What changes will be occured in the volume of a right circualr cone if its radius be increased by `10%` and its height be decreased by `10%` ?

Explanation

Let the radius of the right circular cone be r units and height be h units.
`therefore` the volume of the cone `=(1)/(3)pi r^2h` cubic - units.
Now , if the radius be increased by `10%` , then the radius become `(r+r xx(10)/(100))"units"=(11 r)/(10)"units"`.
Again , if the height of the cone be decreased by `10%`, then its height becomes `(h-hxx(10)/(100))"units"=(9h)/(10)"units"`.
As a result of these increasing and decreasing , the volume of the cone becomes
`pi xx((11r)/(10))^2xx(9h)/(10)`cu-units `=pi xx(12 l r^2)/(100)xx(9h)/(10)` cubic - units `=(1089)/(1000)pi r^2h` cubic - units.
`therefore` the increment in volume `=((1089)/(1000)pi r^2h-(1)/(3)pi r^2h)` cu-units `=(2267)/(3000)pi r^2h` cubic - units.
So, the percentage of increment in volume `=((2267)/(3000)pir^2h)/(pir^2h)xx100%=(2267)/(3000)xx100%=75.57%`
Hence the volume of the cone will be increased by `75.57%` (approx).