一个扇形的周长为30cm,当这个扇形的圆心角a等于多少度时面积最大,并求这个扇形的最大面积

写一下解题过程
2025-03-25 16:59:03
推荐回答(3个)
回答1:

设扇形的半径=R。
则弧长=30-2r=a/360×2πR
解得:a/360=(15-R)/πR
面积S=a/360×πR²
=(15-R)/πR×πR²
=-R²+ 15R
=-(R-15/2)²+225/4
当R-15/2=0,即R=15/2时,S有最大值=225/4
∴a/360=(15-R)/πR
解得:a=360°/π=114.59°
答:当这个扇形的圆心角a等于114.59°时面积最大,最大值为225/4

回答2:

设扇形的半径=r,则弧长=30-2r
圆心角a,面积s=(1/2)(30-2r)r
=-r²+ 15r=-(r-15/2)²+225/4
当r- 15/2=0,即r=15/2时,s有最大值
这时弧长L=15
由弧长公式:L=αr
α=2弧度=2×57°18′=114°36′

回答3:

1.因:扇形周长C=弧长L+2r,所以:L=C-2r
2.据扇形面积计算公式S扇=rL/2得:S扇=rL/2=r(C-2r)/2=rC/2-r^2。可求当S扇最大时:r=C/4=7.5。
3.L=C-2r=30-15=15,扇形角a=L/r=15/7.5=2弧度=2*180/3.14=114.65°
4.S扇=rC/2-r^2=7.5*30/2-7.5*7.5=168.75 .

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