Question

if the radius of base of the base of a cone is doubled and the height is kept the same as before find the change in volume​

Answer 1

Answer:

4 times

Step-by-step explanation:

Answer 2

Answer:

$\pi {r}^{2} h$

Step-by-step explanation:

1st case:

$volume \: of \: cone = \frac{1}{3} \pi {r}^{2} h$

let it be eq1

2nd case:

radius is doubled i.e.r= 2r

$volume = \frac{1}{3} \pi ({2r})^{2} h$

$= \frac{1}{3} \pi4 {r}^{2} h$

let it be eq2

change in volume = final volume- initial volume

= eq2 -eq1

$= (\frac{1}{3} \pi {r}^{2} h) - (\frac{1}{3} \pi4 {r}^{2} h)$

$= ( \frac{1}{3} \pi {r}^{2} h)(4 - 1)$

$= (\frac{1}{3} \pi {r}^{2} h)3$

so the volume is expanded to 3 times that of initial volume

$= \pi {r}^{2} h$

this is the change in volume