Question

if the radiois. of circle is increased by 50/3 the find the percent change in its area​

Answer 1

Step-by-step explanation:

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Answer 2

Answer:

Step-by-step explanation:

Let the radius at first be r.

Then area (A)=πr^2

Now when r is increased by 50/3 then

r'=r+(50/3)=(3r+50)/3

Increased area (A')=π{(3r+50)/3}^2

=π(9r^2+300r+2500)/9

Now change in area = A' - A

=π[(9r^2+300r+2500)/9]-πr^2

=π[9r^2+300r+2500-9r^2]/9

=π[300r+2500]/9

Now % change in area is given by

[π{(300 r+2500)/9}÷πr^2]*100%

=(300r+2500)/9r^2]*100%