1.求1個圓周為24π cm 的圓形之面積。
2.某扇形之半徑為10cm,面積為40cm(2次方)。求扇形之周界
數學牙...幫幫手t_____________t
2007-08-14 7:05 am
回答 (4)
2007-08-14 7:17 am
✔ 最佳答案
1. 圓形半徑 = 24π/2π = 12 cm面積 = 12*12*π = 144π cm^2
2.設扇形角為 theta.
π*10*10*theta/360 = 40
theta = 144/π
所以扇形周界 = 2*10*π*(144/π)/360 + 10 + 10 = 8 + 10 + 10 = 28 cm
參考: me
2007-08-15 2:31 am
1. Radius = 24π / 2π = 12 cm
Area = π (12)^2 = 144π cm^2
2. Let a be the angle.
a / 360 (π10^2) = 40
a = 45.8366
Perimeter = 45.8366 / 360 (2π10) = 7.9999999904
= 8 cm
2007-08-14 18:32:23 補充：
Sorry. I forgot to add the radius.So the total perimeter is:= 8 10 10= 28
Area = π (12)^2 = 144π cm^2
2. Let a be the angle.
a / 360 (π10^2) = 40
a = 45.8366
Perimeter = 45.8366 / 360 (2π10) = 7.9999999904
= 8 cm
2007-08-14 18:32:23 補充：
Sorry. I forgot to add the radius.So the total perimeter is:= 8 10 10= 28
2007-08-14 7:21 am
1. 圓周 = 2π r = 24 π
所以 2r = 24 => r = 12 cm
所以 圓面績 = π r^2 = π (12^2) = 144 π cm^2
2. 扇面績 = 圓面績 x (扇所開的角大少/360度)
所以 (扇所開的角大少/360度) = 扇面績 / 圓面績
扇周界
= 兩個半徑 + 曲線 [曲線 = 圓周界 x (扇所開的角大少/360度) ]
= 兩個半徑 + [圓周界 x (扇所開的角大少/360度)]
= 兩個半徑 + [圓周界 x (扇面績 / 圓面績) ]
= 2r + [ 2 π r x (40 / π r^2) ]
= 2r + (80/r)
= 2x10 + (80/10)
= 28 cm
所以 2r = 24 => r = 12 cm
所以 圓面績 = π r^2 = π (12^2) = 144 π cm^2
2. 扇面績 = 圓面績 x (扇所開的角大少/360度)
所以 (扇所開的角大少/360度) = 扇面績 / 圓面績
扇周界
= 兩個半徑 + 曲線 [曲線 = 圓周界 x (扇所開的角大少/360度) ]
= 兩個半徑 + [圓周界 x (扇所開的角大少/360度)]
= 兩個半徑 + [圓周界 x (扇面績 / 圓面績) ]
= 2r + [ 2 π r x (40 / π r^2) ]
= 2r + (80/r)
= 2x10 + (80/10)
= 28 cm
2007-08-14 7:20 am
1.
圓周 = 24π
半徑 = 24π / 2π = 12
面積 = 12^2 x π - 144 π
2.
扇形佔全個圓形的面積 = 40 / (10^2 x π ) = 2/(5π )
周界 = 2 x 10 x π x (2/(5π )) + 10 + 10
= 28 cm
圓周 = 24π
半徑 = 24π / 2π = 12
面積 = 12^2 x π - 144 π
2.
扇形佔全個圓形的面積 = 40 / (10^2 x π ) = 2/(5π )
周界 = 2 x 10 x π x (2/(5π )) + 10 + 10
= 28 cm
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