题目内容
如图,⊙O′的弦AB是⊙O的直径,点O′在⊙O上,设图中两个阴影部分的面积分别为S和S′,则S′:S=______.
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如图,设小圆半径OO′=R,则△ABO′是等腰直角三角形,大圆半径AO′=
R,
S△ABO′=
AB•OO′=R2,S扇形O′AB=
•π•(
R)2=
πR2,
S小半圆的面积O′AB=
πR2,
∴空白部分的面积=2×
πR2-R2=πR2-R2,
∴S=πR2-(πR2-R2)=R2,
S′=π•(
R)2-(πR2-R2)=πR2+R2,
∴S′:S=(πR2+R2):R2=π+1.
故本题答案为:π+1.
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2 |
S△ABO′=
1 |
2 |
1 |
4 |
2 |
1 |
2 |
S小半圆的面积O′AB=
1 |
2 |
∴空白部分的面积=2×
1 |
2 |
∴S=πR2-(πR2-R2)=R2,
S′=π•(
2 |
∴S′:S=(πR2+R2):R2=π+1.
故本题答案为:π+1.
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