题目内容
15
15
平方厘米.分析:连接BO,S△BOD=S△BOE[等底等高],AB=AD+DB=4+6=10,S△BOD=
S△DOA=1.5S△DOA,[等高];S△ABE=BE×AB×
=6×
=30,S△DOA+S△BOD+S△BOE=30,S△DOA+1.5S△DOA+1.5S△DOA=30,S△DOA=
,同理S△EOC=
,阴影面积=S△DOA+S△EOC=
+
=15(平方厘米);据此解答即可.

| 6 |
| 4 |
| 1 |
| 2 |
| 10 |
| 2 |
| 30 |
| 4 |
| 30 |
| 4 |
| 30 |
| 4 |
| 30 |
| 4 |
解答:解:连接BO,S△BOD=S△BOE,
AB=AD+DB=4+6=10,S△BOD=
S△DOA=1.5S△DOA;
S△ABE=BE×AB×
=6×
=30,
S△DOA+S△BOD+S△BOE=30,S△DOA+1.5S△DOA+1.5S△DOA=30,
S△DOA=
,
同理S△EOC=
,
阴影面积=S△DOA+S△EOC=
+
=15(平方厘米);
答:阴影部分的面积是15平方厘米.
故答案为:15.
AB=AD+DB=4+6=10,S△BOD=
| 6 |
| 4 |
S△ABE=BE×AB×
| 1 |
| 2 |
| 10 |
| 2 |
S△DOA+S△BOD+S△BOE=30,S△DOA+1.5S△DOA+1.5S△DOA=30,
S△DOA=
| 30 |
| 4 |
同理S△EOC=
| 30 |
| 4 |
阴影面积=S△DOA+S△EOC=
| 30 |
| 4 |
| 30 |
| 4 |
答:阴影部分的面积是15平方厘米.
故答案为:15.
点评:解答此题的主要依据是:等高不等底的三角形的面积比等于其对应底的比,等底等高的三角形面积相等.
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