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