答案
解:(1)在直角△BCE中,∠EBC=180°-120°=60°,
则tan∠EBC=
,

∴BE=
=
=
=17
(cm);
(2)在直角△ACF中,∠ADF=90°-∠FAD=90°-45°=45°.
∴∠FAD=∠ADF,
∴FD=AF=EC=51cm.
又∵FC=AE=AB+BE=34+17
(cm).
∴CD=34+17
-51=17
-17(cm).
则S
阴影=
(AB+CD)×EC=
(34+17
-17)×51=
(17
+17)≈1184(cm
2).
解:(1)在直角△BCE中,∠EBC=180°-120°=60°,
则tan∠EBC=
,

∴BE=
=
=
=17
(cm);
(2)在直角△ACF中,∠ADF=90°-∠FAD=90°-45°=45°.
∴∠FAD=∠ADF,
∴FD=AF=EC=51cm.
又∵FC=AE=AB+BE=34+17
(cm).
∴CD=34+17
-51=17
-17(cm).
则S
阴影=
(AB+CD)×EC=
(34+17
-17)×51=
(17
+17)≈1184(cm
2).