1. 求下面各圆的周长与面积。
(1)
$C = 2\pi r$
$= 2×3.14×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
$S=\pi r^{2}$
$= 3.14×\_\_\_\_$
$=\_\_\_\_×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
(2)
$C=\pi d$
$= 3.14×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
$S=\pi(\frac{d}{2})^{2}$
$= 3.14×\_\_\_\_$
$= 3.14×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
(1)
$C = 2\pi r$
$= 2×3.14×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
$S=\pi r^{2}$
$= 3.14×\_\_\_\_$
$=\_\_\_\_×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
(2)
$C=\pi d$
$= 3.14×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
$S=\pi(\frac{d}{2})^{2}$
$= 3.14×\_\_\_\_$
$= 3.14×\_\_\_\_$
$=\_\_\_\_(\ \ \ \ )$
答案:
1.
(1)
C = 2πr
= 2×3.14×6
= 37.68(cm)
S = πr²
= 3.14×6²
= 3.14×36
= 113.04(cm²)
(2)
C = πd
= 3.14×10
= 31.4(cm)
S = π($\frac{d}{2}$)²
= 3.14×($\frac{10}{2}$)²
= 3.14×25
= 78.5(cm²)
(1)
C = 2πr
= 2×3.14×6
= 37.68(cm)
S = πr²
= 3.14×6²
= 3.14×36
= 113.04(cm²)
(2)
C = πd
= 3.14×10
= 31.4(cm)
S = π($\frac{d}{2}$)²
= 3.14×($\frac{10}{2}$)²
= 3.14×25
= 78.5(cm²)
2. 根据下面的条件,求圆柱的底面积和底面周长。
(1)
$S =$
$C =$
(2)
$S =$
$C =$
(1)
$S =$
$C =$
(2)
$S =$
$C =$
答案:
2.
(1)
S = πr²
= 3.14×3²
= 28.26(cm²)
C = 2πr
= 2×3.14×3
= 18.84(cm)
(2)
S = π($\frac{d}{2}$)²
= 3.14×($\frac{4}{2}$)²
= 12.56(cm²)
C = πd
= 3.14×4
= 12.56(cm)
(1)
S = πr²
= 3.14×3²
= 28.26(cm²)
C = 2πr
= 2×3.14×3
= 18.84(cm)
(2)
S = π($\frac{d}{2}$)²
= 3.14×($\frac{4}{2}$)²
= 12.56(cm²)
C = πd
= 3.14×4
= 12.56(cm)
3. 把下面的表格填写完整。

答案:
3.
0.3dm 1.884dm 0.2826dm²
16m 50.24m 200.96m²
14cm 7cm 153.86cm²
0.3dm 1.884dm 0.2826dm²
16m 50.24m 200.96m²
14cm 7cm 153.86cm²
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