CAT Exam 2021 Slot 2 Solved Paper

All the sides of the rhombus are equal.
The area of a rhombus is $\frac{1}{2}\left({\mathit{d}}_1\right).\left({\mathit{d}}_2\right)=12$
${\mathit{d}}_1\times {\mathit{d}}_2=24$
The length of the side of a rhombus is given by $\frac{\sqrt{{\mathit{d}}_1+{\mathit{d}}_2}}{2}$ . This is because the two diagonals and a side from a right-angled triangle with sides $\frac{{\mathit{d}}_1}{2},\frac{{\mathit{d}}_2}{2}$ and the side length.
$\frac{\sqrt{{\mathit{d}}_1+{\mathit{d}}_2}}{2}=5$
Hence,
$\sqrt{{\mathit{d}}_1^2+{\mathit{d}}_2^2}=10$
${\mathit{d}}_1^2+{\mathit{d}}_2^2=100$
Using
${\mathit{d}}_1\times {\mathit{d}}_2=24,2\times {\mathit{d}}_1\times {\mathit{d}}_2=48$
${\mathit{d}}_1^2+{\mathit{d}}_2^2+2.{\mathit{d}}_1.{\mathit{d}}_2=100+48=148$
${\mathit{d}}_1^2+{\mathit{d}}_2^2-2.{\mathit{d}}_1.{\mathit{d}}_2=100-48=52$
${\mathit{d}}_1+{\mathit{d}}_2=\sqrt{148}$
${\mathit{d}}_1-{\mathit{d}}_2=\sqrt{52}$
adding equation (1) and (2), we get,
$2\times {\mathit{d}}_1=2\times \left(\sqrt{37}+\sqrt{13}\right)$
${\mathit{d}}_1=\sqrt{37}+\sqrt{13}$
or
In a rhombus the area of a Rhombus is given by :

The diagonals perpendicularly bisect each other. Considering the length of the diagonal to be 2a, 2b.
The area of a Rhombus is : $\frac{1}{2}.\left(2\mathit{a}\right).\left(2\mathit{b}\right)=12$
$\mathit{a}\mathit{b}=6$ .
The length of each side is :
$\sqrt{{\mathit{a}}^2+{\mathit{b}}^2}=5,{\mathit{a}}^2+{\mathit{b}}^2=25$
${\left(\mathit{a}+\mathit{b}\right)}^2=37,\left(\mathit{a}+\mathit{b}\right)=\sqrt{37}$
${\left(\mathit{a}-\mathit{b}\right)}^2=13,\left(\mathit{a}-\mathit{b}\right)=\sqrt{13}$
$2\mathit{a}=\left(\sqrt{37}+\sqrt{13}\right)$ , $2\mathit{b}=\left(\sqrt{37}-\sqrt{13}\right)$
2a is longer diagonal which is equal to $\sqrt{37}+\sqrt{13}$