题目内容
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分析:先表示出花坛所占面积,再表示出花坛面积减去矩形面积的一半,可得出
S矩形=0.75y2,即得出小明的设计方案符合要求.
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解答:解:解法一:S花坛=mn+(
n)2π,(1分)
=
x•
y+(
×
y)2π,
=
xy+
πy2(2分);
S花坛-
S矩形=
xy+
πy2-
xy,(3分)
=
πy2-
xy,
=
πy•
x-
xy,
=(
-
)xy<0
解法二:S花坛=mn+(
n)2π(1分)
=
x•
y+(
×
y)2π
=(
+
)y2
≈0.572y2(2分)
S矩形=0.75y2(3分),
∴符合要求(5分)(此处取近似值比较扣1分)
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=
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=
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S花坛-
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=
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=
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=(
| π |
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解法二:S花坛=mn+(
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=
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=(
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≈0.572y2(2分)
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∴符合要求(5分)(此处取近似值比较扣1分)
点评:本题是一道综合性的题目,考查了整式的混合运算、扇形的面积、矩形的面积等知识点,难度较大,是一道竞赛题.
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