又∵A2B2∥A3B3,A2B1∥A3B2,
∴∠OB2A2=∠OB3A3,∠A2B1B2=∠A3B2B3,
∴△B1B2A2∽△B2B3A3,
∴
B1B2 |
B2B3 |
1 |
2 |
A2B2 |
A3B3 |
∴
A2A3 |
A3A4 |
1 |
2 |
∵
S△A2B2A3 |
S△B2A3B3 |
1 |
2 |
∴△A2B2A3的面积为=
1 |
2 |
1 |
2 |
同理可得:△A3B3A4的面积=2×S△A3B2B3=2×4=8;
△A1B1A2的面积=
1 |
2 |
1 |
2 |
∴三个阴影面积之和=0.5+2+8=10.5.
故答案为:10.5.
B1B2 |
B2B3 |
1 |
2 |
A2B2 |
A3B3 |
A2A3 |
A3A4 |
1 |
2 |
S△A2B2A3 |
S△B2A3B3 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |
1 |
2 |