CAT 2018

Slot 2

QA

Question & Solution

Let 'a' and 'b' be the length of sides of the rectangle. (a > b)

Area of the rectangle = a*b

Perimeter of the rectangle = 2*(a+b)

$\Rightarrow$ $\dfrac{a*b}{(2*(a+b))^2}=\dfrac{1}{25}$

$\Rightarrow$ $25ab=4(a+b)^2$

$\Rightarrow$ $4a^2-17ab+4b^2=0$

$\Rightarrow$ $(4a-b)(a-4b)=0$

$\Rightarrow$ $a = 4b$ or $\dfrac{b}{4}$

We initially assumed that a > b, therefore a $ eq$ $\dfrac{b}{4}$.

Hence, a = 4b

$\Rightarrow$ b : a = 1 : 4