题目内容
考点:勾股定理,含30度角的直角三角形
专题:规律型
分析:用面积法求出AC1,再根据三角函数求出A1C1,A1C2,A3C3…,AA1,A1A2,A2A3…,从而表示出△AA1C1的面积,△A1A2C2的面积,△A1A2C2的面积,发现为无穷等比数列,用极限求解即可.
解答:解:∵AC=2,∠B=30°,
∴BC=4,
∴AB=2
,
∴BC•AC1=AC•AB,
∴4AC1=2×2
,
∴AC1=
,
则A1C1=
cos30°=
,
A1C2=A1C1cos30°=
,
同理可得,A2C2=
,
A2C3=
,
A3C3=
,
…
AA1=
×tan30°=
,
A1A2=
×tan30°=
,
A2A3=
×tan30°=
,
…
△AA1C1的面积为
AA1•A1C1=
×
×
=
,
△A1A2C2的面积为
A1A2•A2C2=
×
×
=
,
△A1A2C2的面积为
A1A3•A3C3=
×
×
=
,
S阴影=
+
+
+…=
=
.
∴BC=4,
∴AB=2
| 3 |
∴BC•AC1=AC•AB,
∴4AC1=2×2
| 3 |
∴AC1=
| 3 |
则A1C1=
| 3 |
| 3 |
| 2 |
A1C2=A1C1cos30°=
3
| ||
| 4 |
同理可得,A2C2=
| 9 |
| 8 |
A2C3=
9
| ||
| 16 |
A3C3=
| 27 |
| 32 |
…
AA1=
| 3 |
| 2 |
| ||
| 2 |
A1A2=
| 9 |
| 8 |
3
| ||
| 8 |
A2A3=
| 27 |
| 32 |
9
| ||
| 32 |
…
△AA1C1的面积为
| 1 |
| 2 |
| 1 |
| 2 |
| ||
| 2 |
| 3 |
| 2 |
3
| ||
| 8 |
△A1A2C2的面积为
| 1 |
| 2 |
| 1 |
| 2 |
3
| ||
| 8 |
| 9 |
| 8 |
27
| ||
| 128 |
△A1A2C2的面积为
| 1 |
| 2 |
| 1 |
| 2 |
9
| ||
| 32 |
| 27 |
| 32 |
243
| ||
| 2048 |
S阴影=
3
| ||
| 8 |
27
| ||
| 128 |
243
| ||
| 2048 |
| lim |
| n→∞ |
| ||||||
1-
|
6
| ||
| 7 |
点评:本题考查了含30°角的直角三角形,涉及三角形的面积、等比数列、极限等知识,重在探究规律.
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