如图,点A1,A2,A3,A4在射线OA上,点B1,B2,B3在射线OB上,且A1B1∥A2B2∥A3B3,A2B1∥A
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如图,点A1,A2,A3,A4在射线OA上,点B1,B2,B3在射线OB上,且A1B1∥A2B2∥A3B3,A2B1∥A3B2∥A4B3.若△A2B1B2,△A3B2B3的面积分别为1,4,则图中三个阴影三角形面积之和为______.
△A2B1B2,△A3B2B3的面积分别为1,4,
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,△A3B2B3的面积是4,
∴△A2B2A3的面积为=
1
2×S△A3B2B3=
1
2×4=2(等高的三角形的面积的比等于底边的比).
同理可得:△A3B3A4的面积=2×S△A3B2B3=2×4=8;
△A1B1A2的面积=
1
2S△A2B1B2=
1
2×1=0.5.
∴三个阴影面积之和=0.5+2+8=10.5.
故答案为:10.5.
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,△A3B2B3的面积是4,
∴△A2B2A3的面积为=
1
2×S△A3B2B3=
1
2×4=2(等高的三角形的面积的比等于底边的比).
同理可得:△A3B3A4的面积=2×S△A3B2B3=2×4=8;
△A1B1A2的面积=
1
2S△A2B1B2=
1
2×1=0.5.
∴三个阴影面积之和=0.5+2+8=10.5.
故答案为:10.5.
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