Question

the radius of a cone is increased by 25% and H is decreased by 20%.Find percent change in volume.
I want appropriate answer ​

Answer 1

Answer:

Step-by-step explanation:

Volume of cone = 1/3 πr^2 h

let new r = r/4 + r = 5r/4

and new h = h/5 - h = 4h/5

new volume = 1/3π(5r/4)^2 * 4h/5

change in volume = new volume - previous volume.

good day

Answer 2

Answer:

275% INCREASE

Step-by-step explanation:

Original volume of a cone = [$\pi$r²h]÷3

Given that,

• Radius is increased by 25%
• Height is decreased by 20%

New Volume = $\pi$×[r+{(25/100)×r}]²×[h-{(20/100)×h}]

                      = $\pi$×(5r/4)²×(4h/5)

                      = $\pi$×(25r²/16)×(4h/5)

                      = $\pi$×5r²/4×h

                      = [5$\pi$r²h]÷4

Change in  Volume = [(5$\pi$r²h)÷4] - [($\pi$r²h)÷3]

                                = [15$\pi$r²h - 4$\pi$r²h]÷12

                                = [11$\pi$r²h]÷12

Percentage change = [(11$\pi$r²h/12)÷($\pi$r²h/3)]×100

                                 = 275% INCREASE   ←ANSWER

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