Question

If the volume of a sphere is increased by 25%, its area is increased by how much?

Answer 1

@ganeshie8

Answer 2

@mathmath333

Answer 3

@ParthKohli

Answer 4

@Jhannybean

Answer 5

@ganeshie8

Answer 6

\(V=\frac{1}{3}rA\iff A=\frac{3V}r{}\)

Answer 7

yes then

Answer 8

?

Answer 9

so what is the increase, can u provide more steps

Answer 10

\[A = 4\pi r^2\]\[V= \frac{4}{3} \pi r^3\]and radius = r \[A \propto r^2 ~,~ SA \propto r^3\]So if the volume increases by 25% (0.25), then the radius increases by \(\sqrt[3]{1.25}\) then our Area has increased by \((\sqrt[3]{1.25})^2\)

Answer 11

\(V=\dfrac{4}{3}\pi(r)^3\)
let \(r=21\),then \(V=38808\) and
\(V_1=1.25V\\
V_1=\dfrac{4}{3}\pi(r_1)^3
=48510\\
r_1=22.65\\
A=4\pi r^2\\
=4\pi (21)^2\\
=5544\\
A_1=4\pi (r_1)^2\\
=4\pi (22.65)^2\\
=6446.83\\
\%~increase=\dfrac{6446.83-5544}{5544}\times 100\\
\approx 16.28\%
\)

Answer 12

sorry, typo.

Answer 13

\[r=\sqrt[3]{r}\]

Answer 14

r is radius..

Answer 15

@mathmath333 where did you get \(r=21\)?

Answer 16

@Jhannybean what was ur solution

Answer 17

i assumed it to make the calulation with \(\pi\) easier

Answer 18

@CrashOnce calculate my answer and compare it with yours. I've set it up for you.

Answer 19

@mathmath333 we are off just a little bit, around the same answer though :)

Answer 20

ill go with 16% thanks both

Answer 21

@CrashOnce , to find the percentage increase...

Answer 22

you take \(\left(\sqrt[3]{1.25}\right)^2\),and plug it back into your formula for Surface Area,

Answer 23

assumption method is useful when its objective type questions or the time is very less to solve the problem

Answer 24

\[A=4\pi r^2\]\[A=4\pi \left(\left(\sqrt[3]{1.25}\right)^2\right)^2\]

Answer 25

This will give you the percentage increase of Area.

Answer 26

Do you understand @CrashOnce ?

Answer 27

yeas i do

Answer 28

Awesome :)