圆筒的半径增加10%,则周长增加_%,面积增加_%.

问题描述:

圆筒的半径增加10%,则周长增加______%,面积增加______%.

[2π(1+10%)r-2πr]÷2πr
=[2π110%r-2πr]÷2πr
=20%πr÷2πr
=10%.
[π[(1+10%)r]2-πr2]÷πr2
=[π121%r2-πr2]÷πr2
=21%πr2÷πr2
=21%.
即圆筒的半径增加10%,则周长增加10%,面积增加21%
故答案为:10,21.