2020/21 AHSMC Part 1 Problem 3

We can let the first rectangle be $3$ by $1$, and the second rectangle be $4k$ by $3k$ for some $k$. The areas are the same, so $3 = 12k^2$, so $k = \dfrac{1}{2}$. Then the ratio of the perimeter of the first rectangle to the perimeter of the second rectangle is $(3+1+3+1) : \bigl(2 + \dfrac{3}{2} + 2 + \dfrac{3}{2}\bigr)$, that is $8:7$. The answer is (B).