题目内容
已知△ABC是边长为4的正三角形,D、P是△ABC内部两点,且满足
=
(
+
),
=
+
,则△APD的面积为______.
| AD |
| 1 |
| 4 |
| AB |
| AC |
| AP |
| AD |
| 1 |
| 8 |
| BC |
取BC的中点E,连接AE,根据△ABC是边长为4的正三角形

∴AE⊥BC,
=
(
+
)
而
=
(
+
),则点D为AE的中点,AD=
取
=
,以AD,AF为边作平行四边形,可知
=
+
=
+
而△APD为直角三角形,AF=
∴△APD的面积为
×
×
=
故答案为:
∴AE⊥BC,
| AE |
| 1 |
| 2 |
| AB |
| AC |
而
| AD |
| 1 |
| 4 |
| AB |
| AC |
| 3 |
取
| AF |
| 1 |
| 8 |
| BC |
| AP |
| AD |
| 1 |
| 8 |
| BC |
| AD |
| AF |
而△APD为直角三角形,AF=
| 1 |
| 2 |
∴△APD的面积为
| 1 |
| 2 |
| 1 |
| 2 |
| 3 |
| ||
| 4 |
故答案为:
| ||
| 4 |
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