题目内容
(1)如图1中,甲、乙两个图形重叠部分的面积相当于甲面积
,相当于乙面积的
.甲、乙两个图形的面积比是多少?
(2)如图2,AO3=
AB,AO2=
AO3,阴影甲与阴影乙的面积的比多少?
(3)如图3,AB=
AD,EC=
ED,图中阴影部分与空白部分面积的比多少?
(4)如图4,S甲=16,S乙=12,S丙=10,阴影部分的面积是多少?
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解:(1)重叠部分的面积=甲面积×
,则甲面积=重叠部分面积÷
;
重叠部分的面积=乙面积×
,则乙面积=重叠部分的面积÷
;
所以甲面积:乙面积,
=(重叠部分面积÷
):(重叠部分的面积÷
);
=(重叠部分面积×
):(重叠部分的面积×
)
=
:
,
=(
×8):(
×8),
=18:15,
=6:5;
答:甲、乙两个图形的面积比是6:5.
(2)阴影部分甲的面积=大圆面积-中圆的面积=π(
-
)=π[
-
]=
π
=
π×(2AO2)2=3π
;
阴影乙的面积=中圆的面积-小圆的面积=π(
-
)=π[
-
]=
π
;
所以甲面积:乙面积=3π
:
π
=3:
=4:1.
答:阴影甲与阴影乙的面积的比是4:1.
(3)因为AB=
AD,EC=
ED,
所以阴影部分的面积=
三角形ACD的面积;
三角形ACD的面积=
三角形ABC的面积,
所以阴影部分的面积=
×
三角形ABC的面积=
三角形ABC的面积,
所以阴影部分的面积:空白处的面积=1:7.
(4)由题意得:如图所示:
,
甲面积=ax=16,乙面积=ay=12,丙面积=by=10,
因为ay:by=12:10=6:5,所以b=
a,
则阴影三角形的面积=bx÷2,
=
ax÷2,
=甲面积×
÷2,
=16×
×
,
=
.
答:阴影部分的面积是
.
分析:(1)因为重叠部分的面积=甲面积×
,则甲面积=重叠部分面积÷
;重叠部分的面积=乙面积×
,则乙面积=重叠部分的面积÷
,二者求比即可;
(2)由题意得:阴影部分甲的面积=大圆面积-中圆的面积=π(
-
);阴影乙的面积=中圆的面积-小圆的面积=π(
-
),代数计算即可;
(3)因为AB=
AD,EC=
ED,根据高一定时,三角形的面积与底成正比例的性质可得:阴影部分的面积=
三角形ACD的面积;三角形ACD的面积=
三角形ABC的面积,由此可得:阴影部分的面积=
×
三角形ABC的面积=
三角形ABC的面积,由此即可解答.
(4)如图所示:
,甲面积=ax=16,乙面积=ay=12,丙面积=by=10,阴影三角形面积=bx÷2,又因为ay:by=12:10=6:5,所以b=
a,则bx=
ax,则三角形的面积=甲面积×
,计算即可.
点评:(1)解决本题的关键是将两个图形的面积借助中间量表示出来,再求比;
(2)解决本题的关键是借助半径之间的关系求比;
(3)此题考查了高一定时,三角形的面积与底成正比例的性质的灵活应用.
(4)解决本题的关键是表示出各个图形的面积,再转换计算.
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重叠部分的面积=乙面积×
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所以甲面积:乙面积,
=(重叠部分面积÷
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=(重叠部分面积×
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=
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=(
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=18:15,
=6:5;
答:甲、乙两个图形的面积比是6:5.
(2)阴影部分甲的面积=大圆面积-中圆的面积=π(
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阴影乙的面积=中圆的面积-小圆的面积=π(
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所以甲面积:乙面积=3π
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答:阴影甲与阴影乙的面积的比是4:1.
(3)因为AB=
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所以阴影部分的面积=
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三角形ACD的面积=
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所以阴影部分的面积=
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所以阴影部分的面积:空白处的面积=1:7.
(4)由题意得:如图所示:
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甲面积=ax=16,乙面积=ay=12,丙面积=by=10,
因为ay:by=12:10=6:5,所以b=
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则阴影三角形的面积=bx÷2,
=
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=甲面积×
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=16×
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=
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答:阴影部分的面积是
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分析:(1)因为重叠部分的面积=甲面积×
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(2)由题意得:阴影部分甲的面积=大圆面积-中圆的面积=π(
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(3)因为AB=
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(4)如图所示:
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点评:(1)解决本题的关键是将两个图形的面积借助中间量表示出来,再求比;
(2)解决本题的关键是借助半径之间的关系求比;
(3)此题考查了高一定时,三角形的面积与底成正比例的性质的灵活应用.
(4)解决本题的关键是表示出各个图形的面积,再转换计算.
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