2 $?
∴?$A E^2+D E^2=140^2， A E^2+(130-D E)^2=150^2 $?
∴?$D E=\frac {700}{13}， A E=\frac {1680}{13} $?
∴?$S_{\text {四邊形 } A B C D}=S_{\triangle A B C}+S_{\triangle A E D} $?
?$=\frac {1}{2} ·A B ·B C+\frac {1}{2} ·C D ·A E$?
?$=\frac {1}{2} ×90 ×120+\frac {1}{2} ×130 ×\frac {1680}{13} $?
?$=13800(\mathrm{m}^{2})$?
