Question

Of the radius and height of a cone are increased by 10%,then how much is the volume of cone is increased.

Make sure that answer is 33.1%.
Answer it fast plzz!

Answer 1

let original radius be 10r and original height be 10h
therefore if height is increased by 10% new height will be 10h + (10/100)10h = 11h
new radius will be 10r + (10/100)10r = 11r

volume of cone = ⅓πr²h
therefore original volume / new volume
= (⅓π(10r)²10h)/(⅓π(11r)²11h)
= 1000r²h/1331r²h
= 1000/1331

if original area is 100%,
1000/100 = 1331/n where n is percentage increase
n = 1331*100/1000
n = 1331/10 = 133.1%
therefore increase is 133.1 - 1000 = 33.1%

Answer 2

YOUR ANSWER WITH EXPLANATION:-
--------–--------–------------------
volume of the cone =
$\pi \: r {}^{2}h \div 3$
New radius=11r/10
New hight=11h/10

Now,,
New ,,
V=π(11r/10)^2(11h/10)(1/3)
=π(121r/100)(11h/10)(1/3)
=π1331rh/3000
so, percentage increase=(new volume/old volume×100)-100
=(π/3×3/π×1331/1000×rh/rh×100)-100
=(1331/1000×100)-100
=33.1

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