Given : R = 2S(Eq1) and S = 2T (Eq2)

Multiplying Eq2 by 2 on both sides:

2\times S = 2 \times(2T)

2S = 4T (Eq3)

Substituting the value of 2S in Eq1 to that we obtain from Eq3:

R = 4T (Eq4)

now, area of circle = \pi \times (radius)^2 .

using area of square in Eq4 we get

\pi \times(r)^2 = 4 \times(\pi \times(t)^2)

Now, as \pi is common on both side we can cancel it :

r^2 = 4t^2 \text{ or } r= \pm \sqrt{4t^2}

Now, as radius can’t be negative we eliminate the minus aspect of the solution hence,

r = \sqrt{4t^2} \text{ or } r = 2t

ohh, thanks but how did you know that the equation needed to be made equal ?

we, are given that area R is of circle with radius `r`

and and same for all other letters! Now, in the question they’re comparing r(Qty A) to 2t (Qty B) and the relationship given is R = 2S and S= 2T . Thus, we need to somehow make a connection with information provided b/w R and T otherwise the answer will be D (Not enough information provided!). Hence , we have to manipulate R = 2S and S= 2T to make R and T appear in same equation!

we an also do it like this, if i am not wrong

since R=2S–1 and S=2T—2

now we can put S value from 2 in 1

that will be R=2(2T)->R=4T

now, 3.14 x r^2=4 x 3.14 x t^2

r^2=4t^2

r=2t